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measured from the x-axis to the x'-axis, which is attached to the rotor. T h e spin velocity b e c o m e s p = 4' •
Figure 7/20
T h e components of the angular velocity u> of the rotor and the angular velocity fi of the axes x-y-z from Fig. 7/20 become = i>
IAx
= Ó
•ftV = (A sin 8
a)y — if/ s i n 8
Q z = if' cos 0
(t)z — if/ cos 0 + p
It is important to note that the axes and the body have i d e n t i c a l « - and y - c o m p o n e n t s of angular velocity, but that the ¿-components differ by the relative angular velocity p. T h e angular-momentum components f r o m Eq. 7/12 become fi.V
=
Ixi01x ' A l l
Hy — IyyiDy = /(J l/' SLU 8 Hz - lzz«>z = H

if/ — 0
6 = constant,
0 = 0=0
p — constant,
p —0
and Eqs. 7/26 become IMX = if/ sin tt[I(ip cos 0 + p) — 104'
cos
(7/27)
IM y = 0
m - o From these results, we see that the required moment acting on the rotor about O (or about G) must be in the ac-direction since the y- and ¿-components are zero. Furthermore, with the constant values of 0, ip, and p, the moment is constant in magnitude. It is also important to note that the moment axis is perpendicular to the plane defined by the precession axis (Z-axis) and the spin axis (z-axis). We may also obtain Eqs. 7/27 by recognizing that the components of H remain constant as observed in x-y-z so that (H),^ = 0. Because in general ZM = (H) xyz + il x H, we have for the case of steady precession IM = il x II
(7/28)
which reduces to Eqs. 7/27 upon substitution of the values of il and H. By far the most common engineering examples of gyroscopic motion occur when precession takes place about an axis which is normal to the rotor axis, as in Fig. 7/14. Thus with the substitution ft — TT/2, OJz = p, ip = ii, and IMX — M, we have from Eqs. 7/27 M = Iilp
[7/24 j
Article 7/11
Gyroscopic Motion: Steady Precession
which we derived initially in this article from a direct analysis of this special case. Now let us examine the steady precession of the rotor (symmetrical top) of Fig. 7/20 for any constant value o f f other than TT/2. The moment 1MX about the x-axis is due to the weight of the rotor and is mgr sin 8. Substitution into Eqs. 7/27 and rearrangement of terms give us mgr = Ij/p - (Iu - I) if'2 cos 8 We see that if/ is small when p is large, so that the second term on the right-hand side of the equation becomes very small compared with Iifip- If we neglect this smaller term, we have if/ = mgri(Ip) which, upon use of the previous substitution !i = if/ and mk2 = I, becomes
il =
k2p
[7/25]
We derived this same relation earlier by assuming that the angular momentum was entirely along the spin axis.
Steady Precession with Zero Moment Consider now the motion of a symmetrical rotor with no external moment about its mass center. Such motion is encountered with spacecraft and projectiles which both spin and precess during flight. Figure 7/21 represents such a body. Here the Z-axis, which has a fixed direction in space, is chosen to coincide with the direction of the angular momentum H (j -, which is constant since I M C — 0. The x-y-z axes are attached in the manner described in Fig. 7/20. From Fig. 7/21 the three components of momentum are HA — 0, HG — HG sin 8, HG = HU cos 8. From the defining relations, Eqs. 7/12, with the notation of this article, these components are also given by HQ = I()IOR, H0 — IATOY, HG — HOZ. Thus, w x = i l x = 0 so that ft is constant. This result means that the motion is one of steady precession about the constant H , ; vector. With no «-component, the angular velocity u> of the rotor lies in the y-z plane along with the Z-axis and makes an angle ¡3 with the ¿-axis. The relationship between ¡3 and 8 is obtained from tan 8 — HG ¡HG — IOCOYJ(ICOZ), which is
4
tan 0 = ^ t a n £
(7/29)
Thus, the angular velocity U> makes a constant angle ¡3 with the spin axis. The rate of precession is easily obtained from Eq. 7/27 with M — 0, which gives 4r = 77 r, „ Uu - J) cos 8
(7/30)
It is clear from this relation that the direction of the precession depends on the relative magnitudes of the two moments of inertia.
Figure 7 / 1 4
581
582
Chapter 7
Introduction to T h r e e - D i m e n s i o n a l Dynamics of Rigid Bodies
Space cone
Body cone
P Direct precession I0> I
Retrograde precession /„ < I
(a)
(6)
Figure 7/22
I f / u > I, then /j < 0, as indicated in Fig. 7122a, and the precession is said to be direct. Here the body cone rolls on the outside of the space cone. If / > I 0 , then 0 < ¡3, as indicated in Fig. 7/22b, and the precession is said to be retrograde. In this instance, the space cone is internal to the body cone, and >j> a n d p have opposite signs.
This spinning top is an example of fixed-point rotation, and, for large spin rates, is a gyroscopic system.
If / = Iq, then 6 = (3 from Eq. 7/29, and Fig. 7/22 shows that both angles must be zero to be equal. For this case, the body has no precession and merely rotates with an angular velocity p. This condition occurs for a body with point symmetry, such as with a homogeneous sphere.
Article 7/11
Gyroscopic Motion: Steady Precession
Sample Problem 7/8
Port 1 left)
The turbine rotor in a ship's power plant has a mass of 1000 kg, with center of mass at G and a radius of gyration of 200 mm. The rotor shaft is mounted in bearings A and B with its axis in the horizontal fore-and-aft direction and turns counterclockwise at a speed of 500 rev/min when viewed from the stern. Determine the vertical components of the bearing reactions at A and B if the ship is making a turn to port (left) of 400-m radius at a speed of 25 knots (1 knot 0.514 m/s). Docs the bow of the ship tend to rise or fall because of the gyroscopic action?
and
RB = R.Z +
Forward
Starboard (right)
y I ! i
Solution. The vertical component of the bearing reactions will equal the static reactions i? 1 and R 2 due to the weight of the rotor, plus or minus the increment 1R due to the gyroscopic effect. The moment principle from statics easily gives (T) RI 5890 N and R., 3920 N, The given directions of the spiir velocity p and the precession velocity il are shown with the free-body diagram of the rotor. Because the spin axis always tends to rotate toward the torque axis, we see that the torque axis M points in the starboard direction as shown. The sense of the Ai?'s is, therefore, up at B and down at A to produce the couple M. Thus, the bearing reactions at A and B are RA = R1 - 1R
583
C B « I I
Atf
The precession velocity II is the speed of the ship divided by the radius of its turn. 25(0.514) i l = — — — = 0.0321 rad/s 400
[ « = oil]
Equation 7/24 is now applied around the mass center G of the rotor to give [M
/fipl
1.500(AB) = 1 0 0 0 < 0 . 2 0 0 m 0 3 2 1 )
1R
5000(2-rr)' 60
449 N
Helpful Hints © If the ship is making a left turn, the rotation is counterclockwise as viewed from above, and the precession vector ii is up by the righthand rale,
The required bearing reactions become RA •
(2;
5890 - 449
5440 N
and
RB = 3920 + 449 = 4370 N Arcs.
We now observe that the forces just computed are those exerted on the rotor shaft by the structure of the ship. Consequently, from the principle of action and reaction, the equal and opposite forces are applied to the ship by the rotor shaft, as shown in the bottom sketch. Therefore, the effect of the gyroscopic couple is to generate the increments i f ? shown, and the bow will tend to fall and the stern to rise (but only slightly).
(5) After figuring the correct sense of M on the rotor, the common mistake is to apply it to the ship in the same sense, forgetting the actionand-reaction principle. Clearly, the results are then reversed. (Be certain not to make this mistake when operating a vertical gyro stabilizer in your yacht to counteract its roll!)
584
Chapter 7
Introduction to T h r e e - D i m e n s i o n a l Dynamics of Rigid Bodies
Sample Problem 7/9 A proposed space station is closely approximated by Four uniform spherical shells, each of mass m and radius r. The mass of the connect ing structure and internal equipment may be neglected as a first approximation. If the station is designed to rotate about its ¿-axis at the rate of one revolution every 4 seconds, determine (a) the number n of complete cycles of precession for each revolution about the ¿-axis if the plane of rotation deviates only slightly from a fixed orientation, and (6) find the period r of precession if the spin axis z makes an angle of 20° with respect to the axis of fixed orientation about which precession occurs. Draw the space and body cones for this latter condition.
Solution, (a) The number of precession cycles or wobbles for each revolution of the station about the ¿-axis would be the ratio of the precessional velocity ip to the spin velocity p. which, from Eq. 7/30, is ê
I l J0 — I) cos f)
p The moments of inertia are
/ „ » / = 4[§ mr2 + m(2r)2 = Ixz ••
I0
mr2
Helpful Hint
2(J)mr 2 + 2 [ f m r 2 + m(2r)2 = f mr2
With 0 very small, cos 8 — 1, and the ratio of angular rates becomes
p
56 3 32 _ 56 3 3
7
AJ!S.
3
The minus sign indicates retrograde precession where, in the present case, iji and p are essentially of opposite sense. Thus, the station will make seven wobbles for every three revolutions.
(b) For 0 20° and p 2ir/4 rad/s, the period of precession or wobble is t 2ir/
4 ( f ) cos 20°
1.611s
Alis.
The precession is retrograde, and the body cone is external to the space cone as shown in the illustration where the body-cone angle, from Eq. 7/29, is tan jS = f tan 0 ij]
o2/ ¿5

Does The Center Hold 6th Edition Pdf Download Torrent

(0.364) = 0.637
¡i = 32.5°
(T) Our theory is tion that I rs inertia about perpendicular the case here, it to your own
based on the assumpI,,y = the moment of any axis through G to the ¿-axis. Such is and you should prove satisfaction.
A r t i c l e 7/11
PROBLEMS Introductory Problems 7 / 9 5 The jet aircraft at the bottom of an inside vertical loop has a tendency, due to gyroscopic action of the engine rotor, to yaw to the right (as seen by the pilot and as indicated by the dashed orange wingtip movements). Determine the direction of rotation pi or pi of the engine rotor as depicted in the expanded view.
Problems
585
7 / 9 7 The two identical disks are rotating freely on the shaft, with angular velocities equal in magnitude and opposite in direction as shown. The shaft in turn is caused to rotate about the vertical axis in the sense indicated. Prove whether the shaft bends as in A or as in B because of gyroscopic action.
Problem 7/95
Problem 7/97
7 / 9 6 A dynamics instructor demonstrates gyroscopic principles to his students. He suspends a rapidly spinning wheel with a string attached to one end of its horizontal axle. Describe the precession motion of the wheel.
7 / 9 8 The student has volunteered to assist in a classroom demonstration involving a momentum wheel which is rapidly spinning with angular- speed p as shown. The instructor has asked her to hold the axle of the wheel in the horizontal position shown and then attempt to tilt the axis upward in a vertical plane. What motion tendency of the wheel assembly will the student sense?
z I [
Problem 7/96
Problem 7/98
586
Chapter 7
Introduction to T h r e e - D i m e n s i o n a l Dynamics of Rigid Bodies
7 / 9 9 A car makes a turn to the right on a level road. Determine whether the normal reaction under the right rear wheel is increased or decreased as a result of the gyroscopic effect of the precessing wheels. Ans. Decreased 7 / 1 0 0 The special-purpose fan is mounted as shown. The motor armature, shaft, and blades have a combined mass of 2.2 kg with radius of gyration of 60 mm. The axial posit ion b of the 0.8-kg block A can be adjust ed. With the fan turned off, the unit is balanced about the x-axis when b 180 mm. The motor and fan operate at 1725 rev/min in the direction shown. Determine the value of b which will produce a steady precession of 0.2 rad/s about the positive y-axis.
Problem 7/100 7/101 An aii-plane has just cleared the runway with a takeoff speed u. Each of its freely spinning wheels has a mass in, with a radius of gyration k about its axle. As seen from the front of the airplane, the wheel processes at the angular rate il as the landing strut is folded into the wing about its pivot O. As a result of the gyroscopic action, the supporting member A exerts a torsional moment M on B to prevent the tubular member from rotating in the sleeve at B. Determine M and identify whether it is in the sense of M± or MA. Ans. M - Mi = mk2il r
Problem 7/101 7 / 1 0 2 An experimental antipollution bus is powered by the kinetic energy stored in a large flywheel which spins at a high speed p in the direction indicated. As the bus encounters a short upward ramp, the front wheels rise, thus causing the flywheel to precess. What changes occur to the forces between the tires and the road during this sudden change?
Problem 7/102 7 / 1 0 3 The 210-kg rotor of a turbojet aircraft engine has a radius of gyration of 220 mm and rotates counterclockwise at 18 000 rev/min as viewed from the front. If the aircraft is traveling at 1200 km/h and starts to execute an inside vertical loop of 3800-m radius, compute the gyroscopic moment M transmitted to the airframe. What correction to the controls does the pilot have to make in order to remain in the vertical plane? Ans. M 1681 N - m , Left rudder
A r t i c l e 7/11
Representative Problems 7 / 1 0 4 A small air compressor for an aircraft cabin consists of the 3,50-kg turbine A which drives the 2.40-kg blower B at a speed of 20 000 rev/min. The shaft of the assembly is mounted transversely to the direction of flight and is viewed from the rear of the aircraft in the figure. The radii of gyration of A and B are 79.0 and 71.0 mm. respectively. Calculate the radial forces exerted on the shaft by the bearings at C and D if the aircraft executes a clockwise roll (rotation about the longitudinal flight axis) of 2 rad/s viewed from the rear of the aircraft. Neglect the small moments caused by the weights of the rotors. Draw a free-body diagram of the shaft as viewed from above and indicate the shape of its deflected centerline.
BSS
Problems
587
7 / 1 0 6 The blades and hub of the helicopter rotor weigh 140 lb and have a radius of gyration of 10 ft about the ¿-axis of rotation. With the rotor turning at 500 rev/min during a short interval following vertical liftoff, the helicopter tilts forward at the rate 8 - 10 deg/sec in order to acquire forward velocity. Determine the gyroscopic moment M transmitted to the body of the helicopter by its rotor and indicate whether the helicopter tends to deflect clockwise or counterclockwise, as viewed by a passenger facing forward.
D 32Ê
L 150 u mm Problem 7/106 Problem 7/104 7 / 1 0 5 The open-ended thin-wall rectangular box of square cross section is rotating in space about its central longitudinal axis as shown. If the axis has a slight wobble, for what ratios lib will the motion be direct or retrograde precession? Ans. I > b v ; 2, direct precession / < f>v'2j retrograde precession
7 / 1 0 7 The 4-oz top with radius of gyration about its spin axis of 0.62 in. is spinning at the rate p = 3600 rev/min in the sense shown, with its spin axis making an angle 9 ~ 20° with the vertical. The distance from its tip O to its mass center G is r = 2.5 in. Determine the precession !I of the top and explain why 0 gradually decreases as long as the spin rate remains large. An enlarged view of the contact of the tip is shown. Ans. il = 6.67k rad/sec
Enlarged view of tip contact Problem 7/105
Problem 7/107
588
Chapter 7
Introduction to T h r e e - D i m e n s i o n a l Dynamics of Rigid Bodies
7 / 1 0 8 The figure shows a gyro mounted with a vertical axis and used to stabilize a hospital ship against rolling. The motor A turns the pinion which processes the gyro by rotating the large precession gear B and attached rotor assembly about a horizontal transverse axis in the ship. The rotor turns inside the housing at a clockwise speed of 960 rev/min as viewed from the top and has a mass of SO Mg with radius of gyration of 1.45 m. Calculate the moment exerted on the hull structure by the gyro if the motor turns the precession gear B at the rate of 0.320 rad/s. In which of the two directions, (a) or (6), should the motor turn m order to counteract a roll of the ship to port?
I

Does The Center Hold 6th Edition Pdf Download Pc

(a)
Vertical
(6) Problem 7/109
Forward
Starboard (right)
7/110 If the wheel in case (a) of Prob. 7/109 is forced to process about the vertical by a mechanical drive at the steady rate il = 2j rad/s, determine the bending moment in the horizontal shaft at A. In the absence of friction, what torque Mo is applied to the collar- at O to sustain this motion? 7/111
Problem 7/108 7/109 Each of the identical wheels has a mass of 4 kg and a radius of gyration k, = 120 mm and is mounted on a horizontal shaft AB secured to the vertical shaft at O. In case (a), the horizontal shaft is fixed to a collar at O which is free to rotate about the vertical y-axis. In case (6), the shaft is secuied by a yoke hinged about the x-axis to the collar. If the wheel has a large angular velocity p 3600 rev/min about its ¿-axis in the position shown, determine any precession which occurs and the bending moment MA in the shaft at A for each case. Neglect the small mass of the shaft and fitting at O. Ans. (a) No precession, M A = 12.56 N -m (b) il = 0.723 rad/s, M A = 3.14 N - i n
The figure shows the side view of the wheel carriage (truck) of a railway passenger car' where the vertical load is transmitted to the frame in which the journal wheel bearings are located. The lower view shows only one pair of wheels and their axle which rotates with the wheels. Each of the 33-in.-diameter wheels weighs 500 lb, and the axle weighs 300 lb with a diameter of 5 in. Both wheels and axle are made of steel with a specific weight of 4S9 lb/ft 3 . If the train is traveling at SO mi/hr while rounding an 8' curve to the right (radius of curvature 717 ft), calculate the change i f f in the vertical force supported by each wheel due only to the gyroscopic action. As a close approximation, treat each wheel as a uniform circular disk and the axle as a uniform solid cylinder. Also assume that both rails are in the same horizontal plane. A P ! S . IE = 98.1 lb
A r t i c l e 7/11
Side view of carriage
Problems
589
7 / 1 1 3 The uniform 640-mm rod has a mass of 3 kg and is welded centrally to the uniform 160-mm-radius circular disk which has a mass of 8 kg. The unit is given a spin velocity p = 60 rad/s in the direction shown. The axis of the rod is seen to wrobble through a total angle of 30', Calculate the angular velocity

<=' v=' a=' is.=' ip=' -=' ip1='124.2' rad=' s=' (direct=' precession)=' 320='>
IrHlrH
P View of wheels and axle Problem 7/111 7 / 1 1 2 The primary structure of a proposed space station consists of five spherical shells connected by tubular' spokes. The moment of inertia of the structure about its geometric axis A-A is twice as much as that about any axis through O normal to A-A. The station is designed to rotate about its geometric axis at the constant rate of 3 rev/min. If the spin axis A-A precesses about the Z-axis of fixed orientation and makes a very small angle with it, calculate the rate
Problem 7/113 7 / 1 1 4 The electric motor has a total weight of 20 lb and is supported by the mounting brackets A and B attached to the rotating disk. The armature of the motor has a weight of 5 lb and a radius of gyration of 1.5 in. and turns counterclockwise at a speed of 1725 rev/min as viewed from A to B. The turntable revolves about its vertical axis at the constant rate of 48 rev/min in the direction shown. Determine the vertical components of the forces supported by the mounting brackets at A and B.
Problem 7/112
Problem 7/114
590
Chapter 7
Introduction to T h r e e - D i m e n s i o n a l Dynamics of Rigid Bodies
7 / 1 1 5 The spacecraft shown is symmetrical about its 2-axis and has a radius of gyration of 720 mm about this axis. The radii of gyration about the x- and y-axes through the mass center are both equal to 540 nun. When moving in space, the z-axis is observed to generate a cone with a total vertex angle of 4° as it precesses about the axis of total angular momentum. If the spacecraft has a spin velocity about its 2-axis of 1.5 rad/s, compute the period r of each full precession. Is the spin vector in the positive or negative z-direction? Arcs. r = 1.831s spin vector in negative ^-direction
X.x
Problem 7/116 7 / 1 1 7 The housing of the electric motor is freely pivoted about the horizontal x-axis, which passes through the mass center G of the rotor. If the motor is turning at the constant rate tji = p, determine the angular acceleration iti which will result from the application of the moment M about the vertical shaft i f y = iji 0. The mass of the frame and housing is considered negligible compared with the mass m of the rotor. The radius of gyration of the rotor about the ^-axis is k, and that about the jr-axis is kx. Problem 7/115
Ans. 4' =
7 / 1 1 6 The 8-lb rotor with radius of gyration of 3 in. rotates on ball bearings at a speed of 3000 rev/min about its shaft OG. The shaft is free to pivot about the X-axis, as well as to rotate about the Z-axis. Calculate the vector il for precession about the Z-axis. Neglect the mass of shaft OG and compute the gyroscopic couple M exerted by the shaft on the rotor at G.
Problem 7/117
M/m kx2 cos 2 y - k 2 sin 2 y
Article
7/10
Problems
591
7 / 1 1 8 The two identical circular disks, each of mass m and radius r, are spinning as a rigid unit about their common axis. Determine the value of b for which no processional motion can take place if the unit is free to move in space.
Problem 7/120 7 / 1 2 1 The rectangular bar is spinning in space about its longitudinal axis at the rate p 200 rev/min. If its axis wobbles through a total angle of 20° as shown, calculate the period r of the wobble. Ans. T = 0.443 sec
Problem 7/118 7 / 1 1 9 A boy throws a thin circulai- disk (like a Frisbee) with a spin rate of 300 rev/min. The plane of the disk is seen to wobble through a total angle of 10°. Calculate the period r of the wobble and indicate whether the precession is direct or retrograde. Ans. T = 0.0996 s, retrograde
Problem 7/121
Problem 7/119 7 / 1 2 0 The figure shows a football in three common inflight configurations. Case (a) is a perfectly thrown spiral pass with a spin rate of 120 rev/min. Case (6) is a wobbly spiral pass again with a spin rate of 120 rev/min about its own axis, but with the axis wobbling through a total angle of 20°. Case (c) is an end-over-end place kick with a rotational rate of 120 rev/min. For each case, specify the values of p, 0, j8, and if/ as defined in this article. The moment of inertia about the long axis of the ball is 0.3 of that about the transverse axis of symmetry.
592
Chapter 7
Introduction to T h r e e - D i m e n s i o n a l Dynamics of Rigid Bodies
• 7 / 1 2 2 A projectile moving through the atmosphere with a velocity v which makes a small angle 8 with its geometric axis is subjected to a resultant aerodynamic force R essentially opposite in direction to v as shown. If R passes through a point C slightly ahead of the mass center G, determine the expression for the minimum spin velocity p for which the projectile will be spin-stabilized with 8 = 0. The moment of inertia about the spin axis is I and that about a transverse axis through G is I0. (Hint: Determine M x and substitute into Eq, 7/27. Express the result as a quadratic equation in ip and determine the minimum value o f p for which the expression under the radical is positive.) An s. pmiri = —
x
Ri-(J0 - I) cos 8
Problem 7/122 • 7/123 The solid circular disk of mass m and small thickness is spinning freely on its shaft at the rate p. If the assembly is released in the vertical position at 9 = 0 with 8 0, determine the horizontal components of the forces A and B exerted by the respective heal ings on the horizontal shaft as the position 8 rr/2 is passed. Neglect the mass of the two shafts compared with m and neglect all friction. Solve by using the appropriate moment equations. Ans. A = -
= ™Ji ( J^L 2 26 where 8
a
wi/r® — — p + 18 2 21
- i / K
-18j
V r2 + 412
Problem 7/123 • 7/124 The earth-scanning satellite is in a circular orbit of period r. The angular velocity of the satellite about its y- or pitch-axis is to = 2ir/r, and the angular rates about the x- and ¿-axes are zero. Thus, the jc-axis of the satellite always points to the center of the earth. The satellite has a re action-wheel attitude-control system consisting of the three wheels shown, each of which may be variably torqued by its individual motor. The angular rate of the ¿-wheel relative to the satellite is il 0 at time t 0, and the x- and y-wheels are at rest relative to the satellite at t = 0. Determine the axial torques Ms, My, and M. which must be exerted by the motors on the shafts of their respective wheels in order that the angular velocity ui of the satellite will remain constant. The moment of inertia of each reaction wheel about its axis is I. The x and z reaction-wheel speeds are harmonic functions of the time with a period equal to that of the orbit. Plot the variations of the torques and the relative wheel speeds 12,, il,., and i l . as functions of the time during one orbit period. (Hint: The torque to accelerate the .r-wheel equals the reaction of the gyroscopic moment on the ¿-wheel, and vice versa.) Ans. Mx = — IoiilQ cos oit My = 0 M. = — Jiijilj) sin tot
A r t i c l e 7/11
Problem 7/124 • 7 / 1 2 5 The two solid homogeneous right-circular cones, each of mass m, are fastened together at their vertices to form a rigid unit and are spinning about their axis of radial symmetry at the rate p 200 rev/min. (a) Determine the ratio hir for which the rotation axis will not process. (f>) Sketch the space and body cones for the case where hir is less than the critical ratio, (c) Sketch the space and body cones when h r and the processional velocity is iji = 18 rad/s. :
1
Problem 7/126
Problem 7/125
59Ï
• 7 / 1 2 6 The solid cylindrical rotor weighs 64.4 lb and is mounted in bearings A and B of the frame which rotates about the vertical Z-axis. If the rotor spins at the constant rate p 50 rad/sec relative to the frame and if the frame itself rotates at the constant rate ii 30 rad/sec, compute the bending moment M in the shaft at C which the lower portion of the shaft exerts on the upper portion. Also compute the kinetic energy T of the rotor. Neglect the mass of the frame. Ans, M = 97.9i Ib-ft, T = 73.8 ft-lb
~y
A/is. (a) h
Problems
594
Chapter 7
Introduction to T h r e e - D i m e n s i o n a l Dynamics of Rigid Bodies
7/12
CHAPTER REVIEW
In Chapter 7 we have studied the three-dimensional dynamics of rigid bodies. Motion in three dimensions adds considerable complexity to the kinematic and kinetic relationships. Compared with plane motion, there is now the possibility of two additional components of the vectors describing angular quantities such as moment, angular velocity, angular momentum, and angular acceleration. For this reason, the full power of vector analysis becomes apparent in the study of threedimensional dynamics. We divided our study of three-dimensional dynamics into kinematics, which is covered in Section A of the chapter, and kinetics, which is treated in Section B.
Kinematics We arranged our coverage of three-dimensional kinematics in order of increasing complexity of the type of motion. These types are: 1. Translation. As in plane motion, covered in Chapter 5 (Plane Kinematics of Rigid Bodies), any two points on a rigid body have the same velocity and acceleration. 2. Fixed-Axis Rotation. In this case the angular-velocity vector does not change orientation, and the expressions for the velocity and acceleration of a point are easily obtained as Eqs. 7/1 and 7/2, which are identical in form to the corresponding plane-motion equations in Chapter 5. 3. Parallel-Plane Motion. This case occurs when all points in a rigid body move in planes which are parallel to a fixed plane. Thus, in each plane, the results of Chapter 5 hold. 4. Rotation about a Fixed Point. In this case, both the magnitude and the direction of the angular-velocity vector may vary. Once the angular acceleration is established by careful differentiation of the angular-velocity vector, Eqs. 7/1 and 7/2 may be used to determine the velocity and acceleration of a point. 5. General Motion. The principles of relative motion are useful in analyzing this type of motion. Relative velocity and relative acceleration are expressed in terms of translating reference axes by Eqs. 7/4. When rotating reference axes are used, the unit vectors of the reference system have nonzero time derivatives. Equations 7/6 express the velocity and acceleration in terms of quantities referred to rotating axes; these equations are identical in form to the corresponding results for plane-motion, Eqs. 5/12 and 5/14. Equations Ilia and lllb are the expressions relating the time derivatives of a vector as measured in a fixed system and as measured relative to a rotating system. These expressions are useful in the analysis of general motion.
Article 7/12
Kinetics We applied momentum and energy principles to analyze threedimensional kinetics, as follows. 1. Angular Momentum. In three dimensions the vector expression for angular momentum has numerous additional components which are absent in plane motion. The components of angular momentum are expressed hy Eqs. 7/12 and depend on both moments and products of inertia. There is a unique set of axes, called principal axes, for which the products of inertia are zero and the moments of inertia have stationaiy values. These values are called the principal moments of inertia. 2. Kinetic Energy. The kinetic energy of three-dimensional motion can be expressed either in terms of the motion of and about the mass center (Eq. 7/15) or in terms of the motion about a fixed point (Eq. 7/18). 3. Momentum Equations of Motion. By using the principal axes we may simplify the momentum equations of motion to obtain Euler's equations, Eqs. 7/21. 4. Energy Equations. The work-energy principle for three-dimensional motion is identical to that for plane motion.
Applications In Chapter 7 we studied two applications of special interest, namely, parallel-plane motion and gyroscopic motion. 1. Parallel-Plane Motion. In such motion all points in a rigid body move in planes which are parallel to a fixed plane. The equations of motion are Eqs. 7/23. These equations are useful for analyzing the effects of dynamic imbalance in rotating machinery and in bodies which roll along straight paths. 2. Gyroscopic Motion. This type of motion occurs whenever the axis about which the body is spinning is itself rotating about another axis. Common applications include inertial guidance systems, stabilizing devices, spacecraft attitude motion, and any situation in which a rapidly spinning rotor (such as that of an aircraft engine) is being reoriented. In the case wrhere an external torque is present, a basic analysis can be based upon the equation M = H. For the case of torque-free motion of a body spinning about its axis of symmetry, the axis of symmetry is found to execute a coning motion about the fixed angular-momentum vector.
Chapter Review
595
596
Chapter 7
Introduction to T h r e e - D i m e n s i o n a l Dynamics of Rigid Bodies
REVIEW PROBLEMS 7 / 1 2 7 The cylindrical shell is rotating in space about its geometric axis. If the axis has a slight wobble, for wrhat ratios of 1/r will the motion be direct or retrograde precession? Arcs. Direct precession, - >
N
7 / 1 3 0 The wheels of the jet plane are spinning at their angular rate corresponding to a takeoff speed of 150 km/h. The retracting mechanism operates with (t increasing at the rate of 30° per second. Calculate the angular acceleration a of the wheels for these conditions.
6
Retrograde precession, - < Jß
Problem 7/130 7/131
Problem 7/127
The electric fan has a constant speed of 1720 rev/min in the direction indicated with its axis oriented as shown. If the x- and y-components of the velocity of point A on the blade tip are 15 m/s and —20 m/s, respectively, determine the magnitude v of the velocity of the blade tip and the diamet er of the fan blades. Ans. v 25.5 m/s, d = 283 m m
7 / 1 2 8 If the ship of Sample Problem 7/8 is on a straight course but its bow is falling as it drops into the trough of a wrave, determine the direction of the gyroscopic moment exerted by the turbine rotor on the hull structure and its effect on the motion of the ship. 7 / 1 2 9 An experimental car is equipped with a gyro stabilizer to counteract completely the tendency of the car to tip when rounding a curve (no change in norma] force between tires and road). The rotor of the gyro has a mass m() and a radius of gyration k, and is mounted in fixed bearings on a shaft which is parallel to the rear axle of the car'. The center of mass of the car' is a distance h above the road, and the car' is rounding an unbanked level turn at a speed v. At what speed p should the rotor turn and in what direction to counteract completely the tendency of the car to overturn for either a right or a left turn? The combined mass of car and rotor is m. mvh Ans. p = • opposite direction to car wheels m^f-
Problem 7/131
Article 7 / 1 2
7 / 1 3 2 The collais at the ends of the telescoping link AB slide along the fixed shafts shown. During an interval of motion, vA = 5 in./sec and vB = 2 in./sec. Determine the vector expression for the angular velocity ion of the centerline of the link for the position where = 4 in. and yB = 2 in.
Review Problems
597
7 / 1 3 4 The tip of the cone in Prob. 7/133 is, in reality, somewhat rounded, causing the point of contact P with the supporting surface to be sEghtly off the spin axis, as illustrated here. In the presence of kinetic friction, show why the angle 0 will slowly decrease.
Problem 7/134
x
Problem 7/132 7 / 1 3 3 The solid cone of mass m, base radius r, and altitude h is spinning at a high rate p about its own axis and is released with its vertex O supported by a horizontal surface. Friction is sufficient to prevent the vertex from slipping in the x-y plane. Determine the direction of the precession ii and the period r of one complete rotation about the vertical ¿-axis. Aiis. n = ilk, T 4irr'zp/(5gh) z
7 / 1 3 5 The circular disk of radius r is mounted on its shaft which is pivoted at O so that it may rotate about the vertical ¿ u -axis. If the disk rolls at constant speed without slipping and makes one complete turn around the circle of radius if in time r, determine the expression for the absolute angular velocity itJ of the disk. Use axes x-y-z which rotate around the ¿o-axis. (Hint: The absolute angular velocity of the disk equals the angular velocity of the axes plus (vectorially) the angular- velocity relative to the axes as seen by holding x-y-z fixed and rotating the circular disk of radius R at the rate of 2 - J T / T . ) , 2rr Ans. to = —

r
Problem 7/135
Problem 7/133
RT
R
598
Chapter 7
Introduction to T h r e e - D i m e n s i o n a l Dynamics of Rigid Bodies z
7 / 1 3 6 Determine the angular acceleation a for the rolling circular disk of Prob. 7/135. Use the results cited in the answer for that problem. 7 / 1 3 7 Determine the velocity v of point A on the disk of Prob. 7/135 for the position shown.
7 / 1 3 6 Determine the acceleration a of point A on the disk of Prob. 7/135 for the position shown. 7 / 1 3 9 A top consists of a ring of mass m = 0.52 kg and mean radius r 60 mm mounted on its central pointed shaft writh spokes of negligible mass. The top is given a spin velocity of 10 000 rev/min and released on the horizontal surface with the point O remaining in a fixed position. The axis of the top is seen to make an angle of 15' with the vertical as it processes. Determine the number N of precession cycles per minute. Also identify the direction of the precession and sketch the body and space cones. Ans. N 1.9SS cycles/min Z I
7/141
Rework Prob. 7/140 if/i, instead of being constant at 20°, is increasing at the steady rate of 120 rev/min. Find the angular' momentum H<> of the disk for the instant when ¡i 20°. Also compute the kinetic energy T of t he disk. Is T dependent on fil Ans. H ( J = 0.0867i + 0.421j + 1.281k lb-ft-sec T = 11.85 ft-lb, N o
7 / 1 4 2 The dynamic imbalance of a certain crankshaft is approximated by the physical model shown, where the shaft carries three small 1.5-lb spheres attached by rods of negligible mass. If the shaft rotates at the constant speed of 1200 rev/min, calculate the forces R A and Ra acting on the bearings. Neglect the gravitational forces.
Problem 7/139 7 / 1 4 0 The uniform circular disk of 4-in. radius and small thickness weighs 8 lb and is spinning about itsy'-axis at the rate N 300 rev/min with its plane of rotation tilted at a constant angle ¡i = 20° from the vertical x-z plane. Simultaneously, the assembly rotates about the fixed z-axis at the rate p 60 rev/min. Calculate the angular momentum H,, of the disk alone about the origin O of the x-y-z coordinates. Also calculate the kinetic energy T of the disk.
End view Problem 7/142
Article 7 / 1 2
7 / 1 4 3 Each of the two right-angle bent rods weighs 2.80 lb and is parallel to the horizontal x-y plane. The rods are welded to the vertical shaft, which rotates about the ¿-axis with a constant angular speed N = 1200 rev/mm. Calculate the bending moment M in the shaft at its base O. Ans. M = 271 lb-ft
7 / 1 4 4 Each of the quarter-circular plates has a mass of 2 kg and is secured to the vertical shaft mounted in the fixed bearing at O. Calculate the magnitude M of the bending moment in the shaft at O f o r a constant rotational speed N 300 rev/min. Treat the plates as exact quarter-circular shapes. z
Problem 7/144
Review Problems
599
7 / 1 4 5 Calculate the bending moment M in the shaft at O for the rotating assembly of Prob. 7/144 as it starts from rest with an initial angular acceleration of 200 rad/s 2 . Ans.M = 2.70 N - m • 7 / 1 4 6 Derive Eq. 7/24 by relating the forces to the acceler ations for a differential element of the thin ring of mass m. The ring has a constant angular velocity p about the ¿-axis and is given an additional constant angular' velocity i 1 about the y-axis by the application of an external moment M (not shown). [Hint: The acceleration of the element in the ¿-direction is due to (a) the change in magnitude of its velocity component in this direction resulting from ii and (¿0 the change in direction of its ^-component of velocity.)
»
£ s-
This illustration shovus the e l e m e n t s of the left-front s u s p e n s i o n on an a l l - w h e e l - d r i v e a u t o m o b i l e . The spring and shock absorber are coaxial in this McPherson-strut type of s u s p e n s i o n .
V I B R A T I O N AND TIME RESPONSE C H A P T E R OUTLINE 8/1
Introduction
8/2
Free Vibration of Particles
8/3
Forced Vibration of Particles
8/4
V i b r a t i o n o f Rigid B o d i e s
8/5
Energy Methods
8/6
Chapter Review
8/1
INTRODUCTION
An important and special class of problems in dynamics concerns the linear and angular motions of bodies which oscillate or otherwise respond to applied disturbances in the presence of restoring forces. A few examples of this class of dynamics problems are the response of an engineering structure to earthquakes, the vibration of an unbalanced rotating machine, the time response of the plucked string of a musical instrument, the wind-induced vibration of power lines, and the flutter of aircraft wings. In many cases, excessive vibration levels must be reduced to accommodate material limitations or human factors. In the analysis of every engineering problem, we must represent the system under scrutiny by a physical model. We may often represent a continuous or distributed-parameter system (one in which the mass and spring elements are continuously spread over space) by a discrete or lumped-parameter model (one in which the mass and spring elements are separate and concentrated). The resulting simplified model is especially accurate when some portions of a continuous system are relatively massive in comparison with other portions. For example, the physical model of a ship propeller shaft is often assumed to be a massless but twistable rod with a disk rigidly attached to each end—one disk representing the turbine and the other representing the propeller. As a second example, we observe that the mass of springs may often be neglected in comparison with that of attached bodies. Not eveiy system is reducible to a discrete model. For example, the transverse vibration of a diving board after the departure of the diver is
602
Chapter 8
Vibration and Time R e s p o n s e
a somewhat difficult problem of distributed-parameter vibration. In this chapter, we will begin the study of discrete systems, limiting our discussion to those whose configurations may be described writh one displacement variable. Such systems are said to possess one degree of freedom. For a more detailed study which includes the treatment of two or more degrees of freedom and continuous systems, you should consult one of the m a n y textbooks devoted solely to the subject of vibrations. T h e r e m a i n d e r of Chapter 8 is divided into f o u r sections: Article 8/2 treats the free vibration of particles and A r t . 8/3 introduces the forced vibration of particles. Each of these two articles is subdivided into u n d a m p e d - and d a m p e d - m o t i o n categories. In A r t , 8/4 we discuss the vibration of rigid bodies. Finally, an energy approach to the solution of vibration problems is presented in Art. 8/5. T h e topic of vibrations is a direct application of the principles of kinetics as developed in Chapters 3 and 6. In particular, a complete freebody diagram drawn for an arbitrary positive value of the displacement variable, followed by application of the appropriate governing equations of dynamics, will yield the equation of motion. From this equation of motion, which is a second-order ordinary differential equation, y o u can obtain all information of interest, such as the motion frequency, period, or the motion itself as a function of time.
8/2
FREE VIBRATION
OF
PARTICLES
When a spring-mounted body is disturbed from its equilibrium position, its ensuing motion in the absence of any imposed external forces is termed free vibration . In every actual case of free vibration, there exists some retarding or damping force which tends to diminish the motion. C o m m o n damping forces are those due to mechanical and fluid friction. In this article we first consider the ideal case where the damping forces are small enough to be neglected. T h e n we treat the case where the damping is appreciable and must be accounted for.
Equilibrium I7* kx -
mg
T
¥ (a) Harci Lincai' Fs - kx -Soft
(b) Figure B/l
Equation of Motion for Undamped Free Vibration We begin by c o n s i d e r i n g the horizontal vibration of the simple frictionless spring-mass system of Fig. 8 / l a . N o t e that the variable x denotes the displacement of the mass f r o m the equilibrium position, which, f o r this system, is also the position of zero spring deflection. Figure 8/1 b shows a plot of the f o r c e F s necessary to deflect the spring versus the c o r r e s p o n d i n g spring deflection f o r three types of springs. A l t h o u g h nonlinear hard and soft springs are useful in s o m e applications, we will restrict our attention to the linear spring. S u c h a spring exerts a restoring f o r c e —kx on the m a s s — t h a t is, w h e n the mass is displaced to the right, the spring f o r c e is to the left, and vice versa. We must be careful to distinguish b e t w e e n the forces of m a g n i t u d e F a which must be applied to both ends of the massless spring to cause tension or c o m p r e s s i o n and the f o r c e F — —kx of equal m a g n i t u d e which the spring exerts on the mass. T h e constant of proportionality k is called the spring constant, modulus, or stiffness and has the units N / m o r lb/ft.
Article 8/2
The equation of motion for the body of Fig. 8 / l a is obtained by first drawing its free-body diagram. Applying Newton's second law in the form IFX — mx gives or
—kx = mx
mx + kx — 0
(8/1)
The oscillation of a mass subjected to a linear restoring force as described hy this equation is called simple harmonic motion and is characterized by acceleration which is proportional to the displacement but of opposite sign. Equation 8/1 is normally written as x + w„2x — 0
(8/2)
where ojn
=
(8/3)
Jk/m
is a convenient substitution whose physical significance will be clarified shortly.
Solution for Undamped Free Vibration Because we anticipate an oscillatory motion, we look for a solution which gives X as a periodic function of time. Thus, a logical choice is x = A cos o)nt + B sin ojnt
(8/4)
or, alternatively, x — C sin (uj + 0
(8/5)
Direct substitution of these expressions into Eq. 8/2 verifies that each expression is a valid solution to the equation of motion. We determine the constants A and B, or C and 4-', from knowledge of the initial displacement XQ and initial velocity ,r0 of the mass. For example, if we work with the solution form of Eq. 8/4 and evaluate .r and x at time t = 0, we obtain x0 — A
and
i 0 = Bo>,
Substitution of these values of A and B into Eq. 8/4 yields
jc =
cos w„t H
sin ojnt.
(8/6)
The constants C and if/ of Eq. 8/5 can be determined in terms of given initial conditions in a similar manner. Evaluation of Eq. 8/5 and its first time derivative at t — 0 gives 3C0 = C sin i/f
and
,r0 = Co>n cos if/
Free Vibration of Particles
603
604
Chapter 8
Vibration and Time R e s p o n s e
Solving for C and i
if
=
tan _1 (.r 0 iü„/.t 0 )
Substitution of these values into Eq. 8/5 gives x — vStF
+
^xjii)n)2 sin [ojJ + tan' 1 (x 0 iij n /i ü )]
(8/7)
Equations 8/6 and 8/7 represent two different mathematical expressions for the same time-dependent motion. We observe that C — J A 2 + B 2 and ip -• tan _ 1 (A/B).
Graphical Representation of Motion The motion may be represented graphically, Fig. 8/2, where x is seen to be the projection onto a vertical axis of the rotating vector of length C. The vector rotates at the constant angular velocity ton — Jkfrn, which is called the natural circular frequency and has the units radians per second. The number of complete cycles per unit time is the natural frequency f„ = IOJ2TT and is expressed in hertz (1 hertz (Hz) = 1 cycle per second). The time required for one complete motion cycle (one rotation of the reference vector) is the period of the motion and is given by T = 1 //„ = X
+x
Figure 8/2
We also see from the figure that x is the sum of the projections onto the vertical axis of two perpendicular vectors wrhose magnitudes are A and B and whose vector sum C is the amplitude. Vectors A, B, and C rotate together writh the constant angular velocity ion. Thus, as we have already seen, C = + B' and ip = tan ^(AJB).
Equilibrium Position as Reference As a further note on the free undamped vibration of particles, we see that, if the system of Fig. 8 / l a is rotated 90° clockwise to obtain the system of Fig. 8/3 where the motion is vertical rather than horizontal,
Article 8/2
the equation of motion (and therefore all system properties) is i m changed if we continue to define .t as the displacement from the equilibrium position. T h e equilibrium position n o w involves a nonzero spring deflection 5 st . F r o m the free-body diagram of Fig. 8/3, N e w t o n ' s second law gives —&(Sat + x) + nig — mx
Free Vibration of Particles
J £ ^

'A
L Equilibrium position
mg
-kSst + mg = 0 Thus, we see that the pair of forces — kSst and mg on the left side of the motion equation cancel, giving
Figure 8/3
mx + kx = 0 which is identical to Eq. 8/1, T h e lesson here is that by defining the displacement variable to be zero at equilibrium rather t h a n at the position of zero spring deflection, we may ignore the equal and opposite forces associated with equilibrium. 11 '
Equation of Motion for Damped Free Vibration Every mechanical system possesses some inherent degree of friction, which dissipates mechanical energy. Precise mathematical models of the dissipative friction forces are usually complex. T h e dashpot or viscous damper is a device intentionally added to systems for the p u i p o s e of limiting or retarding vibration. It consists of a cylinder filled with a viscous fluid and a piston with holes or other passages by which the fluid can flow from one side of the piston to the other. Simple dashpots arranged as shown schematically in Fig. 8/4« exert a force F,j whose magnitude is proportional to the velocity of the mass, as depicted in Fig. 8/4b. The constant of proportionality c is called the viscous damping coefficient and has units of N *s/m or lb-sec/ft. T h e direction of the damping force as applied to the mass is opposite that of the velocity „i. Thus, the force on the mass is — ex. Complex dashpots with internal flow-rate-dependent one-way valves can produce different damping coefficients in extension and in compression; nonlinear characteristics are also possible. We will restrict our attention to the simple linear dashpot. T h e equation of motion for the body with damping is determined from the free-body diagram as shown in Fig. 8/4a Newton's second law gives mx + cx + kx = 0
É(5rt + x)
I
At the equilibrium position if = 0, the force s u m must be zero, so that
-kx — cx — mx
> Sk <
605
(a)
(8/8)
*For nonlinear systems, all forces, including the static forces associated with equilibrium, should be included in the analysis.
(il) Figure B/4
606
Chapter 8
Vibration and Time R e s p o n s e
In addition to the substitution to,, — Jk/m, it is convenient, for reasons which will shortly b e c o m e evident, to introduce the combination of constants I = ç/(2mù)J T h e quantity ( (zeta) is called the viscous clamping factor or damping ratio and is a measure of the severity of the damping. Y o u should verify that C is nondimensional. Equation 8/8 may n o w be written as x + 2£corlx + u>2x — 0
(8/9)
Solution for Damped Free Vibration In order to solve the equation of motion, Eq. 8/9, we assume solutions of the form X. - AeÀt Substitution into Eq. 8/9 yields A 2 + n^ê +
- 0
which is called the characteristic equation. Its roots are Kx = < o „ ( - f +
- 1)
A 2 = m n ( - C - J? - 1 )
Linear systems have the property of superposition, which means that the general solution is the sum of the individual solutions each of which corresponds to one root of the characteristic equation. Thus, the general solution is x = A]eAl' + A2ex^ - A^
v ' ? 3 ! ) ^ + A2e{ ' f ' - f ^ t o . f
(8/10)
Categories of Damped Motion Because 0 £ f £ the radicand (f — 1) may be positive, negative, or even zero, giving rise to the following three categories of damped motion: I. ( > 1 (overdamped). T h e roots Aj and K 2 are distinct, real, and negative numbers. T h e motion as given by Eq. 8/10 decays so that « approaches zero for large values of time t. There is no oscillation and therefore no period associated with the motion. II, § ~ 1 (critically damped). The roots Aj and A 2 are equal, real, and negative numbers (Ax = A a = — oj n ). T h e solution to the differential equation for the special case of equal roots is given by (Aj
+A2t)e~a,-t
Article 8/2
Conditions: m - 1 kg, k - 9 N/m 3'0 = 30 mm, ,tu = 0
- c = 15 N-s/m (f = 2.5), overdampod - c = 6 N-s/m (f = 1), critically damped
Again, the motion decays with ,t approaching zero for large time, and the motion is nonperiodic. A critically damped system, when excited with an initial velocity or displacement (or both), will approach equilibrium faster than will an overdamped system. Figure 8/5 depicts actual responses for both an overdamped and a critically damped system to an initial displacement x u and no initial velocity ( ¿ 0 = 0). III. I < 1 (underdamped). N o t i n g that the radicand — 1) is negative and recalling that e[a~b> — eaeh, we may rewrite Eq. 8/10 as
where i — J—l. It is convenient to let a n e w variable u>d represent the combination tu_v l — f . Thus, x - {A1e'Ji + A2e-ia<*}e-~i'°»t Use of the Euler formula e = cos .r ± i sin x allows the previous equation to be written as x — {A[(cos iodt + i sin 10dt) + A a ( c o s ojdt — i sin (i)dt)}e~^' — { ( A j + A2) cos codt + ¿(A, - A2) sin w d t}e~t' a »' = {A :J cos wdt + A 4 sin todt}e-^J
(8/11)
where A, = CAi + A a ) and A 4 = i ( A j — A a ). We have shown with Eqs. 8/4 and 8/5 that the sum of two equal-frequency harmonics, such as those in the braces of Eq. 8/11, can be replaced by a single trigonometric function which involves a phase angle. Thus, Eq. 8/11 can be written as x = {C sin (ojdt + 4>) or x =
sin (wdt + < />
(8/12)
Free Vibration of Particles
607
608
Chapter 8
Vibration and Time R e s p o n s e
x, mm Conditions: m = 1 kg, k - 36 N/m c= 1 N-s/m (£ = 0.0833) XQ = 30 mm, x0 = 0
Equation 8/12 represents an exponentially decreasing harmonic function, as shown in Fig. 8/6 for specific numerical values. T h e frequency '>d
=
' J i - C2
is called the damped natural frequency. T h e damped period is given by r d - 2TT¡
- (2).
It is important to note that the expressions developed for the constants C and if/ in terms of initial conditions for the case of no damping are not valid for the case of damping. To find C and d/ if damping is present, y o u must begin anew, setting the general displacement expression of Eq. 8/12 and its first time derivative, both evaluated at time t — 0, equal to the initial displacement ,ru and initial velocity x 0 , respectively.
-20
-30 Figure 8/6
Determination of Damping by Experiment We often need to experimentally determine the value of the damping ratio C f o r an underdamped system. T h e usual reason is that the value of the viscous damping coefficient c is not otherwise well known. To determine the damping, we may excite the system by initial conditions and obtain a plot of the displacement x versus time t, such as that shown schematically in Fig. 8/7. We then measure two successive amplitudes and x 2 a full cycle apart and compute their ratio Ce-fr^i
= pi<'nTd
T h e logarithmic decrement S is defined as
S
sin icOfft +i¡/)
=
111
I ? I = I f c Ä t = &>n f= ' i) û» nv I - f 2
F r o m this equation, we m a y solve for £ and obtain
i ~ Figure 8/7
7 = = 5 v'l - i
y(2rr) z +• Sa
For a small damping ratio, = x2 and 8 < < 1, so that f = S/2TT. If and x 2 are so close in value that experimental distinction between them is impractical, the above analysis may be modified by using two observed amplitudes which are n cycles apart.
Article 8/2
Free Vibration of Particles
609
Sample Problem 8/1 A body weighing 25 lb is suspended from a spring of constant k = 160 lb/ft. At time t
0, it has a downward velocity of 2 ft/sec as it passes through the posik = 160 lb/ft
tion of static equilibrium. Determine Co) the static spring deflection Sji (b) the natural frequency of the system in both rad/sec (UJ,,) and cycles/sec ( / „ )
W = 25 lb
(c) the system period t (d) the displacement x as a function of time, where x is measured from the position of static equilibrium (e) the maximum velocity ¡ j m „ attained by the mass
klSst
( / ) the maximum acceleration a m . „ attained by the mass.
Solution,
(a) From the spring relationship F s mg
mg
S st
kßa
WW;
(f>)
'f-
T =
t
=
Ans.
14.36 rad/sec
Ans.
= 2.28 cycles/sec
2.k=
Ans.
Ans.
0 438sec
-
id) From Eq. 8/6: x
Xu cos uj,,t -
sin 10,t
(0) cos 14.36f +
14.36
sin 14.36f
0.1393 sin 14,36?
Ans.
As an exercise, let us determine JC from the alternative Eq. 8/7:
x ••
+ (xu/ion)'2 sin to J + tan
= ' UmJ™
L
V ¿0 /_
14.36/ + tan
/'{OKI4.36) j
0.1393 sin 14.36i ie) The velocity is x 14.36(0.1393) cos 14.36/ 2 cos 14.36f. Because the cosine function cannot be greater than 1 or less than —1, the maximum velocity i' lllas is 2 ft/sec, which, in this case, is the initial velocity. Ans. if)
The acceleration is x = - 1 4 . 3 6 ( 2 ) sin 14.36f
The maximum acceleration
9
kx, we see that at equilibrium
0.1562 ft or 1.875 in.
J
mg
+$
kx
I
I
T mg
Helpful Hints 160 25/32.2
/„ = < 1 4 . 3 6 ) ^ )
(c)
25
Equilibrium ium I position
is 28.7
- 2 8 . 7 sin 14.36i ft/sec 2 .
Ans.
(T) You shoidd always exercise extreme caution in the matter of units. In the subject of vibrations, it is quite easy to commit errors due to mixing of feet and inches, cycles and radians, and other pairs which frequently enter the calculations. (2) Recall that when we refer the motion to the position of static equilibrium, the equation of motion, and therefore its solution, for the present system is identical to that for the horizontally vibrating system.
i
610
Chapter 8
V i b r a t i o n and T i m e R e s p o n s e
Sample Problem 8/2 The 8-kg body is moved 0.2 m to the right of the equilibrium position and released from rest at time f = 0. Determine its displacement at time t 2 s. The viscous damping coefficient c is 20 N • s/m, and the spring stiffness k is 32 N/m. Solution. We must first determine whether the system is underdamped, critically damped, or overdamped. For that purpose, we compute the damping ratio Ç. io„ = Jhjm = v '32/8 = 2 rad/s
£=
20 = 0.625 2(8X2)
C
2m,
Equilibrium position
Since f < 1, the system is underdamped. The damped natural frequency is uid = t o j l - ¿® = 2V5 - (0.62512 = 1.561 rad/s. The motion is given by Eq. 8/12 and is x
*
mg
1 cx - 20x -
Ce-e'' sin goi J I i/i) = C e ' ^ sin (1.561i + if.) 1
1I
kx = 32r •
The velocity is then x = -1.25Ce
[ aSi
sin (1.561i + f>) + 1.561& 'l aSf cos (1.56If + #
Evaluating the displacement and velocity at time t = C sin $ = 0.2
= - 1 . 2 5 C sin f + 1.561C cos if.
Solving the two equations for C and ift yields C Therefore, the displacement in meters is x Evaluation for time t
0 gives
0.250c
1/251
0
Helpful Hint
0.256 in and ip = 0.896 rad.
sin (1.561f + 0.896)
2 s gives x2 = —0.01616 m.
Ans.
(T) We note that the exponential factor e ' l z5t is 0.0821 at t ^ - 2 s. Thus, C 0.625 represents severe damping, although the motion is still oscillatory.
Sample Problem 8/3 The two fixed count errotating pulleys are driven at the same angular speed (Do- A round bar' is placed off center on the pulleys as shown. Determine the natural frequency of the resulting bar motion. The coefficient of kinetic friction between the bar and pulleys is ¡¡.¡¡.
Central position
?
Solution. The free-body diagram of the bar is constructed for an arbitrary displacement x from the central position as shown. The governing equations are [XFj. = tnx] = 0] |XAfA = 0]
Na — p.j,
= mx A
Na + Nb - mg = 0
N, 11
cNß- I (f' • + xjrng = 0
G
lf
B i • T
mg
I JV b
Helpful Hints (T) Because the bar is slender and does not rotate, the use of a moment equilibrium equation is justified.
Eliminating N A and iVfi from the first equation yields x + —— - x = 0 We recognize the form of this equation as that of Eq. 8/2, so that the natural frequency in radians per second is o>„ x 2/j,/.g/a and the natural frequency in cycles per second is % lUkgfa
1A T
Aj!S.
© We note that the angular speed ui0 does not enter the equation of motion. The reason for this is our assumption that the kinetic friction force does not depend on the relative velocity at the contacting surface.
Article
PROBLEMS (Unless otherwise indicated, all motion variables are referred to the equilibrium position.)
Introductory Problems—Undamped, Free Vibrations 8/1 When a 3-kg collar is placed upon the pan which is attached to the spring of unknown constant, the additional static deflection of the pan is observed to be 42 mm. Determine the spring constant k in N/m, lb/in., and lb/ft. Ans. k 701 N/m k 4.00 lb/in. k 48.0 lb/ft
7/6
Problems 55Ï
8 / 4 For the system of Frob. 8/3, determine the position x of the mass as a function of time if the mass is released from rest at time t 0 from a position 2 inches to the left of the equilibrium position. Determine the maximum velocity and maximum acceleration of the mass over one cycle of motion. 8 / 5 For the system of Prob. 8/3, determine the position x as a function of time if the mass is released at time / = 0 from a position 2 inches to the right of the equilibrium position writh an initial velocity of 9 in./sec to the left. Determine the amplitude C and period r of the motion. A/is. * = 2.06 sin {18/ + 1.816) in. C = 2.06 in., T = 0.349 sec 8 / 6 For the spring-mass system shown, determine the static deflection S st , the system period r, and the maximum velocity which result if the cylinder is displaced 100 mm downward from its equilibrium position and released.
k = 98 N/m 42 mm
Equilibrium position
Problem 8/1 8 / 2 Show that the natural frequency of a vertically oriented spring-mass system, such as that of Frob. 8/1, may be expressed as w„ = Jgitist where 3 i ( is the static deflection. 8 / 3 Determine the natural frequency of the spring-mass system in both radians per second and cycles per second (Hz). Arts. ton 18 rad/sec, /„ = 2.86 Hz I k = 54 lb/in.
H/WWVH
64.4 lb
Problem 611/51
m = 2 kg
Problem 8/6 8 / 7 The cylinder of the system of Prob. 8/6 is displaced 100 mm downward from its equilibrium position and released at time t = 0. Determine the position y, velocity v, and acceleration a when t 3 s. What is the maximum acceleration? Arts, y = - 0 . 0 5 4 8 m, v - 0.586 m/s a = 2.68 m/s = 4,9 m/s 2
612
Chapter 8
V i b r a t i o n and T i m e R e s p o n s e
8 / 8 In the equilibrium position, the 30-kg cylinder causes a static deflection of 50 mm in the coiled spring. li the cylinder is depressed an additional 25 mm and released from rest, calculate the resulting natural frequency /„ of vertical vibration of the cylinder in cycles per second (Hz).
8/11 If the 100-kg mass has a downward velocity of 0.5 m/s as it passes through its equilibrium position, calculate the magnitude a mal: of its maximum acceleration. Each of the two springs has a stiffness k - • 180 kN/m.
Problem 8/11
Problem 8/8 8 / 9 For the cylinder of Prob. 8/8, determine the vertical displacement x, measured positive down in millimeters from the equilibrium position, in terms of the time t in seconds measured f r o m the instant of release from the position of 25 mm added deflection. Alls, x 25 cos 14.01i mm
Representative Problems-Undamped, Free Vibrations 8 / 1 2 Prove that the natural frequency /,, of oscillation for the mass m is independent of 0.
8/10 The vertical plunger has a mass of 2.5 kg and is supported by the two springs, which are always in compression. Calculate the natural frequency /„ of vibration of the plunger if it is deflected from the equilibrium position and released from rest. Friction in the guide is negligible.
Problem 8/12 k1 = 3.6 kN/m Fixed
2.5 kg Problem 8/10
8 / 1 3 An old car being moved by a magnetic crane pickup is dropped from a short distance above the ground. Neglect any damping effects of its worn-out shock absorbers and calculate the natural frequency /„ in cycles per second (Hz) of the vertical vibration which occurs after impact with the ground. Each of the four springs on the 1000-kg car has a constant of 17.5 kN/m. Because the center of mass is located midway between the axles and the car is level when dropped, there is no rotational motion. State any assumptions. Ans. f„ = 1.332 Hz
Article 8/2
Electromagnet
613
8 / 1 6 Explain how the values of the mass m l and the spring constant k may be experimentally determined if the mass m 2 is known. Develop expressions for mj and k in terms of specified experimental results. Note the existence of at least three ways to solve the problem. m
Problem 8/13
Problems
ij-—^^
k
8 / 1 4 During the design of the spring-sup port system for the 4000-kg weighing platform, it is decided that the frequency of free vertical vibration in the unloaded condition shall not exceed 3 cycles per second. (a) Determine the maximum acceptable spring constant k for each of the three identical springs. (b) For this spring constant, what would be the natural frequency /„ of vertical vibration of the platform loaded by the 40-Mg truck?
Problem 8/16 8 / 1 7 A 90-kg man stands at the end of a diving board and causes a vertical oscillation which is observed to have a period of 0.6 s. What is the static deflection £>st at the end of the board? Neglect the mass of the board.
Ans. Sat
89.5 mm
Problem 8/14 8 / 1 5 Replace the springs in each of the two cases shown by a single spring of stiffness k (equivalent spring stiffness) which will cause each mass to vibrate with its original frequency. A»s. (a) k • k-, + k2, ib) f = + ~ k ¿j k2
ki
8 / 1 8 With the assumption of no slipping, determine the mass m of the block which must be placed on the top of the 6-kg cart in order that the system period be 0.75 s. What is the minimum coefficient of static friction |L',: for which the block will not slip relative to the cart if the cart is displaced 50 mm from the equilibrium position and released?
6 kg (a)
(tt Problem 8/15
Problem 8/18
614
Chapter 8
V i b r a t i o n and T i m e R e s p o n s e
8 / 1 9 Calculate the natural frequency C
I/-M
5m
8 / 2 2 The large cement bucket suspended from the crane by an elastic cable has a mass of 4000 kg. When the bucket is disturbed, a vertical oscillation of period 0.5 s is observed. What is the static deflection i3st of the bucket? Neglect the mass of the cable and assume that the crane is rigid for the inboard support position shown. IXIX1X X I X I X I X I X I X I X I X I X !
Problem 8/19 8 / 2 0 An energy-absorbing car bumper with its springs initially undeformcd has an equivalent spring constant of 3000 lb/in. If the 2500-lb car approaches a massive wall with a speed of 5 mi/hr, determine fa) the velocity v of the car as a function of time during contact with the wrall, where t = 0 is the beginning of the impact , and (i>) the maximum deflection xmgl of the bumper.
Problem 8/22 8 / 2 3 The cylindrical buoy floats in salt water (density 1030 kg/m ) and has a mass of 800 kg with a low center of mass to keep it stable in the upright position. Determine the frequency f„ of vertical oscillation of the buoy. Assume that the water level remains undisturbed adjacent to the buoy. Ans. fn = 0.301 Hz 0.6 m
M 5 mi/hr
Problem 8/20 8 / 2 1 A small particle of mass m is attached to two highly tensioned wires as shown. Determine the system natural frequency ai„ for small vertical oscillations if the tension T in both wires is assumed to be constant. Is the calculation of the small static deflection of the particle necessary? j2T Ans. to,. - 1 -ml V»
o
Problem 8/21
I
Problem 8/23
Article 7/10 Problems 8/24 The cylinder of mass in is given a vertical displacement y0 from its equilibrium position and released. Write the differential equation for the vertical vibration of the cylinder and find the period r of its motion. Neglect the friction and mass of the puEey.
Problem 8/24
8/26 Derive the differential equation of motion for the system shown in terms of the variable x1. The mass of the linkage is negligible. State the natural frequency 0n' in rad/s for the case hj k2 — k and mi m-, = m. Assume small oscillations throughout.
Problem 8/26
8/25 Shown in the figure is a model of a one-story building. The bar of mass m is supported by two light elastic upright columns whose upper and lower ends are fixed against rotation. For each column, if a force P and corresponding moment M were applied as shown in the right-hand part of the figure, the deflection ¡5 would be given by 5 = PL:iil2EI, where L is the effective column length, E is Young's modulus, and I is the area moment of inertia of the column cross section with respect to its neutral axis. Determine the natural frequency of horizontal oscillation of the bar' when the columns bend as shown in the figure.
8/27 A 3-kg piece of putty is dropped 2 m onto the initially stationary 28-kg block, which is supported by four springs, each of which has a constant k 800 N/m. Determine the displacement ,v as a function of time during the resulting vibration, where x is measured from the initial position of the block as shown. Arc.s. * = 9.20(10 :i )(l - cos 10.16f) + 59.7(10 sin 10.16f m 3 kg
* „ IsEI Alls. OJ„ = 2 / mL3
2m
28 kg
Problem 8/25
615
Problem 8/27
616
Chapter 8
Introductory
Vibration and Time Response
Problems-Damped,
Free
Vibrations
8/28 Determine the value of the damping ratio C for the simple spring-mass-das hp ot system shown. h
c = 2.5 lb-sec/ft
8/32 A linear harmonic oscillator having a mass of 1.10 kg is set into motion with viscous damping. If the frequency is 10 Hz and if two successive amplit udes a full cycle apart are measured to be 4.65 mm and 4.30 mm as shown, compute the viscous damping coefficient e. mm
8 1b k = 3 lb/in. Time
Problem 8/28 8/29 Determine the value of the viscous damping coefficient c for which the system shown is critically damped. Ans. c 2050 N-s/m Problem 8/32
' 30 kN/m
8/33 Determine the value of the viscous damping coefficient c for which the system shown is critically damped. Ans. c = 154.4 lb-sec/ft
35 kg 200 lb/in.
80 lb
Problem 8/29 8/30 The 8-lb body of Frob. 8/28 is released from rest a distance .r0 to the right of the equilibrium position. Determine the displacement x as a function of time t, where t = 0 is the time of release. 8/31 The addition of damping to an undamped springmass system causes its period to increase by 25 percent. Determine the damping ratio g. Ans.
Problem 8/33
Article 7/10 Problems 8/34 The 2.5-kg spring-supported cylinder is set into free vertical vibration and is observed to have a period of 0.75 s in part (a) of the figure. The system is then completely immersed in an oil batli in part (ft) of the figure, and the cylinder is displaced from its equilibrium position and released. Viscous damping ensues, and the ratio of two successive positive-displacement amplitudes is 4. Calculate the viscous damping ratio C, the viscous damping constant c, and the equivalent spring constant k.
t XN
1
617
A
. ÎjV
J^N+1
Problem 8/35 8/36 For the damped spring-mass system shown, determine the viscous damping coefficient for which critical damping will occur.
lb)
(a) Problem 8/34
Representative
Problems-Damped,
Free
Vibrations
8/35 The figure represents the measured displacementtime relationship for a vibration with small damping where it is impractical to achieve accurate results by measuring the nearly equal amplitudes of two successive cycles. Modify the expression for the viscous damping factor I based on the measured amplitudes a'd and wrhich are N cycles apart. Ans. f =
SN
J&TTN}2 + V
where Sfj = In In I( — —) W /
k - 2 kN/m
Problem 8/36
618
Chapter 8
Vibration and Time Response
8/37 A damped spring-mass system is released from rest from a positive initial displacement x0. If the succeeding maximum positive displacement is xJ2, determine the damping ratio C of the system. Arcs. C 0.1097

x
c - 42 N-s/m 2 kg
M W — k = 98 N/m
x
Problem 8/40 8/41 The system shown is released from rest from an initial position ,ï0. Determine the overshoot displacement Xi. Assume transnational motion in the.r-direction. Ans. = —0.1630.ru x
Problem 8/37 8/38 If the amplitude of the eighth cycle of a linear oscillator with viscous damping is sixteen times the amplitude of the twentieth cycle, calculate the damping ratio if. 8/39 Further design refinement for the weighing platform of Frob. 8/14 is shown here where two viscous dampers are to be added to limit the ratio of successive positive amplitudes of vertical vibration in the unloaded condition to 4. Determine the necessary viscous damping coefficient c for each of the dampers. A/is. C = 16.24U03) N-s/m 4000 kg
Problem 8/39 8/40 The 2-kg mass is released from rest at a distance ,r0 to the right of the equilibrium position. Determine the displacement r a s a function of time.
Problem 8/41 8/42 The mass of a given critically damped system is released at time t = 0 from the position JC0 > 0 with a negative initial velocity. Determine the critical value (,vu)t, of the initial velocity below which the mass will pass through the equilibrium position. 8/43 The mass of the system shown is released from rest at A'o = 6 in. when t 0. Determine the displacement x at t = 0.5 sec if (a) c = 12 lb-sec/ft and (b)c 18 Ib-sec/ft. Arcs, ta) x 4.42 in., (6) x = 4.72 in. x
c
k - 1 lb/in Problem 8/43
Article 8/2
8 / 4 4 The owner of a 3400-lb pickup truck tests the action of his rear-wheel shock absorbers by applying a steady 100-lb force to the rear bumper and measuring a static deflection of 3 in. Upon sudden release of the force, the bumper rises and then falls to a maximum of V 2 in. below the unloaded equilibrium position of the bumper on the first rebound. Treat the action as a one-dimensional problem with an equivalent mass of half the truck mass. Find the viscous damping factor £ for the real' end and the viscous damping coefficient c f o r each shock absorber assuming its action to be vertical.
_L 8=
I Equil. • position
8 / 4 5 Derive the differential equation of motion for the system shown in its equilibrium position. Neglect the mass of link AB and assume small oscillations. Ans.
m, +
—
m
m-j x + — ci + kx = 0 b2
Problem 8/45
X 5 L
Problem 8/46
Problem 8/44
619
8 / 4 6 Develop the equation of motion in terms of the variable x for the system shown. Determine an expression for the damping ratio L, in terms of the given system properties. Neglect the mass of the crank AB and assume small oscillations about the equilibrium position shown.
n
100 lb
Problems
620
Chapter 8
Vibration and Time Response
8 / 3
FORCED
VIBRATION
OF
PARTICLES
Although there are many significant applications of free vibrations, the most important class of vibration problems is that where the motion is continuously excited by a disturbing force. The force may be externally applied or may be generated within the system by such means as unbalanced rotating parts. Forced vibrations may also be excited by the motion of the system foundation.
Harmonic Excitation Various forms of forcing functions F = F(t.) and foundation displacements .tg = Xyitl are depicted in Fig. 8/8. The harmonic force showrn in part a of the figure occurs frequently in engineering practice, and the understanding of the analysis associated with harmonic forces is a necessary first step in the study of more complex forms. For this reason, we will focus our attention on harmonic excitation. We first consider the system of Fig. 8/9a, where the body is subjected to the external harmonic force F = Fa sin ojt, in which F{] is the force amplitude and w is the driving frequency (in radians per second). Be sure to distinguish between ojn = Jkijm, which is a property of the system, and oj, which is a property of the force applied to the system. We also note that for a force F — F0 cos tot, one merely substitutes cos lot for sin ojt in the results about to be developed.
F(t) or xB(t)
WWW
An automobile undergoing vibration testing of its suspension system.
(a) Harmonic
Ft) or xgit)
P=Fb A / W W Square
Triangle
Saw tooth
Fit) or xB(t) r
r
r
¡ A ^ a
Half sine
General
(6) Periodic Nonharmonic F(t) or xB(t)
l^Vv, Step
Ramp
Cycloidal Impulse (e) Nonperiodic Figure 8/16
Random
A r t i c l e B/3
Forced V i b r a t i o n of Particles
621
F,, sin tot
From the free-body diagram of Fig. 8/9a, we may apply Newton's second law to obtain
• kx —kx — cx + Fa sin ojt — m'x In standard form, with the same variable substitutions made in Art. 8/2, the equation of motion becomes
x + 2 £tonx + ion 2,r =
F0 sin tot
k h'VYVVWH
(«)
(8/13)
m
Base Excitation In many cases, the excitation of the mass is due not to a directly applied force but to the movement of the base or foundation to which the mass is connected by springs or other compliant mountings. Examples of such applications are seismographs, vehicle suspensions, and structures shaken by earthquakes. Harmonic movement of the base is equivalent to the direct application of a harmonic force. To show this, consider the system of Fig. 8/9b where the spring is attached to the movable base. The free-body diagram shows the mass displaced a distance X from the neutral or equilibrium position it would have if the base were in its neutral position. The base, in turn, is assumed to have a harmonic movement xb — b sin
x + 2 l
kb sin lot
(8/14)
We see immediately that Eq. 8/14 is exactly the same as our basic equation of motion, Eq. 8/13, in that Fg is replaced by kb. Consequently, all the results about to be developed apply to either Eq. 8/13 or 8/14.
Undamped Forced Vibration First, we treat the case where damping is negligible (c = 0). Our basic equation of motion, Eq. 8/13, becomes n x + to„2x — — sin tot m
(8/15)
The complete solution to Eq. 8/15 is the sum of the complementary solution xL., which is the general solution of Eq. 8/15 with the right side set to zero, and the particular solution xp, which is any solution to the complete equation. Thus, i = x c + xp. We developed the complementary solution in Art. 8/2. A particular solution is investigated by assuming
• k(x — xB >
'r —
1
c
—
jn
VVWAW^FT Neutral position
(b) Figure 8/9
J
j i
xB = b sin at
622
Chapter 8
Vibration and Time Response that the form of the response to the force should resemble that of the force term. To that end, we assume xp — X sin lot
(8/16)
where X is the amplitude (in units of length) of the particular solution. Substituting this expression into Eq. 8/15 and solving for X yield X =
F0i& 1 ~ (afuj*
(8/17)
Thus, the particular solution becomes
x„ — p
n/* l - Uokonr
. , sin cot
(8/18)
The complementary solution, known as the transient solution, is of no special interest here since, with time, it dies out with the small amount of damping which is always unavoidably present. The particular solution^, describes the continuing motion and is called the steady-state solution. Its period is r — 2rr:a>, the same as that of the forcing function. Of primary interest is the amplitude X of the motion. If we let <5^ stand for the magnitude of the static deflection of the mass under a static load F[h then Sst = FQ/k, and we may form the ratio M
5 4 3
(8/19)
Damped Forced Vibration
2
We now reintroduce damping in our expressions for forced vibration. Our basic differential equation of motion is
1• 0„
—S., 1 - (o)/oıJ 2
The ratio M is called the amplitude ratio or magnification factor and is a measure of the severity of the vibration. We especially note that M approaches infinity as OJ approaches ton. Consequently, if the system possesses no damping and is excited by a harmonic force whose frequency co approaches the natural frequency ojn of the system, then M, and thus X, increase without limit. Physically, this means that the motion amplitude would reach the Emits of the attached spring, which is a condition to be avoided. The value con is called the resonant or critical frequency of the system, and the condition of co being close in value to o)„ with the resulting large displacement amplitude X is called resonance. For os < ion. the magnification factor M is positive, and the vibration is in phase with the force F. For co > co„, the magnification factor is negative, and the vibration is 180° out of phase with F. Figure 8/10 shows a plot of the absolute value of M as a function of the driving-frequency ratio co/co,,.
6
M
-
1
2 co/ma
Figure 8/16
x + 2Çion.i + wn2x —
F0 sin lot
[8/131
Again, the complete solution is the sum of the complementary solution xc, which is the general solution of Eq. 8/13 writh the right side equal to
A r t i c l e 8/3 zero, and the particular solution xp, which is any solution to the complete equation. We have already developed the complementary solution x c in Art, 8/2. When damping is present, we find that a single sine or cosine term, such as we were able to use for the undamped case, is not sufficiently general for the particular solution. So we try X„ —
cos cot + X2 sin < />
xp — X sin (cot — )
or
Substitute the latter expression into Eq. 8/13, match coefficients of sin cot and cos cot, and solve the resulting two equations to obtain F0m
X =
(8/20)
{[l - (W^) ]2 + E ^ K l 2 } 1 7 2 2
0 = tan
U'th'n
1
1 - (wK) J
(8/21)
2
The complete solution is now known, and for underdamped systems it can be written as x = Ce
sin (codt + cp) + X sin (cot - A)
(8/22)
Because the first term on the right side diminishes with time, it is known as the transient solution . The particular solution xp is the steadystate solution and is the part of the solution in which we are primarily interested. All quantities on the right side of Eq. 8/22 are properties of the system and the applied force, except for C and ip (which are determinable from initial conditions) and the running time variable t.
Magnification Factor and Phase Angle Near resonance the magnitude X of the steady-state solution is a strong function of the damping ratio C and the nondimensional frequency ratio colcon. It is again convenient to form the nondimensional ratio M — XtiFJk), which is called the amplitude ratio or magnification factor M =
{[1 - (co/co„nZ + [2Cco!coJ2) 1/2
(8/23)
An accurate plot of the magnification factor M versus the frequency ratio co/co,, for various values of the damping ratio ( i s shown in Fig. 8/11. This figure reveals the most essential information pertinent to the forced vibration of a single-degree-of-freedom system under harmonic excitation. It is clear from the graph that, if a motion amplitude is excessive, two possible remedies would be to (a) increase the damping (to obtain a larger value of () or (b) alter the driving frequency so that co is farther from the resonant frequency wJ(. The addition of damping is most effective near resonance. Figure 8/11 also shows that, except for I = 0, the magnification-factor curves do not actually peak at colcon — 1. The peak for any given value of I can be calculated by finding the maximum value of M from Eq. 8/23.
Forced Vibration of Particles
623
624
Chapter 8
Vibration and Time Response
Figure 8/11
Figure 8/12
The phase angle 4>, given by Eq. 8/21, can vary from 0 to TT and represents the part of a cycle (and thus the time) by which the response x p lags the forcing function F. Figure 8/12 shows how the phase angle 4' varies with the frequency ratio for various values of the damping ratio C Note that the value of 4> when <.okon — 1 is 90° for all values of C To further illustrate the phase difference between the response and the forcing function, we show in Fig. 8/13 two examples of the variation of F and xp with cot. In the first example, co < con and 4> IS taken to be TT/4. In the second example, co > corl and d> is taken to be 3tt/4.
A r t i c l e 8/3
Forced Vibration of Particles
Applications Vibration-measuring instruments such as seismometers and accelerometers are frequently encountered applications of harmonic excitation. The elements of this class of instruments are shown in Fig. 8/14a.. We note that the entire system is subjected to the motion XR of the frame. Letting x denote the position of the mass relative to the frame, we may apply Newton's second law and obtain d2 —ex — kx — m — Oe + at») dt2
•• , c_ • ,k mx mx
-
*lf
x
where (x + xB) is the inertial displacement of the mass. If xB = b sin cot, then our equation of motion with the usual notation is | + 2{tonx + ojn2x = bor sin tot which is the same as Eq. 8/13 if bio2 is substituted for FJm. Again, we are interested only in the steady-state solution xp. Thus, from Eq. 8/20, we have bico/coj2
sin (od — ) 2il t% {[1 - (wiwn)2Y + [2(w/w„V) If X represents the amplitude of the relative response xp, then the nondimensional ratio XJb is X!b = (oikon2M where M is the magnification ratio of Eq. 8/23. A plot of X/b as a function of the driving-frequency ratio co/co,, is shown in Fig. 8/14/). The similarities and differences between the magnification ratios of Figs. 8/146 and 8/11 should be noted. 6 5
I kx
4 X/b 3
J 5
m y I—L. •
lex
Equilibrium Y position f%= = b sin cot
Neutral position
| ai
2
1 0
1
2 tot to,, lb)
(a) Figure 8/14
625
626
Chapter 8
Vibration and Time Response If the frequency ratio w/w„ is large, then X/b = 1 for all values of the damping ratio Under these conditions, the displacement of the mass relative to the frame is approximately the same as the absolute displacement of the frame, and the instrument acts as a displacement meter. To obtain a high value of ti>/mn, we need a small value of d)n — Jkjm, which means a soft spring and a large mass. With such a combination, the mass will tend to stay inertially fixed. Displacement meters generally have very light damping. On the other hand, if the frequency ratio io/ton is small, then M approaches unity (see Fig. 8/11) andX/ft = (to/ioj2 or X = b(w/w„)2. But bio2 is the maximum acceleration of the frame. Thus, X is proportional to the maximum acceleration of the frame, and the instrument may be used as an accelerometer. The damping ratio is generally selected so that M approximates unity over the widest possible range of w/w„. From Fig. 6/11, we see that a damping factor somewhere between f = 0.5 and ( — 1 would meet this criterion.
Electric Circuit Analogy An important analogy exists between electric circuits and mechanical spring-mass systems. Figure 8/15 shows a series circuit consisting of a voltage E which is a function of time, an inductance L, a capacitance C, and a resistance R. If we denote the charge by the symbol q, the equation which governs the charge is
R'
(8/24)
L 'q + Rq + ~ q = E E
This equation has the same form as the equation for the mechanical system. Thus, by a simple interchange of symbols, the behavior of the electrical circuit may be used to predict the behavior of the mechanical system, or vice versa. The mechanical and electrical equivalents in the following table are worth noting:
Figure 8/3 5
MECHANICAL-ELECTRICAL
EQUIVALENTS
MECHANICAL
ELECTRICAL
QUANTITY
SYMBOL
SI UNIT
QUANTITY
SYMBOL
Mass Spring stiffness Force Velocity Displacement Viscous damping constant
m k F
kg N/m N m/s m N-s/m
Inductance 1/Capacitance Voltage Current Charge Resistance
L VC E I
X X C
ci R
SI UNIT H 1 IF V A C n
henry 1/farad volt ampere coulomb ohm
A r t i c l e 8/3
Forced Vibration of Particles
Sample Problem 8/4
r
A 50-kg instrument is supported by four springs, each of stiffness 7500 N/m. If the instrument foundation undergoes harmonic motion given in meters by Xg 0.002 cos 50i, determine the amplitude of the steady-state motion of the instrument. Damping is negligible.
0
627
i
u? 0
Û
XB
Solution. For harmonic oscillation of the base, we substitut e kb for F„ in our- particular's olution results, so that, from Eq. 8/17, the steady-state amplitude becomes X =
Helpful Hints
1 - («>/«>„ )2
The resonant frequency is ui„ = Jkjm v;4(7500J/50 pressed frequency io 50 rad/s is given. Thus, X =
0.002 - = -6.32(10 1 - (50/24.5)
24.5 rad/s, and the im-
) m
-0.632 mm
Ans.
Note that the frequency ratio w/ui,, is appr oximately 2, so that the condition of resonance is avoided.
(T) Note that either sin 50i or cos 50f can be used for the forcing function with this same result. © The minus sign indicates that the motion is 180' out of phase with the applied excitation.
Neutral position f xB = b cos oit
Sample Problem 8/5 The spring attachment point B is given a horizontal motion xB = b cos ait. Determine the critical driving frequency uiL for which the oscillations of the mass m tend to become excessively large. Neglect the friction and mass associated with the pulleys. The two springs have the same stiffness k.
Solution. The free-body diagram is drawn for arbitrary positive displacements x and xB. The motion variable .r is measured downward from the position of static equilibr ium defined as that which exists when xB 0. The additional stretch in the upper spring, beyond that which exists at static equilibrium, is 2x — xB. Therefore, the dynamic spring force in the upper spring, and hence the dynamic tension T in the cable, is k(2x — xB). Summing forces in the A--direction gives [ZF t = mXj
— 2k(2x —
kx (Dynamic forces only)
— kx = mx
which becomes
Helpful Hints »
+
bk m
(T) If a review of the kinematics of constrained motion is necessary, see Art . 2/9,
2kb cos mi m
The natural frequency of the system is io„
v 5 k/m.
„=
Thus, Ans.
(?) We learned from the discussion in Art. 8/2 that the equal and opposite forces associated with the position of static equilibrium may be omitted from the analysis. Our use of the terms dynamic spring force and dynamic tension stresses that only the force increments in addition to the static values need be considered.
628
Chapter 8
V i b r a t i o n and T i m e R e s p o n s e
Sample Problem 8/6
n - n. flin /jif
The 100-lb piston is supported by a spring of modulus k 200 lb/in. A dashpot of damping coefficient c 85 Ib-sec/ft acts in parallel with the spring. A fluctuating pressure p 0.625 sin 3Of in lb/in. 2 acts on the piston, whose top surface area is 80 in.' Determine the steady-state displacement as a function of time and the maximum force transmitted to the base.
Solution. rati or
We begin by computing the system natural frequency and damping
,J
'
(2001(12)
=
c =
J t
=
1
F = pA sin cot
27.8 rad/sec
V 100/32.2 85
0.492 (underdamped)
2 mm,,
Equilibrium position
The steady-state amplitude, from Eq. 8/20, is X =
F0/k
{[i - w a ö 2 ] 2 + (Dynamic forces only)
(0.625)(80)/[(200)(12)| {[1 - (30/27. SI2]3 + [2!0,492)!30/27.8)]-}
Helpful Hints
= 0.01938 ft
(T) You are encouraged to repeat these calculations with the damping coefficient c set to zero so as to obseive the influence of the relatively large amount of damping present.
The phase angle, from Eq. 8/21, is é = tan
1
= tan
2(ÙJ/OJ„
1 - iw/w J 3 .492X30/2 7.8) - (30/27.8) 2
H
1.724 rad The steady-state motion is then given by the second term on the right side of Eq. 8/22: XB
X sin (od - )
0.01938 sin (30i - 1.724) ft
AP!S.
The force Ft]. transmitted to the base is the sum of the spring and damper forces, or Ftr = kXp + cip = kX sm iiot — d>) + cioX cos (ojt — é) The maximum value of Flr is = ,/(kX)2 + (cwX)2
Xjk2 + c'W
0.01938 v [(200H12)] 2 + (S5) 2 (30) z 67.91b
Aiis.
© Note that the argument of the inverse tangent expression for é has a positive numerator and a negative denominator for the case at hand, thus placing tii in the second quadrant. Recall that the defined range of
Article 7/10 Problems
PROBLEMS
629
k = 6 lb/in. F - 5 cos cot lb
(Unless otherwise instructed, assume that the damping is light to moderate so that the amplitude of the forced response is a maximum at ut/p„ s 1-) Problem 8/50
Introductory
Problems
8/47 A spring-mounted machine with a mass of 24 kg is observed to vibrate harmonically in the vertical direction with an amplitude of 0.30 mm under the action of a vertical force which varies harmonically between Fu and -Fn with a frequency of 4 Hz. Damping is negligible. If a static force of magnitude Fu causes a deflection of 0.60 mm, calculate the equivalent spring constant k for the springs wliich support the machine. Ans. k 5050 N/m 8/48 Determine the amplit ude X of the steady-state motion of the 10-kg mass if (a) c = 500 N- s/m and ib) c 0.
k - 100 kN/m
r
S
m - 10 kg
T
F = 1000 cos 1201 N
8/51 If the viscous damping coefficient of the damper in the system of Prob. 8/50 is c 2.4 lb-sec/ft, determine the range of the driving frequency to for which the magnitude of the steady-state response is less than 3 in. Arcs, to < 5.32 rad'sec, to > 6.50 rad/sec 8/52 If the driving frequency for the system of Prob. 8/50 is to = 6 rad/sec, determine the required value of the damping coefficient c if the steady-state amplitude is not to exceed 3 in. 8/53 The 4-lb body is attached to two springs, each of which has a stiffness of 6 lb/in. The body is mounted on a shake table which vibrates harmonically in the horizontal direction with an amplitude of 0.5 in. and a frequency f which can be varied. Power to the shake table is turned off when electrical contact is made at A or B. Determine the maximum value of the frequency / at which the shake table may be operated without turning the power off as it starts from rest and increases its frequency gradually. Damping may be neglected. The equilibrium position is centered between the fixed contacts. Arcs. / = 3.83 Hz
Problem 8/48
8/49 A viscously damped spring-mass system is excited by a harmonic force of constant amplitude F0 but varying frequency ui. If the amplitude of the steady-state motion is observed to decrease by a factor of 8 as the frequency ratio w/(u„ is varied from 1 to 2, determine the damping ratio ( of the system. Arcs. £ 0.1936
8/50 The 64.4-lb cart is acted upon by the harmonic force shown in the figure. If c 0, determine the range of the driving frequency ui for which the magnitude of the steady-stat e response is less than 3 in.
Problem 8/53
630
Chapter 8
Vibration and Time Response
8/54 The block of weight W = 100 lb is suspended by two springs each of stiffness k = 200 lb/ft and is acted upon by the force F = 75 cos 15f lb where t is the time in seconds. Determine the amplitude X of the steady-state motion if the viscous damping coefficient c is (a) 0 and (b) 60 lb-sec/ft. Compare these amplitudes to the static spring deflection Sit.
8/57 It was noted in the text that the maxima of the curves for the magnification factor JVf are not located at t,iitu„ 1. Determine an expression in terms of the damping ratio I for the frequency ratio at which the maxima occur. Ans. — = , 1 - 2? 8/58 The circular disk of mass m is secured to an elastic shaft which is mounted in a rigid beaiing at A. With the disk at rest a lateral force P applied to the disk produces a lateral deflection A, so that the equivalent spring constant is k = PI A. If the center of mass of the disk is off center by a small amount e from the shaft ccnterline, determine the expression for the lateral deflection S of the shaft due to unbalance at a shaft speed in/ in terms of the natural frequency ui„ = Jk/m oflateral vibration of the shaft. At what critical speed mc would the deflection tend to become large? Neglect damping.
Problem 8/54 8/55 A viscously damped spring-mass system is forced harmonically at the undamped natural frequency (ut/at„ 1). Ifthe damping ratio f is doubled from 0.1 to 0.2, compute the percentage reduction Ri in the steady-state amplitude. Compare with the result R2 of a similar calculation for the condition oi/ui„ = 2. Verify your results by inspecting Fig. 8/11. Ans. Ri = 50%,R 2 = 2.52%
Representative
Problems
8/56 A s ingle-cylinder four-stroke gasoline engine writh a mass of 90 kg is mounted on four stiff spring pads, each with a stiffness of 30( 10;i) kN/m, and is designed to run at 3600 rev/min. The mounting system is equipped with viscous dampers which have a large enough combined viscous damping coefficient c so that the system is critically damped when it is given a vertical displacement and then released while not running. When the engine is running, it fires every other revolution, causing a periodic vertical displacement modeled by 1.2 cos wt mm with t in seconds. Determine the magnification factor M and the overall damping coefficient c.
Problem 8/58 8/59 Each 0.5-kg ball is attached to the end of the light elastic rod and deflects 4 mm when a 2-N force is statically applied to the ball. If the central collar is given a vertical harmonic movement with a frequency of 4 Hz and an amplitude of 3 mm, find the amplitudey 0 of vertical vibration of each ball. Ans. y0 • 8.15 mm
Article 7/10 Problems
It
0.5 kg
631
0.5 kg
Problem 8/59 8/60 Derive the equation of motion for the inertial displacement x, of the mass of Fig. 8/14. Comment on, but do not cariy out, the solution to the equation of motion.
8/61 The motion of the outer cart B is given by xB
Problem 8/62 8/63 When the person stands in the center of the floor system shown, he causes a static deflection Ss( of the floor under his feet. If he walks (or runs quickly!) in the same area, how many steps per second would cause the floor to vibrate with the greatest vertical amplitude?
b sin oit. For what range of the driving frequency to is the amplitude of the motion of the mass in relative to the cart less than 26? /2 Ans. — < f' ~ > 2 V 3* ton 'J,, • xB - b sin at
Problem 8/61 8/62 The 20-kg variable-speed motorized unit is restrained in the horizontal direction by two springs, each of which has a stiffness of 2.1 kN/m. Each of the two dashpots has a viscous damping coefficient c 58 N-s/m. In what ranges of speeds N can the motor be run for which the magnification factor M will not exceed 2?
Problem 8/63
632
Chapter 8
Vibration and Time Response
8/64 The instrument shown has a mass of 43 kg and is spring-mounted to the horizontal base. If the amplitude of vertical vibration of the base is 0.10 mm, calculate the range of frequencies f n of the base vibration which must be prohibited if the amplitude of vertical vibration of the instrument is not to exceed 0.15 mm. Each of the four identical springs has a stiffness of 7.2 kN/m.
A 0 ® 0
8/67 The equilibrium position of the mass m occurs where y 0 and yB = 0. When the attachment B is given a steady vertical motion yB = b sin oit, the mass m will acquire a steady vertical oscillation. Derive the deferential equation of motion for in and specify the circular frequency for which the oscillations of m tend to become excessively large. The stiffness of the spring is k, and the mass and friction of the pulley are negligible. , .. 4k 2kb . . . rr,— Ans. y -l y sin dit, uic = 2 Jk/m m m
. 0 . 0
Equilibrium position
Problem 8/64 8/65 Attachment B is given a Horizontal motion xg = b cos uit. Derive the equation of motion for the mass m and state the critical frequency toc for which the oscillations of the mass become excessively large. Ans. m 'x + cx + (Ax + k2)x = k2b cos uit
W-
4- k a
Problem 8/67 ' xB = b cos oit
«
•wwww 1 kt
_1
nUUV-j—
8/68 Derive an expression for the transmission ratio T for the system of the figure. This ratio is defined as the maximum force transmitted to the base divided by the amplitude Fa of the forcing function. Express your answer in terms o f t , hi, oitt, and the magnification factor M.
Problem 8/65 8/66 Attachment B is given a horizontal motion xB = b cos uit. Derive the equation of motion for the mass m and state the critical frequency mc for which the oscillations of the mass become excessively large. What is the damping ratio I for the system? I xB = 6 cos ait I I
I
Base Problem 8/68
WvyV k Problem 8/66
Article
8 / 6 9 A device to produce vibrations consists of the two counter-rotating wheels, each carrying an eccentric mass mo = 1 kg with a center of mass at a distance e 12 mm from its axis of rotation. The wheels are synchronized so that the vertical positions of the unbalanced masses are always identical. The total mass of the device is 10 kg. Determine the two possible values of the equivalent spring constant k for the mounting which will permit the amplitude of the periodic force transmitted to the fixed mounting to be 1500 N due to the imbalance of the rotors at a speed of 1800 rev/min. Neglect damping. Ans. k = 227 kN/m or 823 kN/m
8/2
Problems
633
8 / 7 1 The seismic instrument is mounted on a structure which has a vertical vibration writh a frequency of 5 Hz and a double amplitude of 18 mm. The sensing element has a mass m = 2 kg, and the spring stiffness is k = 1.5 kN/m. The motion of the mass relative to the instrument base is recorded on a revolving drum and shows a double amplitude of 24 mm during the steady-state condition. Calculate the viscous damping constant c. Ans. c = 44.6 N-s/m
Problem 8/71 Problem 8/69 8 / 7 0 Derive and solve the equation of motion for the mass which is subjected to the suddenly applied force F that remains constant after application. The displacement and velocity of the mass are both zero at time t - 0. Plot x versus t for several motion cycles. Force F
• 8 / 7 2 Determine the amplitude of vertical vibration of the spring-mounted trailer as it travels at a velocity of 25 km/h over the corduroy road whose contour may be expressed by a sine or cosine term. The mass of the trailer is 500 kg and that of the wheels alone may be neglected. During tire loading, each 75 kg added to the load caused the trailer to sag 3 mm on its springs. Assume that the wheels are in contact with the road at all times and neglect damping. At what critical speed u c is the vibration of the trailer greatest? Ans. X = 14.75 mm, = 15.23 km/h
Problem 8/70
1.2 m Problem 8/45
634
Chapter 8
Vibration and Time Response
8 / 4
cos at I
1
1
;
3
3
T o
3
1
J
m
c|l|
VIBRATION
OF
RIGID
BODIES
The subject of planar rigid-body vibrations is entirely analogous to that of particle vibrations. In particle vibrations, the variable of interest is one of translation (x), while in rigid-body vibrations, the variable of primary concern may be one of rotation i f f ) . Thus, the principles of rotational dynamics play a central role in the development of the equation of motion. We will see that the equation of motion for rotational vibration of rigid bodies has a mathematical form identical to that developed in Arts. 8/2 and 8/3 for translational vibration of particles. As was the case with particles, it is convenient to draw the free-body diagram for an arbitrary positive value of the displacement variable, because a negative displacement value easily leads to sign errors in the equation of motion. The practice of measuring the displacement from the position of static equilibrium rather than from the position of zero spring deflection continues to simplify the formulation for linear systems hecause the equal and opposite forces and moments associated with the static equilibrium position cancel from the analysis. Rather than individually treating the cases of (a) free vibration, undamped and damped, and (i>) forced vibrations, undamped and damped, as was done with particles in Aits. 8/2 and 8/3, we will go directly to the damped, forced problem.
Rotational Vibration of a Bar As an illustrative example, consider the rotational vibration of the uniform slender bar of Fig. 8/16a. Figure 8/166 depicts the free-body diagram associated with the horizontal position of static equilibrium. Equating to zero the moment sum about O yields
iai
4 1
t
P
I '
1
2
Lo;
6
1
o;= o
where P is the magnitude of the static spring force. Figure 8/'16c depicts the free-body diagram associated with an arbitrary positive angular displacement B. Using the equation of rotational motion I M q — I 0 8 as developed in Chapter 6, we write
rng
(b>
cos 6U - i y 6 cos
cos i) j - ( p + k y sin g Y y cos p j
+ (FuCoswi)^|cos ej
where 10 = 1 + md2 = ml2/12 + m(l!6)2 = ml2¡9 is obtained from the parallel-axis theorem for mass moments of inertia. For small angular deflections, the approximations sin 8 = 6 and cos 0=1 may be used. With P ~ mg/4, the equation of motion, upon rearrangement and simplification, becomes (c)
Figure 8/16
r
•
h
(F,d!3) COS lot
f) + — 0 + 4 — 0 = — — T m m ml 19
(8/25)
A r t i c l e 8/4 The right side has been left unsimplified in the form M u (cos cot)/I0, where Mj = F0l/3 is the magnitude of the moment about point O of the externally applied force. Note that the two equal and opposite moments associated with static equilibrium forces canceled on the left side of the equation of motion. Thus, it is not necessary to include the static-equilibrium forces and moments in the analysis.
Rotational Counterparts of Translational Vibration At this point, we observe that Eq. 8/25 is identical in form to Eq. 8/13 for the translational case, so wre may write Mn COS (lit 0 + 2 ( w n 0 +
(8/26)
Thus, we may use all of the relations developed in Arts. 8/2 and 8/3 merely by replacing the translational quantities with their rotational counterparts. The following table shows the results of this procedure as applied to the rotating bar of Fig. 8/16:
TRANSLATIONAL
ANGULAR (for current problem) e •
m
£=
X
mX
= Jkfra c
c
2 mai,,
2xkm
% = «?«V 1 - P =
4k
8 + — 0 + —6 = m m
m
M„ cos ait Ir,
>„ = v;'4kim = Ujk/m c c i = 2mti 4 J km
1
- c-
'd = *>J1 ~ f:= 2m A6km ~ €-2
xc = Ce~1'' sin (toJ + Ip)
f)c = Ce'^1 sin (ü)dt + tfi)
Xp = iX cos (tut — if>)
9p = B cos (alt — tjr)
Ml
I» /
M-
FM/ 3) 3 F„ ° = M — UP W
In the preceding table, the variable kv in the expression for (H) represents the equivalent torsional spring constant of the system of Fig. 8/16 and is determined by writing the restoring moment of the spring. For a small angle 0, this moment about O is Mk = -[kmm sin 0JK2//3) cos 0 = Thus, k„ = ^kl2. Note that M0/ku is the static angular deflection which would be produced by a constant external moment M0. We conclude that an exact analogy exists between particle vibration and the small angular vibration of rigid bodies. Furthermore, the utilization of this analogy can save the labor of complete rederivation of the governing relationships for a given problem of general rigid-body vibration.
Vibration of Rigid Bodies
635
636
Chapter 8
Vibration and Time Response
Sample Problem 8/7 A simplified version of a pendulum used in impact tests is shown in the figure. Derive the equation of motion and determine the period for small oscillations about the pivot. The mass center G is located a distance r 0.9 m from O, and the radius of gyration about O is k0 0.95 m. The friction of the bearing is negligible.
Solution. We draw the free-body diagram for an arbitrary, positive value of the angular-displacement variable 6, which is measured counterclockwise for the coordinate syst em chosen. Next we apply the governing equation of motion to obtain [1M0
Io0]
—mgr sin 0
mk08
gr 8 + ^ r sin 8 k*
0
Ans.
Note that the governing equation is independent of the mass. When 8 is small, sin 8 = 8, and our equation of motion may be written as gr 8+^—8
(T) With our choice of point O as the moment center, the bearing reactions O and O never enter the equation of motion.
0
=
The frequency in cycles per second and the period in seconds are Ans.
For the given properties;
Helpful Hints
V (9.81K0.9)
(?) For large angles of oscillation, determining the period for the pendulum requires the evaluation of an elliptic integral.
Sample Problem 8/8 The uniform bar of mass m and length I is pivroted at its center. The spring of constant k at the left end is attached to a stationary surface, but the right-end spring, also of constant k. is attached to a support which undergoes a harmonic motion given by b sin toi. Determine the driving frequency w,. which causes resonance.
Solution. obtain
We use the moment equation of motion about the fixed point O to
-(ft I sin e j l- cos
sin
8 -
8 - yi:
yB = b sin iüt
B
Helpful Hints (T) As previously, we consider only the changes in the forces due to a movement away from the equilibrium position.
j — ~ cos tl Y2.'
Assuming small deflections and simplifying give us - , 6k „ 6kb . . 8 H 8 = sill ilit m ml The natural frequency should be recognized from the now-familiar form of the equation to be b)n
x'Gkim
Thus, a: = (jlu J 6k I in will result in resonance (as well as violation of our smallangle assumption!). Ans.
({sine-yj
(2) The standard form, here is 8 + tan20 M u sin ¡0! ktb —, where M0 - -— and Iff IQ I -•^ml1. The natural frequency
A r t i c l e 8/4
Vibration of Rigid Bodies
637
Sample Problem 8/9 x
Derive the equation of motion for the homogeneous circular cylinder, wliich rolls without slipping. If the cylinder mass is 50 kg, the cylinder radius 0.5 m, the spring constant 75 N/in, and the damping coefficient 10 N-s/m, determine Co) the undamped natural frequency (i>) the damping ratio (c) the damped natural frequency (d) the period of the damped system. In addition, determine x as a function of time if the cylinder is released from rest at the position x —0.2 m when t = 0.
Solution. We have a choice of motion variables in that either x or the angular displacement H of the cylinder may be used. Since the problem statement involves A', we draw the free-body diagram for an arbitrary, positive value of A' and write the two motion equations for the cylinder as © [XFj = mx]
-ex — kx + F = mx
XMg = I0]
-Fr=y2rm*
0
The condition of rolling with no slip is x = rii. Substitution of this condition into the moment equation gives F —^mx. Inserting this expression for the friction force into the force equation for the A;-direction yields 1 .. .. —cx — kx — — mx •--•• mx 2
. 2 c . 2 k a x + — x —- x = 0 3m 3 m
or
Comparing the above equation with that for the standard damped oscillator, Eq. 8/9, allows us to state directly ' i
2t(.j ' '
(b) ' '
= -3m
~
-
M
-
£ = i 3 mot,,
M
-
1
-
»
= — = 0.0667 3(50)(1)
Ans.
Hence, the damped natural frequency and the damped period are (c)
0Jd = M b V 5 - P = ( l ) v 1 - (0.0667)' = 0.998 rad/s
Ans.
(d)
rd = 2-niatd = 2^/0.998 = 6.30 s
Ans.
From Eq. 8/12, the undcrdamped solution to the equation of motion is sin ImJ +
x = Ce The velocity is
x
Ce • [0t)ee7,<1 ' sin (0.998i + tp)
-0.0667Ce' tti>tifi7 ' sin (0.998i 4- <$ + 0.998Ce °-U6ti7i cos (0,998? + tp)
At time t = 0, x and A- become JC(i = C sin if/ = —0.2 i0 = -0.0667C sin i/i + 0.998C cos ifr = 0 The solution to the two equations in C and iIt gives C = -0.200 m
ij/ = 1.504 rad
Thus, the motion is given by a: = -0.200c ' °-uefi7i sin (0.998i 4- 1.504) m
Ans.
Helpful Hints © The angle H is taken positive clockwise to be kinematically consistent with x. © T h e friction force F may be assumed in either direction. We will find that the actual direction is to the right for x > 0 and to the left for x < 0; F = 0 when x = 0.
638
Chapter 8
Vibration and Time Response
PROBLEMS Introductory
Problems
8/73 The light rod and attached small spheres of mass m each are shown in the equilibrium position, where all four springs are equally precompressed. Determine the natural frequency OJ„ and period r for small oscillations about the frictionless pivot O. A
Ans. iii,, =
¡2k
m
• r = ir
8/75 The uniform rod of length I and mass m is suspended at its midpoint by a wire of length L. The resistance of the wire to torsion is proportional to its angle of twist 0 and equals (JG/L)(t where J is the polar moment of inertia of the wire cross section and G is the shear modulus of elasticity. Derive the expression for the period r of oscillation of the bar when it is set into rotation about the axis of the wrire.
j2m

k
M s
-
T
=
HiiJd)
Problem 8/73 8/74 Derive the differential equation for small oscillations of the spring-loaded pendulum and find the period r. The equilibrium position is vertical as shown. The mass of the rod is negligible.
Problem 8/75 8/76 A uniform rectangular' plate pivots about a horizontal axis through one of its corners as shown. Determine the natural frequency IO,, of small oscillations.
Problem 8/74
A r t i c l e 8/4 8/77 The thin-walled cylindrical shell of radius r and height h is welded to the small shaft at its upper end as shown. Determine the natural circular frequency îo| for small oscillations of the shell about the y-axis. Ans.
Mi & ' ~ V '2 / V 2
ffi 3
Problems
639
8/80 The thin square plate is suspended from a socket (not shown) which fits the small ball attachment at O. If the plate is made to swing about axis A-A, determine the period for small oscillations. Neglect the small offset, mass, and friction of the ball.
Problem 8/80 Problem 8/77
8/81 If the square plate of Prob. 8/80 is made to oscil-
8/78 Determine the natural frequency fn for small oscillations in the vertical plane about the bearing O for the semicircular disk of radius r.
late about axis B-B, determine the period of small oscillations. 50 Ans. T = 2 - h V eBg 8/82 The homogeneous 250-kg rectangular block is pivoted about a horizontal axis through O and supported by two springs, each of stiffness k. The base of the block is horizontal in the equilibrium position with each spring under a compressive force of 250 N. Determine the minimum stiffness k of the springs which will ensure vibration about the equilibrium position.
Problem 8/78 8/79 The uniform rod of mass m is freely pivoted about a horizontal axis through point O. Assume small oscillations and determine an expression for the damping ratio C- For what value ccr of the damping coefficient c will the system be critically damped? .
.
cb
I 3
2a
km
375 mm 1 375 mm
Ï
B3= b Problem 8/79
Problem 8/82
640
Chapter 8
Representative
Vibration and Time Response
Problems
8/83 The circulai' ring of radius r is suspended fl ora, a socket (not shown) which fits the small ball attachment at O. Determine the ratio if of the period of small oscillations about axis B-B to that about axis A-A. Neglect the small offset, mass, and friction of the ball. Ans, R =
8/85 The mass of the uniform slender rod is 3 kg. Determine the position x for the 1.2-kg slider such that the system period is 1 s. Assume small oscillations about the horizontal equilibrium position shown. Ans. x 0.558 m
: 250 N/m
2_ V3
O
1.2 kg
3 kg 0.4 m 0.8 m Problem 8/85
8/86 The uniform square plate is suspended in a horizon-
Problem 8/83
tal plane by its four corner cables from fixed points A and B on a horizontal line a distance b above the plate. Determine an expression for the frequency /„ of small oscillations of the plate about the axis A-B.
8/84 The mechanism shown oscillates in the vertical plane about the pivot O. The springs of equal stiffness k are both compressed in the equilibrium position 8 0. Determine an expression for the period t of small oscillations about O. The mechanism has a mass in with mass center G, and the radius of gyration of the assembly about O is ko-
Problem 8/86
Vertical Problem 8/102
Article 8/87 When the motor is slowly brought up to speed, a rather large vibratory oscillation of the entire motor about 0 - 0 occurs at a speed of 360 rev/min, which shows that this speed corresponds to the natural frequency of free oscillation of the motor. If the motor has a mass of 43 kg and a radius of gyration of 100 mm about O-O, determine the stiffness k of each of the four identical spring mounts. Ans. k = 3820 N/m
8/2
Problems
641
8/89 The system of Prob. 8/45 is repeated here. If the link AB now has mass m;, and radius of gyration k0 about point O, determine the equation of motion in terms of the variable x. Assume small oscillations. The dampiirg coefficient for the dashpot is c. X +
'A2
Problem 8/87 8 / 8 8 Determine the value of the mass of system (6) so that the frequency of system (6) is equal to that of system (a). Note that the two springs are identical and that the wheel of system (a) is a solid homogenous cylinder of mass The cord does not slip on the cylinder.
Problem 8/89 8/90 The system of Prob, 8/46 is repeated here. If the crank AB now has mass m., and a radius of gyration k0 about point O, determine expressions for the undamped natural frequency ain and the damping ratio C in terms of the given system properties. Assume small oscillations. The damping coefficient for the damper is c.
(a) Problem 8/88
Problem 8/45
642
Chapter 8
Vibration and Time Response
8/91 Two identical uniform bars are welded together at a right angle and are pivoted about a horizontal axis through point O as shown. Determine the critical driving frequency OJl of the block B which will result in excessively large oscillations of the assembly. The mass of the welded assembly is m. Ans. i a2
=
4 (m
+
i)
1/2 w
f
I
Problem 8/93
Î
1
xB = b sin at
8/94 The homogeneous solid cylindrical pulley has mass m i and radius r. If the attachment at B undergoes the indicated harmonic displacement, determine the equation of motion of the system in terms of the variable x. The cord which connects mass m2 to the upper spring does not slip on the pulley. xB = b cos cot | h 1 I
Problem 8/91
8/92 Determine the natural frequency fn for small oscillations of the composite body in the vertical plane about the bearing O. Approximate the body as a slender bar of mass mi5 and a semicircular disk of mass m, both with the dimension r as shown.
'T
R f Problem 8/94
Problem 8/92 8/93 The uniform solid cylinder of mass m and radius r rolls without slipping during its oscillation on the circular surface of radius R. If the mot ion is confined to small amplitudes 8 8U. determine the period r of the oscillations. Also determine the angular velocity i
T =27R
3 (R - r) v
2g
'
>
- ¿2g(R - rj/3
8/95 The circular disk of mass m and moment of inertia I about its central axis is welded to the steel shaft which, in turn, is welded to the fixed block. The disk is given an angular displacement 80 and then released, causing a torsional vibration of the disk with 8 changing between +80 and — 80. The shaft resists the twist with a moment M = JG8IL, where J is the polar moment of inertia of the cross section of the shaft about the rotation axis, G is the shear modulus of elasticity of the shaft (resistance to shear stress), 8 is the angle of twist in radians, and L is the lengt h of the twisted shaft. Derive the expression for the natural frequency /„ of the torsional vibration. •
4 / , s
-
f
'
=
t
IJG s IL i r
Article 7/10 Problems
Problem 8/95 8/96 The segmented 'dummy' of Prob. 6/107 is repeated here. The hip joint O is assumed to remain fixed to the car, and the torso above the hip is treated as a rigid body of mass m. The center of mass of the torso is at G and the radius of gyration of the torso about O is UQ. Assume that muscular response acts as an internal torsional spring which exerts a moment M - K0 on the upper torso, where K is the torsional spring constant and 0 is the angular deflection from the initial vertical position. If the car is brought to a sudden stop with a constant deceleration a, derive the differential equation for the motion of the torso prior to its impact with the dashboard.
643
• 8/97 The elements of the 'swing-axle' type of independent rear- suspension for automobiles are depicted in the figure. The differential D is rigidly attached to the car' frame. The half-axles are pivoted at their inboard ends (point O for the half-axle shown) and are rigidly attached to the wheels. Suspension elements not shown constrain the wheel motion to the plane of the figure. The weight of the wheel-tire assembly is W = 100 lb, and its mass moment of inertia about a diametral axis passing through its mass center G is 1 Ib-ft-sec^. The weight of the half-axle is negligible. The spring rate and shock-absorber damping coefficient are k 50 lb/in. and c = 200 lb-sec/ft, respectively. If a static tire imbalance is present, as represented by the additional concentrated weight w 0.5 lb as shown, determine the angular velocity OJ which results in the suspension system being driven at its undamped natural frequency. What would be the corresponding vehicle speed vl Determine the damping ratio I. Assume small angular deflections and neglect gyroscopic effects and any car frame vibration. In order to avoid the complications associated with the varying normal force exerted by the road on the tire, treat the vehicle as being on a lift with the wheels hanging free. Ans. iii 10.24 rad/sec, v = 11.95 ft/sec £ = 1.707
l t = 27' —12 = 36' — Problem 8/96
Problem 8/97 • 8/98 For the automobile suspension system, of Prob. 8/97, determine the amplitude X of the vertical motion of point G if the angular velocity of the tire corresponds to (a) the undamped natural frequency of the system and (6) a vehicle speed of 55 mi/hr. Reconcile the two results. Ans. (a) X = 0.0198 in. ( b ) X = 0.0614 in.
644
Chapter 8
Vibration and Time Response
8 / 5
ENERGY
METHODS
In Arts. 8/2 through 8/4 we derived and solved the equations of motion for vibrating bodies by isolating the body with a free-body diagram and applying Newton's second law of motion. With this approach, we were able to account for the actions of all forces acting on the body, including frictional damping forces. There are many problems where the effect of damping is small and may be neglected, so that the total energy of the system is essentially conserved. For such systems, we find that the principle of conservation of energy may frequently be applied to considerable advantage in establishing the equation of motion and, wrhen the motion is simple harmonic, in determining the frequency of vibration.
Determining the Equation of Motion
5 J Ë Equilibrium position
To illustrate this alternative approach, consider first the simple case of the body of mass m attached to the spring of stiffness k and vibrating in the vertical direction without damping, Fig. 8/17. As previously, we find it convenient to measure the motion variable x from the equilibrium position. With this datum, the total potential energy of the system, elastic plus gravitational, becomes
5* V' = Ve +
= k(x + S s t ) 2 - kù32 - mgx
where fi3l= mg/k is the initial static displacement. Substituting kô^= mg and simplifying give V = kx2
Figure 8/17
Thus, the total energy of the system becomes T+ V=
l2mk2
+ kx2
Because T + V is constant for a conservative system, its time derivative is zero. Consequently,
dt
(T + V) = mxx + ltxx — 0
Canceling x gives us our basic differential equation of motion m'x + kx — 0
which is identical to Eq. 8/1 derived in Art. 8/2 for the same system of Fig. 8/3.
Determining the Frequency of Vibration Conservation of energy may also be used to determine the period or frequency of vibration for a linear conservative system, without having to derive and solve the equation of motion. For a system wrhich oscillates with simple harmonic motion about the equilibrium position, from which the
A r t i c l e 8/5 displacement .r is measured, the energy changes from maximum kinetic and zero potential at the equilibrium position x = 0 to zero kinetic and maximum potential at the position of maximum displacement if = ,rmax. Thus, we may write
The maximum kinetic energy is ^ nlax ) 2 , and the maximum potential energy is ^ H x ^ ) 2 . For the harmonic oscillator of Fig. 8/17, we know that the displacement may be written a s r = ,rmaj. sin (m^i + iff), so that the maximum velocity is x maj( = wnxmax. Thus, we may write = 2k(x™**}2
where _i'max is the maximum displacement, at which the potential energy is a maximum. From this energy balance, we easily obtain vin — Jk/tn
This method of directly determining the frequency may be used for any linear undamped vibration. The main advantage of the energy approach for the free vibration of conservative systems is that it becomes unnecessaiy to dismember the system and account for all of the forces which act on each member. In Art. 3/7 of Chapter 3 and in Arts. 6/6 and 6/7 of Chapter 6, we learned for a system of interconnected bodies that an active-force diagram of the complete system enabled us to evaluate the work U' of the external active forces and to equate it to the change in the total mechanical energy T + V of the system. Thus, for a conservative mechanical system of interconnected parts with a single degree of freedom where U' — 0, we may obtain its equation of motion simply by setting the time derivative of its constant total mechanical energy to zero, giving 4 (T + V) = 0
at
Here V = V? + Vg is the sum of the elastic and gravitational potential energies of the system. Also, for an interconnected mechanical system, as for a single body, the natural frequency of vibration is obtained by equating the expression for its maximum total kinetic energy to the expression for its maximum potential energy, where the potential energy is taken to be zero at the equilibrium position. This approach to the determination of natural frequency is valid only if it can be determined that the system vibrates with simple harmonic motion.
Energy Methods
645
646
Chapter 8
Vibration and Time Response
Sample Problem 8/10 The small sphere of mass m is mounted on the light rod pivoted at O and supported at end A by the vertical spring of stiffness k. End A is displaced a small distance y0 below the horizontal equilibrium position and released. By the energy method, derive the differential equation of motion for small oscillations of the rod and determine the expression for its natural frequency tu,r of vibration. Damping is negligible.
h
>
O
5A
Solution. With the displacement y of the end of the bar' measured from the equilibrium position, the potential energy in the displaced position for small values o f y becomes V=Ve
+
Vg = l k(y +
- I kSst2 - mg ~
where 3at is the static deflection of the spring at equilibrium. But the force in the spring in the equilibrium position, from a zero moment sum about O, is (b/l)mg kSgt. Substituting this value in the expression for V and simplifying yield V=|
where we see that the vertical displacement of in is (b/'Dy. Thus, with the energy sum constant, its time derivative is zero, and we have ( r + v ) =
l
2
m
[ ï V
+
Z
=
k y 2
0
which yields •• i I2 'i o AJIS. + = 0 y b m when y is canceled. By analogy with Eq. 8/2, we may write the motion frequency directly as Alis.
i)„ = - Jklm b
Alternatively, we can obtain the frequency by equating the maximum kinetic energy, which occurs at y 0, to the maximum potential energy, which occurs at y = ,Vo = ,)'ma», where the deflection is a maximum. Thus, T = V mav rn
gives
1 lb. V ^'ly^'n^J
=
1, 2 y'
Knowing that we have a harmonic oscillation, which can be expressed as y = ylnaJ( sin u>nt, we have y11MX yraiisf^„- Substituting this relation into our energy balance gives us 1 (b Y 1 ^ [ ¡ y ^ n j =2 as before.
so that
= 'b v'A/m
Alis.
jpst T
Equilibrium position Helpful Hints (T) For large motion of cause our tion of the
ky*
The kinetic energy in the displaced position is
I
1
values of y, the circular' the end of the bar would expression for the deflecspring to be in error.
© Here again, we note the simplicity of the expression for potential energy when the displacement is measured from the equilibrium position.
A r t i c l e 8/5
Energy Methods
647
Sample Problem 8/11 Determine the natural frequency uj,, of vertical vibration of the 3-kg collar to which are attached the two uniform 1.2-kg links, which may be treated as slender bars. The stiffness of the spring, which is attached to both the collar and the foundation, is k 1.5 kN/m, and the bars are botli horizontal in the equilibrium position. A small roller on end B of each link permits end A to move with the collar. Frictional retardation is negligible.
k = 1.5 kN/m
Solution. In the equilibrium position, the compression P in the spring equals the weight of the 3-kg collar, plus half the weight of each link or P 3(9.81} + J!{|)(1.2){9.81) 41.2 N. The corresponding static deflection of the spring is Sat P/k 4 1.2/1.5(10s) 27.5(10 a) m. With the displacement variabley measured downward from the equilibrium position, which becomes the position of zero potential energy, the potential energy for each member in the displaced position is (Spring)
F, = k(y + S J2 - ^kSj = ky2 + kSsty = ^ (1.5)(10'V + 1.5(10'J)(27.5)(10 = 7 5 0 / + 41.2y J
(Collar)
©
(Each link)
Vj = ~mcgy = Vg =
3(9.81
= -29.4y J
Helpful Hints © Note that the mass center of each link moves down only half as far' as the collar,
= -1.2(9.81) * = -5.89y J
The total potential energy of the system then becomes V = 750y2 + 41.2y - 29.4y - 2(5.89)y = 750y2 J The maximum kinetic energy occurs at the equilibrium position, where the velocity y of the collar has its maximum value. In that position, in which links (3; AB are horizontal, end B is the instantaneous center of zero velocity for each link, and each link rotates with an angular' velocity y/0.3. Thus, the kinetic energy of each part is (Collar)
T
i?^2 = | (-m^W)2
© To appreciate the advantage of the work-energy method for this and similar problems of interconnected systems, you are encouraged to explore the steps required for solution by the force and moment equations of motion of the separate parts.
* m,y3
= |(1.2)y 2 = 0.2y 2 Thus, the kinetic energy of the collar and both links is T=
©
© Our knowledge of rigid-body kinematics is essential at this point.
T = |mty2 = |y2J
(Each link)
©
© We note again that measurement of the motion variable y from the equilibrium position results in the total potential energy being simply V = ky2,
+ 2(0.2y 2 ) = 1,9y2
With the har'monic motion expressed by y = yraax sin wnt, we have y™« that the energy balance = Vma![ with y V M A X becomes
=
ymaJiujfi,
SO
l - S i y ^ w « ) 2 - 750yn
or
u>n = v<'750/1.9
19.87 Hz
A/is.
© If the oscillations were large, we would find that the angular velocity of each link in its general position would equal y/v'0.09 — y2, which would cause a nonlinear response no longer described by y = y max sin uit.
648
Chapter 8
Vibration and Time Response
PROBLEMS (Solve the following problems by the energy method of Art, 8/5.)
8 / 1 0 1 Determine the natural frequency f„ of the inverted pendulum. Assume small oscillations, and note any restrictions on your solution. A,!S-
Introductory
Problems
2kb2 À = h V ml2 mgl h > 2 b2
S I
8/99 Derive the equation of motion for the pendulum which consists of the slender uniform rod of mass m and the bob of mass M. Assume small oscillations, and neglect the radius of the bob. '3g(m + 2M) Ans. 0 + 0= 0 2 Km + 3Mi
E
Li
Problem 8/101
8/102 The 1.5-kg bar OA is suspended vertically from the
Problem 8/99 8/100 The potential energy V of a linear spring-mass system is given in inch-pounds by 64r2, where x is the displacement in inches measured from the neutral equilibrium position. The kinetic energy T of the system in inch-pounds is given by Si', Determine the differential equation of motion for the system and find the period r of its oscillation. Neglect energy loss.
bearing O and is constrained by the two springs each of stiffness h 120 N/m and both equally procompressed with the bar in the vertical equilibrium position. Treat the bar as a uniform slender rod and compute the natural frequency /„ of small oscillations about O.
Problem 8/102
Article 7/10 Problems 8/103 Determine the period r for the uniform circular hoop of radius r as it oscillates with small amplitude about the horizontal knife edge. Ans.
T
= 2n
—
V S
Problem 8/103 8/104 The spoked wheel of radius r, mass m, and centroidal radius of gyration k rolls without slipping on the incline. Determine the natural frequency of oscillation and explore the limiting cases of k 0 and h r.
649
8/105 Determine the period r of small oscillations of the cylindrical shell of Prob. 8/77, repeated here, about the y-axis. I 2 Ii* Ans. r = 2tt / — — V gh V 2
h2
Problem 8/105 8/106 The length of the spring is adjusted so that the equilibrium position of the arm is horizontal as shown. Neglect the mass of the spring and the arm and calculate the natural frequency /„ for small oscillations.
Problem 8/104 Problem 8/106
650
Chapter 8
Representative
Vibration and Time Response
Problems
8/107 Calculate the frequency fr, of vertical oscillation of the system shown. The 40-kg pulley has a radius of gyration about its center O of 200 mm. Ans. / „ = 1.519 Hz
8/109 By the method of this article, determine the period of vertical oscillation. Each spring has a stiffness of 6 lb/in., and the mass of the pulleys may be neglected. Ans. T = 0.326 sec
!
f , *
f
50 lb
Problem 8/107
Problem 8/109
8/108 The disk has mass moment of inertia IQ about O and is acted upon by a torsional spring of constant K. The position of the small sliders, each of which has mass m, is adjustable. Determine the value o f x for which the system has a given period T.
8/110 The rotational axis of the turntable is inclined at an angle a from the vertical. The turntable shaft pivots freely in bearings which are not showir. If a small block of mass m is placed a distance r from point O, determine the natural frequency iii„ for small rotational oscillations through the angle $. The mass moment of inertia of the turntable about the axis of its shaft is I.
Problem B/108
Problem 8/110
Article 7/10 Problems 8/111 The homogeneous circular cylinder of Prob. 8/93, repeated here, rolls without slipping on the track of radius R. Determine the period r for small oscillations. Aus. r =
IT J
/6(H - r)
V
g
651
8/113 The uniform slender rod of length / and mass m2 is secured to the uniform disk of radius l!5 and mass mj. If the system is shown in its equilibrium position, determine the natural frequency u>n and the maximum angular velocity to for small oscillations of amplitude Bu about the pivot O.
O
6k •is. (.j„ = 3 Ait » 13mj + 26 m 2
V
, = 3 f S / ; 3m
6k j + 26m,
Problem 8/111 8/112 The ends of the uniform bar of mass m slide freely in the vertical and horizontal slots as shown. If the bar is in static equilibrium when f) = 0, determine the natural frequency ain of small oscillations. What condition must be imposed on the spring constant k in order that oscillations take place'.'
• 115 » • 1/5 >|*
21/5
Problem 8/113 8/114 Derive the natural frequency f n of the system composed of two homogeneous circular cylinders, each of mass M, and the connecting link AB of mass m. Assume small oscillations.
Problem 8/114
Problem 8/112
»I« 1/5
652
Chapter 8
Vibration and Time Response
8/115 Each of the two uniform. 1,5-kg slender bars is hinged freely at A with its small upper-end guide roller free to move in the horizontal guide. The bars are supported in their 45' equilibrium positions by the vertical spring of stiffness 1050 N/m. If point A is given a very small vertical displacement and then released, calculate the natural frequency of the resulting motion. Ans. f„ = 3.65 Hz
8/117 The semicylinder of mass m and radius r rolls without slipping on the horizontal surface. By the method of this article, determine the period r of small oscillations. Ans. r = 7.78vfrig
Vertical I
Problem 8/117 8/118 Each of the two slider blocks A has a mass in and is constrained to move in one of the smooth radial slots of the flywheel, which is driven at a constant angular speed OJ. Each of the four springs has a stiffness k. Is it correct to state that the system, composed of the flywheel, blocks, and springs possesses a constant energy? Explain your answer. Problem B/115 8/116 The 12-kg block is supported by the two 5-kg links with two torsion springs, each of constant K = 500 N-m/rad, arranged as shown. The springs are sufficiently stiff so that stable equilibrium is established in the position shown. Determine the natural frequency f n for small oscillations about this equilibrium position. 12 kg Problem B/118 5 kg

O.f m
5 kg
K
J
Problem 8/116
ih-
Article 7/10 Problems 8/119 The front-end suspension of an automobile is shown. Each of the coil springs has a stiffness of 270 lb/in. If the weight of the front-end frame and equivalent portion of the body attached to the front end is 1800 lb, determine the natural frequency J„ of vertical oscillation of the frame and body in the absence of shock absorbers. {Hint: To relate the spring deflection to the deflection of the frame and body, consider the frame fixed and let the ground and wheels move vertically.) Ans. f„ 1.142 Hz
Problem 8/119
653
8/121 The semicircular cylindrical shell of radius r with small but uniform wall thickness is set into small rocking oscillation on the horizontal surface. If no slipping occurs, determine the expression for the period r of each complete oscillation. . Ans. r = 2TT
[Ut - 2)r
V
g
• 8/122 A hole of radius RI4 is drilled through a cylinder of radius if to form a body of mass m as shown. If the body rolls on the horizontal surface without slipping, determine the period r for small oscillations.
8/120 The uniform slender rod of length 2b is supported in the horizontal plane by a bifilar suspension. The rod is set into small angular oscillation about the vertical axis through its center O. Derive the expression for the period t of oscillation. [Hint: From the auxiliary sketch note that the rod rises a distance h corresponding to an angular twist 0. Also note that Ifi = bt) for small angles and that cos fj may be replaced by the first two terms of its series expansion. A simple harmonic solution of the form $ 80 sin to„t may be used for small angles.)
Ans. r = 41.4
Problem 8/122
—
V£
654
Chapter 8
Vibration and Time Response
8 / 6
CHAPTER
REVIEW
In studying the vibrations of particles and rigid bodies in Chapter 8, we have observed that the subject is simply a direct application of the fundamental principles of dynamics as presented in Chapters 3 and 6. However, in these previous chapters, we determined the dynamic behavior of a body only at a particular instant of time or found the changes in motion resulting from only finite intervals of displacement or time. Chapter 8, on the other hand, has treated the solution of the defining differential equations of motion, so that the linear or angular displacement can be fully expressed as a function of time.
Particle Vibration We divided our study of the time response of particles into the two categories of free and forced motion, with the further subdivisions of negligible and significant damping. We saw that the damping ratio ( is a convenient parameter for determining the nature of unforced but viscously damped vibrations. The prime lesson associated with harmonic forcing is that driving a lightly damped system with a force whose frequency is near the natural frequency can cause motion of excessively large amplitude—a condition called resonance, wrliich usually must be carefully avoided.
Rigid-Body Vibration In our study of rigid-body vibrations, we observed that the equation of small angular motion has a form identical to that for particle vibrations. Whereas particle vibrations may be described completely by the equations governing translational motion, rigid-body vibrations usually require the equations of rotational dynamics.
Energy Methods In the final article of Chapter 8, we saw how the energy method can facilitate the determination of the natural frequency it),, in free vibration problems where damping may be neglected. Here the total mechanical energy of the system is assumed to be constant. Setting its first time derivative to zero leads directly to the differential equation of motion for the system. The energy approach permits the analysis of a conservative system of interconnected parts without dismembering the system.
Degrees of Freedom Throughout the chapter, we have restricted our attention to systems having one degree of freedom, where the position of the system can be specified by a single variable. If a system possesses n degrees of freedom, it has n natural frequencies. Thus, if a harmonic force is applied to such a system which is lightly damped, there are n driving frequencies which can cause motion of large amplitude. By a process called modal analysis, a complex system with n degrees of freedom can be reduced to n singledegree-of-freedom systems. For this reason, the thorough understanding of the material of this chapter is vital for the further study of vibrations.
A r t i c l e 8/6
REVIEW PROBLEMS
h / W W H
655
8/125 The uniform circular disk is suspended by a socket
8/123 The 0.1-kg projectile is fired into the 10-kg block which is initially at rest with no force in the spring. The spring is attached at both ends. Calculate the maximum horizontal displacement X of the spring and the ensuing period of oscillation of the block and embedded projectile. Ans. X = 0.287 m, t 0.365 s k = 3 kN/m
Review Problems
0.1 kg 10 kg
(not shown) which fits over the small ball attachment at O. Determine the period of small motion if the disk swings freely about (a) axis A-A and (b) axis B-B. Neglect the small offset, mass, and friction of the ball. Ans. (aJ a>A..,
• 500 m/s
r r Problem 8/123
8/124 A 20-m. I-beam is being hoisted by the cable arr angement shown. Determine the period r of small oscillations about the junction O, which is assumed to remain fixed and about which the cables pivot freely. Treat the beam as a slender rod.
Problem 8/125 8/126 The uniform triangular plate pivots freely about a horizontal axis through point O. Determine the natural frequency of small oscillations.
Problem 8/124 Problem 8/126
656
Chapter 8
Vibration and Time Response
8/127 Determine the period r for small oscillations of the assembly composed of two light bars and two particles, each of mass m. Investigate your expression as the angle a approaches values of 0 and 180°. AJIS.
T
=
2-tt
I
8/129 A slender rod is shaped into the semicircle of radius r as shown. Determine the natural frequency /„ for small oscillations of the rod when it is pivoted on the horizontal knife edge at the middle of its length. A,1S f =
' '
———
V g cos (a/2)
Problem 8/127 8/128 Determine the natural frequency /„ of vertical oscillations of the cylinder of mass m. The mass and frict ion of the st epped drum are negligible.
h /|
8/130 Determine the largest amplitude A'0 for which the uniform circular disk will roll without slipping on the horizontal surface. x
Problem 8/130
Problem B/128
8/131 Calculate the damping ratio t of the system shown if the weight and radius of gyration of the stepped cylinder are IV 20 lb and k = 5.5 in., the spring constant is k = 15 lb/in., and the damping coefficient of the hydraulic cylinder is c 2 lb-sec/ft. The cylinder rolls without slipping on the radius r = 6 in. and the spring can support tension as well as compression. Ans. i = 0.0697
Problem 8/131
A r t i c l e 8/6
8/132 A linear oscillator with mass m, spring constant k, and viscous damping coefficient c is set into motion when released from a displaced position. Derive an expression for the energy loss Q during one complete cycle in terms of the amplitude ij at the start of the cycle. (See Fig. 8/7.)
8/133 The cylinder A of radius r, mass m, and radius of gyration k is driven by a cable-spring system, attached to the drive cylinder B, which oscillates as indicated. If the cables do not slip on the cylinders, and if both springs are stretched to the degree that they do not go slack during a motion cycle, determine an expression for the amplitude i9nlaI of the steady-state oscillation of cylinder A.
Review Problems
657
8/135 A 60-g bullet is fired with a velocity of 300 m/s at the 5-kg block mounted on a stiff but light cantilever beam. The bullet is embedded in the block, which is then observed to vibrate with a frequency of 4 Hz. Compute the maximum displacement A in the vibration and find the damping constant c in N • s/m if the ratio of two amplitudes ten full cycles apart is 0.6. Neglect any energy loss during the first quarter cycle. Ans. A = 0.1415 m , c = 2.07 N-s/m.
1.5 m
Ans. 0
60 g
I .
. 5 kg
Problem 8/133
8/134 The seismic instrument shown is secured to a ship's deck near' the stern where propeller-induced vibration is most pronounced. The ship has a single three-bladed propeller which turns at 180 rev/mm and operates partly out of water, thus causing a shock as each blade breaks the surface. The damping ratio of the instrument is 1 - 0.5, and its undamped natural frequency is 3 Hz. If the measured amplitude of A relative to its frame is 0.75 mm, compute the amplitude ¿¡y of the vertical vibration of the deck.
300 m/s
Problem 8/135 8/136 An experimental engine weighing 480 lb is mounted on a test stand with spring mounts at A and B, each with a stiffness of 600 lb/in. The radius of gyration of the engine about its mass center G is 4.00 in. With the motor not running, calculate the natural frequency (/„) y of vertical vibration and (/„)« of rotation about G. If vertical motion is suppressed and a slight rotational imbalance occurs, at what speed N should the engine not be run?
-
- 10'
-
-10'
Problem 8/136 Problem 8/134
658
Chapter 8
Vibration and Time Response
• 8/137 A 200-kg machine rests on four floor mounts, each of which has an effective spring constant k = 250 kN/m and an effective viscous damping coefficient c 1000 N • s/m. The floor is known to vibrate vertically with a frequency of 24 Hz. What would be the effect on the amplitude of the absolute machine oscillation if the mounts were replaced with new ones which have the same effective spring constant but twice the effective damping coefficient? Ans. Amplitude increases by 28.9%!
*Computer-Oriented
*8/140 The 4-kg mass is suspended by the spring of stiffness k 350 N/m and is initially at rest in the equilibrium position. If a downward force F Ct is applied to the body and reaches a value of 40 N when t = 1 s, derive the differential equation of motion, obtain its solution, and plot the displacement y in millimeters as a function of time during the first second. Damping is negligible.
; k = 350 N/m
Problems
*8/l 38 Plot the response x of the 50-lb body over the time interval 0 £ t 1 second. Determine the maximum and minimum values of x and their respective times. The initial conditions are jq, = 0 and i u = 6 ft/sec. I I
F
J
-v = (160 cos 6Of) lb
100 lb/in.
Y F
18 lb-sec/ft 501b
— 3 0 —
m as 4
kg
-ct
Problem 8/140
*B/141 Shown in the figure are the elements of a displacement meter used to study the motion vp — b sin at of the base. The motion of the mass relative to the frame is recorded on the rotating drum. If LL = 1.2 ft, LZ = 1.6 ft, L3 = 2 ft, W 2 lb, c 0.1 lb-sec/ft, and to = 10 rad/sec, determine the range of the spring constant k over which the magnitude of the recorded relative displacement is less than 1,56. It is assumed that the ratio OJ!W„ must remain greater than unity. Am. 0 < k < 1.895 lb/ft
Problem 8/138 *8/139 The mass of a critically damped system having a natural frequency OJ„ 4 rad/s is released from rest at an initial displacement XQ. Determine the time t required for the mass to reach the position x — O.LKQ.
Am; t = 0.972 s
h -h—H
w y B = b sin tat
Í
Problem 8/141
Ne Neutral position
A r t i c l e 8/6 *8/142 The 4-kg cylinder is attached to a viscous damper and to the spring of stiffness k 800 N/m. If the cylinder is released from rest at time t 0 from the position where it is displaced a distance y 100 mm from its equilibrium position, plot the displacement y as a function of time for the first second for the two cases where the viscous damping coefficient is (a ) c = 124 N-s/m and (6) c = 80 N-s/m.
Equilibrium position ^
^
3T $ ^ k - 800 N/m
Review Problems
*B/144 The damped linear' oscillator of mass m 4 kg, spring constant k = 200 N/m, and viscous damping factor £ 0.1 is initially at rest in a neutral position when it is subjected to a sudden impulsive loading F over a very short period of time as shown. If the impulse I = $F dt = 8 N-s, determine the resulting displacement x as a function of time and plot it for the first two seconds following the impulse. x
1
F
4 kg
m /
c

Problem B/144
Problem 8/142 *8/l 43 Determine and plot the response x fit) for the undamped linear oscillator subjected to the force F which varies linearly with time for the first % second as shown. The mass is initially at rest with a: = 0 at time t = 0. Ans. x = 0.0926(i - 0.0913 sin 10.95i) m F, N
Problem 8/143
659
A
A R E A MOMENTS OF INERTIA
See Appendix A of Vol. 1 Statics for a treatment of the theory and calculation of area moments of inertia. Because this quantity plays an important role in the design of structures, especially those dealt with in statics, we present only a brief definition in this Dynamics volume so that the student can appreciate the basic differences between area and mass moments of inertia. The moments of inertia of a plane area A about the x- and y-axes in its plane and about the «-axis normal to its plane, Fig. A/'l, are defined by 4 = J y2 dA
Iy = J
dA
Iz = J r2 dA
where dA is the differential element of area and r2 = x2 + y2. Clearly, the polar moment of inertia Iz equals the sum Ix + Iy of the rectangular moments of inertia. For thin flat plates, the area moment of inertia is useful in the calculation of the mass moment of inertia, as explained in Appendix B. The area moment of inertia is a measure of the distribution of area about the axis in question and, for that axis, is a constant property of the area. The dimensions of area moment of inertia are (distance)4 expressed in m4 or mm4 in SI units and ft 4 or in.4 in U.S. customary units. In contrast, mass moment of inertia is a measure of the distribution of mass about the axis in question, and its dimensions are (mass)) distance) which are expressed in kg • m2 in SI units and in lb-ft-sec2 or lb-in.-sec2 in U.S. customary units.
Figure A/1
661
B
M A S S MOMENTS OF INERTIA
A P P E N D I X OUTLINE B/1
Mass Moments of Inertia about an Axis
B/2
Products of Inertia
B/1
MASS
MOMENTS
OF
INERTIA
ABOUT A N A X I S
The equation of rotational motion about an axis normal to the plane of motion for a rigid body in plane motion contains an integral which depends on the distribution of mass with respect to the moment axis. This integral occurs whenever a rigid body has an angular acceleration about its axis of rotation. Thus, to study the dynamics of rotation, you should be thoroughly familiar with the calculation of mass moments of inertia for rigid bodies. Consider a body of mass m, Fig. B/1, rotating about an axis O-O with an angular acceleration a. All particles of the body move in parallel planes which are normal to the rotation axis O-O. We may choose any one of the planes as the plane of motion, although the one containing the center of mass is usually the one so designated. An element of mass dm has a component of acceleration tangent to its circular path equal to ra, and by Newton's second law of motion the resultant tangential force on this element equals ra dm. The moment of this force about the axis 0-0 is r2a dm, and the sum of the moments of these forces for all elements is j r 2 « dm. For a rigid body, a is the same for all radial lines in the body and we may take it outside the integral sign. The remaining integral is called the mass moment of inertia I of the body about the axis O-O and is '• dm
(B/1)
This integral represents an important property of a body and is involved in the analysis of any body which has rotational acceleration about a
663
664
Appendix B
Mass M o m e n t s of Inertia given axis. Just as the mass m of a body is a measure of the resistance to translational acceleration, the moment of inertia 7 is a measure of resistance to rotational acceleration of the body. The moment-of-inertia integral may be expressed alternatively as I
—
('Lr-rti: B/la)
where r, is the radial distance from the inertia axis to the representative particle of mass m, and where the summation is taken over all particles of the body. If the density p is constant throughout the body, the moment of inertia becomes
where dV is the element of volume. In this case, the integral by itself defines a purely geometrical property of the body. When the density is not constant but is expressed as a function of the coordinates of the body, it must be left within the integral sign and its effect accounted for in the integration process. In general, the coordinates which best fit the boundaries of the body should be used in the integration. It is particularly important that we make a good choice of the element of volume dV. To simplify the integration, an element of lowest possible order should be chosen, and the correct expression for the moment of inertia of the element about the axis involved should be used. For example, in finding the moment of inertia of a solid right-circular cone about its central axis, we may choose an element in the form of a circular slice of infinitesimal thickness, Fig. B/2a. The differential moment of inertia for this element is the expression for the moment of inertia of a circular cylinder of infinitesimal altitude about its central axis. (This expression will be obtained in Sample Problem B/l.) Alternatively, we could choose an element in the form of a cylindrical shell of infinitesimal thickness as shown in Fig. B/26. Because all of the mass of the element is at the same distance r from the inertia axis, the differential moment of inertia for this element is merely r 2 dm where dm is the differential mass of the elemental shell. From the definition of mass moment of inertia, its dimensions are (mass)(distance) and are expressed in the units kg-m~ in SI units and lb-ft-sec a in U.S. customaiy units.
Figure B/2
A r t i c l e B/1
Mass M o m e n t s of Inertia a b o u t an Axis
Radius of Gyration The radius of gyration k of a mass m about an axis for which the moment of inertia is I is defined as
k -
v m
or
(B/2)
I — k2m
>
V
Thus, k is a measure of the distribution of mass of a given body about the axis in question, and its definition is analogous to the definition of the radius of gyration for area moments of inertia. If all the mass m of a body could be concentrated at a distance k from the axis, the moment of inertia would be unchanged. The moment of inertia of a body about a particular axis is frequently indicated by specifying the mass of the body and the radius of gyration of the body about the axis. The moment of inertia is then calculated from Eq. B/2.
Transfer of Axes If the moment of inertia of a body is known about an axis passing tlu'ough the mass center, it may be determined easily about any parallel axis. To prove this statement, consider the two parallel axes in Fig. B/3, one being an axis through the mass center G and the other a parallel axis through some other point C. The radial distances from the two axes to any element of mass dm are ru and r, and the separation of the axes is d. Substituting the law of cosines r2 — r02 + d2 + 2r0d cos 0 into the definition for the moment of inertia about the axis through C gives I = j r'2 dm = J (r 0 2 + d2 + 2r u d cos 0) dm
—j
rt)2 dm + d2
j
dm + 2 d
j
u dm
Figure B/3
The first integral is the moment of inertia / about the mass-center axis, the second term is md2, and the third integral equals zero, since the incoordinate of the mass center with respect to the axis through G is zero. Thus, the parallel-axis theorem is / = / + md2
(B/3)
Remember that the transfer cannot be made unless one axis passes through the center of mass and unless the axes are parallel. When the expressions for the radii of gyration are substituted in Eq. B/3, there results k2 = k2 + d2
(B/3a)
Equation B/3a is the parallel-axis theorem for obtaining the radius of gyration k about an axis which is a distance d from a parallel axis through the mass center, for which the radius of gyration is k.
665
666
Appendix B
Mass M o m e n t s of Inertia
yI
For p l a n e - m o t i o n problems w h e r e rotation occurs about an axis normal to the plane of m o t i o n , a single subscript for I is sufficient to designate the inertia axis. Thus, if the plate of Fig. B/4 has plane motion in the .r-y plane, the m o m e n t of inertia of the plate about the 2-axis through O is designated I 0 . For three-dimensional motion, however, where components of rotation may occur about more than one axis, we use a double subscript to preserve notational s y m m e t i y with product-ofinertia terms, which are described in Art. B/2. Thus, the m o m e n t s of inertia about the x-, y-, and 2-axes are labeled I x x , I yy , and /,,, respectively, and f r o m Fig. B/5 we see that they become
I Figure B/4
These integrals are cited in Eqs. 7/10 of A i t . 7/7 on angular m o m e n t u m in three-dimensional rotation. T h e defining expressions f o r mass moments of inertia and area m o ments of inertia are similar. An exact relationship between the two m o ment-of-inertia expressions exists in the case of flat plates. Consider the flat plate of uniform thickness in Fig. B/4. If the constant thickness is t and the density is p. the mass m o m e n t of inertia /„ of the plate about the z-axis normal to the plate is
Thus, the mass moment of inertia about the 2-axis equals the mass per unit area pt times the polar moment of inertia I z of the plate area about the 2-axis. If t is small compared with the dimensions of the plate in its
y
Figure B/2
A r t i c l e B/1
Mass M o m e n t s of Inertia about an Axis
plane, the mass moments of inertia I xs and I,.y of the plate about the x- and y-axes are closely approximated by 4v = J y z dm — pt J y2dA = ptIx
(B/6) Iyy — J x2 dm — pt J x2dA = ptly
Thus, the mass moments of inertia equal the mass per unit area pt times the corresponding area moments of inertia. The double subscripts for mass moments of inertia distinguish these quantities from area moments of inertia. Inasmuch as l t — I x + I y for area moments of inertia, we have l^X^+lyy
(B/7)
which holds only for a thin flat plate. This restriction is observed from Eqs. B/6, which do not hold true unless the thickness t or the ¿-coordinate of the element is negligible compared with the distance of the element from the corresponding x- or y-axis. Equation B/7 is very useful when dealing with a differential mass element taken as a flat slice of differential thickness, say, dz. In this case, Eq. B/7 holds exactly and becomes dlzz - dlxx + dlyy
(B/7a)
for axes .t and y in the plane of the plate.
Composite Bodies As in the case of area moments of inertia, the mass moment of inertia of a composite body is the sum of the moments of inertia of the individual parts about the same axis. It is often convenient to treat a composite body as defined by positive volumes and negative volumes. The moment of inertia of a negative element, such as the material removed to form a hole, must be considered a negative quantity. A summary of some of the more useful formulas for mass moments of inertia of various masses of common shape is given in Table D/4, Appendix D.
667
668
Appendix 8
Mass M o m e n t s of Inertia
Sample Problem B / l Determine the moment of inertia and radius of gyration of a homogeneous right-circular cylinder of mass m and radius r about its central axis O-O.
U
Solution. An element of mass in cylindrical coordinates is dm p dV ptra dra dIK where p is the density of the cylinder. The moment of inertia about the axis of the cylinder is 1
f ru2 dm
pt |
J
Jü
|
r03dr0d6
Jo
pt
= 'mr2 I
Ans.
i
Tlic radius of gyration is
V
Ajis.
¿2
Helpful Hints (T) If we liad started with a cylindrical shell of radius r0 and axial length t as our mass element dm, then dl ru dm directly. You should evaluate the integral. (2) The result I = ^mr2 appUes ordy to a solid homogeneous circular cylinder and cannot be used for any other wheel of circular periphery.
Sample Problem B/2 -J U— dx
Determine the moment of inertia and radius of gyration of a homogeneous solid sphere of mass m and radius r about a diameter.
Solution. A circular slice of radius y and thickness dx is chosen as the volume element. From the results of Sample Problem B/l, the moment of inertia about the jr-axis of the elemental cylinder is dl„ = {dm )y' = {-npy2 dx)y2
(r2 - x2)2 dx Helpful Hint
where p is the constant density of the sphere. The total moment of inertia about the.r-axis is 4t
y J
2 a (r2 - X2)2 dx = jjr 7rprs = B m
AJIS.
The radius of gyration is
k
m
2r
:
Ans.
(T) Here is an example where we utilize a previous result to express the moment of inertia of the chosen element, which in this case is a rightcircular cylinder of differential axial length dx. It would be foolish to start with a third-order element, such as p dx dy dz, when we can easily solve the problem with a firstorder element.
A r t i c l e B/1
Mass M o m e n t s of Inertia about an Axis
669
Sample Problem B/3 Determine the moments of inertia of the homogeneous rectangular parallelepiped of mass m about the centroidal A'0- and 2-axes and about the JR-axis through one end.
Solution. A transverse slice of thickness dz is selected as the element of volume. The moment of inertia of this slice of infinitesimal thickness equals the moment of inertia of the area of the section times the mass per unit area p dz. Thus, the moment of inertia of the transverse slice about they'-axis is dJyy ~ (fi dzïlhab*) and that about the.*:'-axis is
Helpful Hint
©
dlx.x,
=
(pdz)(hasb)
As long as the element is a plate of differential thickness, the principle given by Eq. B/7 a may be applied to give = dlx.x, + dlyy. = (p dz) || to' + b2) These expressions may now be integrated to obtain the desired results. The moment of inertia about the ¿-axis is %= J
=
Ans.
+ i>2) f dz = l 2 m(a 2 + br
where m is the mass of the block. By interchange of symbols, the moment of inertia about the ,r0-axis is W
I - z ^
Ans.
+ l2)
The moment of inertia about the jc-axis may be found by the parallel-axis theorem, Eq. B/3. Thus,
+
£ (2)
¿9m(a2 + 4l2)
Ans.
i
This last result may be obtained by expressing the moment of inertia of the elemental slice about the A'-axis and integrating the expression over the length of the bar. Again, by the parallel-axis theorem dlxx
dlx x- + z2 dm ~ (p dzil^a-'b) + z2/xib dz
pai)^— +
dz
Integrating gives the result obtained previously;
-^(FH^K)-1-
2 2 12mXa + 41 )
The expression for Ixx may be simplified for a long prismatic bar or slender rod whose transverse dimensions are small compared with the length. In this case, a2 may be neglected compared with 412, and the moment of inertia of such a slender bar about an axis through one end normal to the bar becomes I |m/ 3 . By the same approximation, the moment of inertia about a centroidal axis normal to the bar is I = ^ m ! 2 .
© Refer to Eqs. B/6 and recall the expression for the area moment of inertia of a rectangle about an axis through its center parallel to its base.
670
Appendix
Mass Moments of Inertia
8
PROBLEMS Introductory
Problems
B/L Follow the suggestion of Helpful Hint 1 in Sample Problem B / l and use the differential element shown in the figure to show that the mass moment of inertia of the homogeneous right-circular cylinder about its central axis O-O is I ~ mr2.
Problem B/1 B/2 From the results of Sample Problem B/2, state without computation the moments of inertia of the solid homogeneous hemisphere of mass if! about the x- and if-axes.
B/3 Use the mass element dm p dx, where p is the mass per unit length, and determine the mass moments of inertia I yy and 7 l v of the homogeneous slender rod of mass m and length I. A / I S . Iyy = ly'y' ~
Problem B/3 B/4 State without calculation the moment of inertia about the ¿-axis of the thin conical shell of mass /)! and radius r from the results of Sample Problem B / l applied to a circular disk. Observe the radial distribution of mass by viewing the cone along the z-axis.
Problem B/2 Problem B/4
A r t i c l e B/1
Problems
671
B/5 The moment of inertia of a solid homogeneous cylinder of radius r about an axis parallel to the central axis of the cylinder may be obtained approximately by multiplying the mass of the cylinder by the square of the distance d between the two axes. What percentage error e results if la) d lOr and (6) d = 2rl Ans. (a) e = 0.498% (6) e = 11.11%
Problem B/7 B/8 Every 'slender' rod has a finite radius r. Refer to Table D/4 and derive an expression for the percentage error e which results if one neglects the radius r of a homogeneous solid cylindrical rod of length I when calculating its moment of inertia /.,, Evaluate your expression for the ratios r/l 0.01, 0.1, and 0.5. Problem B/5 B/6 Determine the moment of inertia of the uniform slender rod about the x-axis. Make use of your result to write the moment of inertia about the y-axis by inspection. Use these two results to determine the moment of inertia about the ¿-axis and check your result with that of Prob. B/3 and Sample Problem B/3 for a « l. Problem B/8 B/9 Determine the moment of inertia of the thin equilateral triangular- plate of mass m about the z-z axis normal to the plate through its mass center G. Solve by using the results for the triangular area in Table D/3, the relations developed for thin flat plates, and the transfer-of-axis theorem. A/is. = Problem B/6 B/7 The moment of inertia of a solid homogeneous sphere of radius r about any noncentroidal axis x may be obtained approximately by multiplying the mass of the sphere by the square of the distance d between the jc-axis and the parallel centroidal axis. What percentage error e results if (a) d 2r and (f>) d = lOr? Ans. (a) |e| 9.097*, (6) e = 0.398% Problem B/9
672
Appendix
8
Mass Moments of Inertia
B/L 0 In order to better appreciate the greater ease of integration with ¡ower-order elements, determine the mass moment of inertia IJX. of the homogeneous thin plate by using the square clement (a) and then by using the rectangular element (i>). The mass of the plate is m. Then by inspection state and finally, determine
Problem B/12 B/13 Determine IXI for the cylinder with a centered circular hole. The mass of the body is m. Alls. Ixx = 2>n( + '¡2)
dx Problem B/10 B/L 1 The rectangular metal plate has a mass of 15 kg. Compute its moment of inertia about the y-axis. What is the magnitude of the percentage error e introduced by using the approximate relation J, mP for IXT? Arts. lyy 0.5 kg - m 2 , e = 0,25%
Problem B/13 B/14 Calculate the radius of gyration about axis O-O for the steel disk with the hole.
Problem B / l 1 B/12
Calculate the mass moment of inertia about the axis O-O for the uniform 10-in. block of steel with crosssection dimensions of 6 and 8 in.
Problem B/14 B/L 5 The molded plastic block has a density of 1300 kg/m l Calculate its moment of inertia about the y-y axis. What percentage error e is introduced by using the approximate relation for /,.,? Ans. I = 1.201 kg-m 2 , e = 1.538%
A r t i c l e B/1
Problems
673
B/18 Det ermine the length L of each of the slender rods of mass mi2 which must be centrally attached to the faces of the thin homogeneous disk of mass ii! in order to make the mass moments of inertia of the unit about the x- and ¿-axes equal. z I
Problem B/15
Representative
Problems
B/16 Determine the moment of inertia of the half-ring of mass m about its diametral axis a-a and about axis b-b through the midpoint of the arc normal to the plane of the ring. The radius of the circular cross section is small compared with r. b
Problem B/18 B/19 A badminton racket is constructed of uniform slender rods bent into the shape shown. Neglect the strings and the built-up wooden grip and estimate the mass moment of inertia about the y-axis through O, which is the location of the player's hand. The mass per unit length of the rod material is p. Ans.I}y
b
=
f^
+
y^^pV-
y Problem B/16
B/17 The semicircular disk weighs 5 lb, and its small thickness may be neglected compared with its 10-in. radius. Compute the moment of inertia of the disk about the x-, y-, y'-, and ¿-axes. Ans. = Iyy = 0.0270 lb-ft-sec2 / v > . = 0.0433 lb-ft-sec2, = 0.0539 lb-ft-sec2
B/20 Calculate the moment of inertia of the tapered steel rod of circular cross section about an axis normal to the rod through O. Note that the rod diameter is small compared with its length. 100 mm
Problem B/17 5 mm Problem B/20
674
Appendix B
M a s s M o m e n t s of I n e r t i a
B/21 Calculate the moment of inertia of the steel control wheel, shown in section, about its central axis. There are eight spokes, each of which has a constant cross-sectional area of 200 mm 2 . What percent n of the total moment of inertia is contributed by the outer rim? Ans. I = 1.031 k g - m 2 , n = 97.8%
B/23 The uniform circular cylinder has mass m. radius r, and length I. Derive the expression for its moment of inertia about the end _v-axis. Ir2 l2 Arts. IXI = m|-- + -1
Y 75 --*-
Ü
200 mm 2
120 M
È
M
T 50 _ J L
W_ „
! 100 300 400 I
r-t-J
Problem B/23
m
Dimensions in millimeters Problem B/21 B/22 In the study of high-speed reentry into the earth's atmosphere, small solid cones are fil ed at high velocities into low-density gas. A condition of critical stability occurs when the moment of inertia of the cone about its axis of generation a-a equals that about a transverse axis b-b through the mass center. Determine the critical value of the cone angle a for this condition.
B / 2 4 Use the results cited f o r Prob. B/23 and derive an expression for the percentage error e in calculating the moment of inertia of the cylinder about the end axis by neglecting the term mr 2 /4. Consider the range 0 £ ril £ 1. Plot your results and cite the error for ril - 0.2. B/25 For what length I of the solid homogeneous cylinder are the moments of inertia about the three coordinate axes through the mass center G equal? Refer to the results of Prob. B/23 as necessary. Ans. I
Problem B/25
P r o b l e m B/22
rJ 3
A r t i c l e B/1
Problems
675
B/26 The uniform coiled spring weighs 4 lb. Approximate its moments of inertia about the x-, y-, and z-axes from the analogy to the properties of a cylindrical shell.
Problem B/28
Problem B/26 B/27 The uniform rod of length 4b and mass m is bent into the shape shown. The diameter of the rod is small compared with its length. Determine the moments of inertia of the rod about the three coordinate axes. Ans. IXI = I,, = lmb2,lyy = |mb2 y
B/29 The clock pendulum consists of the slender rod of lengt h / and mass m and the bob of mass 7m. Neglect the effects of the radius of the bob and determine In in terms of the bob position x. Calculate the ratio R of I0 evaluated for x to IQ evaluated for x = I. Ans, lo = m(lxl +
R = 0.582
Problem B/29 B/30 Determine the mass moments of inertia of the thin parabolic plate of mass m about the x-, y-, and z-axes. y
Problem B/27 B/28 Calculate the moment of inertia of the solid steel semicylinder about the x-x axis and about the parallel ro-JCo axis. (See Table D / l for the density of steel.)
Problem B/30
676
Appendix B
Mass M o m e n t s of Inertia
B/31 The varying radius y of the solid of revolution is proportional to the square of its x-coordinate. If the mass of the body is if!, determine I1X. Ans. Ixx = f^mr2
•
—y
Problem B/33 B/34 Determine the moment of inertia about the y-axis for the paraboloid of revolution of Prob. B/33. Problem B/31 B/32 A square plate with a quarter-circular sector removed has a net mass m. Determine its moment of inertia about axis A-A normal to the plane of the plate.
B/35 Determine the moment of inertia about the tangent jc-.r axis for the full ring of mass mj and the half-ring of mass mi. A/is. Full ring; Ixx = | m ^ Half-ring: I xx = m 2 r 2 j| - ^-j
Problem B/35 Problem B/32 B/33 Determine the radius of gyration about the ¿-axis of the paraboloid of revolution shown. The mass of the homogeneous body is m.
B/36 Calculate the moment of inertia of the homogeneous right-circular cone of mass m, base radius r, and altitude h about the cone axis x and about the y-axis through its vertex.
Ann. k. = -— v'3
Problem B/36
Article 7/10 P r o b l e m s
677
B/J7 Determine the moment of inertia about the jc-axis of the homogeneous solid semiellipsoid of revolution having mass m. Ans. IXI = lm(a2 + b2)
Problem B/39 B/40 The slender metal rods are welded together in the configuration shown. Each 6-in. segment weighs 0.30 lb. Compute the moment of inertia of the assembly about the y-axis. Problem B/37 B/38 A preliminary model for a spacecraft consists of a cylindrical shell and two flat panels as shown. The shell and the panels have the same thickness and density. It can be shown that, in order for the spacecraft to have a stable spin about axis 1-1, the moment of inertia about axis 1-1 must be less than the moment of inertia about axis 2-2. Determine the critical value of I which must be exceeded to ensure stable spiir about axis 1-1. Problem B/40 B/41 The welded assembly shown is made from a steel rod which weighs 0.667 lb per foot of length. Calculate the moment of inertia of the assembly about the x-x axis. AJ;S. = 0.410 lb-in.-sec2
Problem B/38 B/39 Determine by integration the moment of inertia of the half-cylindrical shell of mass m about the axis a-a. The thickness of the shell is small compared with r.
- Î'^[ M+ )
Ans.Iaa =
Problem B/41
678
Appendix
8
Mass Moments of Inertia
B/42 The thickness of the homogeneous triangular plate of mass m varies linearly with the distance from the vertex toward the base. The thickness a at the base is small compared with the other dimensions. Determine the moment of inertia of the plate about the y-axis along the centerline of the base.
B/45 Determine the moments of inertia of the half-spherical shell with respect to the x- and ¿-axes. The mass of the shell is m, and its thickness is negligible compared with the radius r. Aiis. Ixx - /„ = ^ mrl x
Problem B/42 B/43 Determine the moment of inertia of the triangular plate described in Prob. B/42 about the ¿-axis. Arcs.Ij,
-~m[-- + hA 10 2 /
B/44 Determine the moment of inertia, about the generating axis, of the hollow circular tube of mass m obtained by revolving the thin ring shown in the sectional view completely around the generating axis.
R
Problem B/44
Problem B/45 B/46 Determine I r t for the cone frustum, which has base radii r, and r., and mass m.
I
*
Problem B/46
Article 7/10 P r o b l e m s *B/47 A preliminary design model to ensure rotational stability for a spacecraft consists of the cylindrical shell and the two square panels as shown. The shell and panels have the same thickness and density. It can be shown that rotational stability about the 2-axis can be maintained i f i s less than and 7vl. For a given value of r, determine the limitation for L. Ans. L > 4.54r
679
• B/49 Compute the moment of inertia of the mallet about the O-O axis. The mass of the head is 0.8 kg, and the mass of the handle is 0.5 kg. Ans. I00 = 0.0671 kg-m 2
Problem B/49
Problem B/47 • B/48 The cube with semicircular grooves in two opposite faces is cast of lead. Calculate the moment of inertia of the solid about the axis a-a. Aits. Iaa = 0.367 kg'in 2
• B/50 By direct integration, determine the moment of inertia about the Z-axis of the thin semicircular disk of mass m and radius R inclined at an angle 6 from, the X-y plane. Ans. Izz = mR2l% + cos 2 6)
Problem B/50
Dimensions in millimeters Problem B/48
680
Appendix
8
Mass M o m e n t s of Inertia
B / 2
PRODUCTS
OF
INERTIA
For problems in the rotation of three-dimensional rigid bodies, the expression for angular momentum contains, in addition to the momentof-inertia terms, product of inertia terms defined as /
^xy
f xy dm
^yx
xz dm
(B/8)
yz dm
These expressions were cited in Eqs. 7/10 in the expansion of the expression for angular momentum, Eq. 7/9. The calculation of products of inertia involves the same basic procedure which we have followed in calculating moments of inertia and in evaluating other volume integrals as far as the choice of element and the limits of integration are concerned. The only special precaution we need to observe is to be doubly watchful of the algebraic signs in the expressions. Whereas moments of inertia are always positive, products of inertia may be either positive or negative. The units of products of inertia are the same as those of moments of inertia. We have seen that the calculation of moments of inertia is often simplified by using the parallel-axis theorem. A similar theorem exists for transferring products of inertia, and we prove it easily as follows. In Fig. B/6 is shown the ,r-y view of a rigid body with parallel axes .ru-y0 passing through the mass center G and located from the .r-y axes by the distances dx and dv. The product of inertia about the .r-y axes by definition is
= J ay dm = J U'0 + dx)(ya + dy)I dm
— j x u y 0 dm + dxdy J dm + dx J' y0 dm + dy J x0 dm
Figure B/6
— Ir .. + mdxx.dv y
The last two integrals vanish since the first moments of mass about the mass center are necessarily zero. Similar relations exist for the remaining two product-of-inertia terms. Dropping the zero subscripts and using the bar to designate the mass-center quantity, we obtain
••
s
+ ht =
ty* V
mdxdy
4 + mdxdz +
mdyd.
(B/9)
A r t i c l e B/2
Products of Inertia
These transfer-of-axis relations are valid only for transfer to or from parallel axes through the mass center. With the aid of the product-of-inertia terms, we can calculate the moment of inertia of a rigid body about any prescribed axis through the coordinate origin. For the rigid body of Fig. B/7, suppose we must determine the moment of inertia about axis OM. The direction cosines of OM are I, m, n, and a unit vector A along OM may be written A = /i + mj + jik. The moment of inertia about OM is hi
= J'
h2 dm -
J
/
(r x A)-(r x A) dm
where |r x A| = r sin 0 — h. The cross product is (r x A) = (yn — zm)i + (zl - xn)j + (xin — y/)k

and, after we collect terms, the dot-product expansion gives Figure B/7 (r x A M r x A) = h2 =
Thus, with the substitution of the expressions of Eqs. B/4 and B/8, we have
hi = hJ2 + ly/n2 + h/'2 ~ 2IJm ~ 21Jn - 21
inn
(B/10)
This expression gives the moment of inertia about any axis OM in terms of the direction cosines of the axis and the moments and products of inertia about the coordinate axes.
Principal Axes of Inertia As noted in Art. 7/7, the array ^XX -hy
-I - I
y* V zx
I
yy
-1
-y
-I yz I zz
whose elements appeal' in the expansion of the angular-momentum expression, Eq. 7/11, for a rigid body writh attached axes, is called the inertia matrix or inertia tensor. If we examine the moment- and product-of-inertia terms for all possible orientations of the axes with respect to the body for a given origin, we will find in the general case an orientation of the x-y-z axes for which the product-of-inertia terms vanish and the array takes the diagonalized form ^xx
0 0
0
0
1 0 yy 0 u
M
681
682
Appendix
8
Mass M o m e n t s of Inertia Such axes x-y-z are called the principal axes of inertia, and IXI, /,,,., and 1„ are called the principal moments of inertia and represent the maximum, minimum, and intermediate values of the moments of inertia for the particular origin chosen. It may be shown* that for any given orientation of axes x-y-z the solution of the determinant equation I XX -I
-I
-1
yx
Iyy
-I zx
-I
xy
-I xz
-I
-I yz
= 0
(B/ll)
Izz - I
for I yields three roots 11,I 2 , and I :i of the resulting cubic equation which are the three principal moments of inertia. Also, the direction cosines I, m, and n of a principal inertia axis are given by % - ni - V '
-
-lyj + (Jyy ~
=
0
~ Iy Jl = 0
(B./12)
- I J - I z y m + (I zz - l ) n = 0
These equations along with I2 + m2 + n2 — 1 will enable a solution for the direction cosines to be made for each of the three Fs. To assist with the visualization of these conclusions, consider the rectangular block. Fig. B/8, having an arbitrary orientation with respect to the x-y-z axes. For simplicity, the mass center G is located at the origin of the coordinates. If the moments and products of inertia for the block about the x-y-z axes are known, then solution of Eq. B / l l would give the three roots, T, I2, and wrhich are the principal moments of inertia. Solution of Eq. B/12 using each of the three Fs, in turn, along with I + m2 + n2 — 1, would give the direction cosines I, m, and n for each of the respective principal axes, which are always mutually perpendicular. From the proportions of the block as drawn, we see that Ij is the maximum moment of inertia, I 2 is the intermediate value, and /;J is the minimum value.
2
Figure B/8 KS- ;
for example, the first a u t h o r D y n a m i c s . SI Version, 1975, John Wiley & Sons, Art. 41.
Article
B/2
P r o d u c t s of Inertia
683
Sample Problem B/4 The bent plate has a uniform thickness t which is negligible compared with its other dimensions. The density of the plate material is p. Determine the products of inertia of the plate with respect to the axes as chosen.
Each of the two parts is analyzed separat ely.
Solution.
Rectangular part. In the separate view of this part, we introduce parallel axes .vu-y„ through the mass center G and use the transfer-of-axis theorem. By symmetry, we see that I' = i = 0 so that fry =
Uxy = Ity + '»W
0
+
I - I = - - ipta2b2 ( -- !)(!)
Because the ¿-coordinate of all elements of the plate is zero, it follows that
Helpful Hints (T) We must be careful to preserve the same sense of the coordinat es. Thus, plus and yo must agree with plus x andy.
= h, = 0.
Triangular part In the separate view of this part, we locate the mass center G and construct XQ-, yo-, and ¿0-axes through G. Since the -coordinate of all elements is zero, it follows that I xy = 7I;V| = 0 and I r , = = 0. The transfer-ofaxis theorems then give us [i™ = Ixy + mdxdy
I« =
I
= 0 + fit -ci —a)I —
= - - ptab2c
to-i
+
I** = 0 + pi l Ci -ai
— - ptabc
We obtain I y l by direct integration, noting that the distance a of the plane of the triangle from the y-z plane in no way affects the y- and ¿-coordinates. With the mass element dm fit dy dz, we have
[>=/ DM] **=PTIJ0 YZ DZ DY=PT I pÎ cylb
dy
Adding the expressions for the two parts gives Ixy = -|ptazb2-ptab% = - l2ptatiA{3a + 4c)
Ans.
Ix, =
-^ptabc2 = — g ptabc2
Ans.
+lptb2c2 = +18ptb2c2
Ans.
& =
0 0
© We choose to integrate with respect to z first, where the upper limit is the variable height ¿ = cylb. If we were to integrate first with respect to y, the limits of the first integral would be from the variable y bz/c to b.
684
Appendix 8
Mass M o m e n t s of Inertia
Sample Problem B/5 The angle bracket is made from aluminum plate with a mass of 13.45 kg per square mete®; Calculate the principal moments of inertia about the origin O and the direction cosines of the principal axes of inertia. The thickness of the plate is small compared with the other dimensions.
The masses of the three parts are
Solution.
110
m1 = 13.45(0.21X0.1) = 0.282 kg m 2 = -13.45tt(0.
035)2
Dimensions in millimeters
= -0.0518 kg
= 13.45(0.121(0.11) = 0.1775 kg
Helpful Hints © Note that the mass of the hole is treated as a negative number.
Part 1 Ixx
mt}2
^(0.2821(0.1)2 = 9.42(10
4)
kg-m 2
ira fa2 + b2) = *(0.282)1(0.21)2 + (0.1) 2 ] J„ Jma2 % = 0
*(0.282)(0.21)2 = 41.5(10 ^ = 0
% = jjI +
mdxdt
50.9(10 4) kg-m 2
4)kg-m2
= 0 + m || = 0.282(0.105)10.05) = 14.83(10' 4 ) kg-m 2
Part 2 Ixx
jrar 5 + mds2 = -0.0518 -1.453(10
4)
(0.035)2
3
+ (0.050):
kg-m 2
= jfflr + mid/ + d'l -0.0518
(0.035)2
-14.86(10
4)
+ (0.16)2 + (0.05)*
kg-m 2
-mr 2 + md2 = -0.0518
(0.035)2
+ (0.16)2
- 1 3 . 4 K 1 0 ' 4 ) kg-m 2 Isy = 0 = 7r.+
Iv, = 0 0 -0.0518(0.16)(0.05) = -4.14(10 4 ) kg-m 2
© You can easily derive this formula. Also check Table D/4.
A r t i c l e B/2
Products of Inertia
685
Sample Problem B/5 (Continued) 6 = 100
Part 3 -„rnd2 = k0,1775){0.12l 2 ¡j(0.1775)(0.11)2
Iyy=i3mc2
8.52(10 7.16(10
4) 4)
kg-m 2 kg-m 2
L = jBile2 + d2) = J ( 0 . 1 7 7 5 ) 1 X 0 . 1 + (0.12)2] = 15.68(10 ^ = fry = 0 7^ = 0
+
4)
kg-m 2
mdxdy
M-
CI — m
2 7
X 1775(0.055)(-0.06)
:
-5.86(10
4)
kg-m 2
= 0
Totals Z„ = 16.48(10
4)
kg-m 2
Jr,, = -5.86(10' 4 ) kg-m 2
43.2(10' 4 ) kg-m 2
J„*0
43.8(10' 4 )
7 „ = 10.69(10
kg-m 2
4)
kg-m 2
Substitution into Eq. B / l l , expansion of the determinant, and simplification yield 73 - 103.5(10
4 )7 2
+ 3180(10 'B)7 - 24 800(10 ' 12 ) = 0
(3) Solution of this cubic equation yields the following roots, which are the principal moments of inertia 7i = 48.3(10' 4) kg'm 2 I 2 = 11.82(10 7a = 43.4(10'
4) 4)
kg-m 2
Ans.
kg'm 2
The direction cosines of eacli principal axis are obtained by substituting each root, in turn, into Eq. B/12 and using I2 + in2 + n2 1. The results are Î! = 0.357 % = 0.410
l2 = 0.934 m 2 = -0.1742
m 3 = 0.895
Ans.
n1 = - 0 . 8 3 9 The bottom figure shows a pictorial view of the bracket and the orientation of its principal axes of inertia.
(3) A computer program for the solution of a cubic equation may be used, or an algebraic solution using the formula cited in item 4 of Ait . C/4, Appendix C, may be employed.
686
Appendix B
Mass M o m e n t s of Inertia
PROBLEMS Introductory
Problems
B/53 Determine the products of inertia of the uniform slender rod of mass m about the coordinate axes shown. Ans. I
B/51 Determine the products of inertia about the coordinate axes for the unit which consists of three smali spheres, each of mass m, connected by the light but rigid slender rods. Ans. Ixy = 0, ij, = Iy, = -2ml2
= -mab
Iy, = - ^mbh = mah
Problem B/53 B/54 Determine the products of inertia about the coordinate axes for the thin plate of mass m which has the shape of a circular' sector of radius a and angle fi as shown.
Problem B/51 B/52 Determine the products of inertia about the coordinate axes for the unit which consists of four small particles, each of mass m, connected by the light but rigid slender rods. Problem B/54 B/55 Determine the products of inertia about the coordinate axes for the thin square plate with two circular holes. The mass of the plate material per unit area is p. pnb4 Ans. Ixy = - Ixz = Iy. = 0
Problem B/52
Article
1 Ö 1 4 '
8/2
Problems
687
1 6 1 4
iEK cS : > 4 8
Wi 1 ' 5 1
b 4 b 4 b 4 b 4
Problem B/5S
Problem B/57 B/58 Determine the product of inertia I xy for the slender rod of mass m.
B/56 The slender rod of mass m is formed into a quartercircular arc of radius r. Determine the products of inert ia of the rod with respect to the given axes.
Problem B/58 B/59 The semicircular disk of mass m and radius if, inclined at an angle 0 from the X-y plane, of Prob. B/50 is repeated here. By the methods of this article, determine the moment of inertia about the Z-axis. Aits. Izz mR2[ 1 + cos 2 «) Problem B/56 B/57 The uniform rectangular block weighs 50 lb. Calculate its products of inertia about the coordinate axes shown. Aits. Iyy - —1.553 lb-m.-sec' l y . = 0.776 lb-in.-sec2 I„ = -1.035 lb-in.-sec2
Problem 8/45
688
Appendix B
M a s s M o m e n t s of I n e r t i a
B/60 Determine the products of inertia for the rod of Prob. B/27, repeated here.
Problem B/62 B/63 For the slender rod of mass m bent into the configuration shown, determine its products of inertia IXI, and I,,,. , j nib2 Ans. Ixv = —— 4 J2 Ixz = - ^ mb*
Problem B/60
mb~ 4/2
Representative
Problems
B/61 The S-shaped piece is formed from a rod of diameter d and bent into the two Semicircular shapes. Determine the products of inertia for the rod, for which d is small compared with r. Ans. Ixy 2m?J/Tt, = Iy. = 0
Problem B/63 B/64 Determine the moment of inertia of the solid cube of mass m about the diagonal axis A-A through opposite corners.
Problem B/61 B / 6 1 Determine the three products of inertia writh respect to the given axes for the uniform rectangular plate of mass in.
Problem B/64
Article B/65 The steel plate with two right-angle bends and a central hole has a thickness of 15 mm. Calculate its moment of inertia about the diagonal axis through the corners A and B. Ans. Iw = 2.58 kg-nr-
500
300
*Computer-Oriented
8/4
Problems
689
Problems
*B/67 Each sphere of mass m has a diameter which is small compared with the dimension b. Neglect the mass of the connecting struts and determine the principal moments of inertia of the assembly with respect to the coordinates shown. Determine also the direction cosines of the axis of maximum moment of inertia. Ans. % = 7 . 5 3 m I , = 6Mmb4 = 1.844mb2 h = 0.521, ml = -0.756, n.j = 0.397
500
Dimensions in millimeters Problem B/65 B/66 Prove that the moment of inertia of the rigid assembly of three identical balls, each of mass m and radius r, has the same value for all axes through O. Neglect the mass of the connecting rods.
4,
Problem B/67
m
y.
Problem B/66
*B/68 Determine the moment of inertia I about axis OM for the uniform slender rod bent into the shape shown. Plot I versus 6 from 9 0 to 0 = 90° and determine the minimum value of I and the angle a which its axis makes with the JI-direction. [Note: Because the analysis does not involve the ¿-coordinate, the expressions developed for area moments of inertia, Eqs. A'9, AJ10, and A/11 in Appendix A of Vol. 1 Statics, may be utilized for this problem in place of the three-dimensional relations of Appendix B. l The rod has a mass p per unit length.
Problem 8/79
690
Appendix B
M a s s M o m e n t s of I n e r t i a
* B / 6 9 The assembly of three small spheres connected by light rigid bars of Prob. B/51 is repeated here. Determine the principal moments of inertia and the direction cosines associated with the axis of maxim u m moment of inertia. Ans. Il = 9ml 2 , I.s = 7.37m? 2 , l:i = 1.628m^ I. = 0.816, m.< = 0.408, = 0.408
*B/71 T h e thin plate has a mass p per unit area and is formed into the shape shown. Determine the principal moments of inertia of the plate about axes through O. Ans. = 3.78pf>4,12 = 0.612pf>4, JK = 3 . 6 1 ^ 4
Problem B/71 * B / 7 2 The slender rod has a mass p per unit length and is formed into the shape shown. Determine the principal moments of inertia about axes through O and calculate the direction cosines of the axis of minimum moment of inertia. Problem B/69 * B / 7 0 The bent rod of Probs. B/27 and B/60 is repeated here. Its mass is m, and its diameter is small compared with its length. Determine the principal moments of inertia of the rod about the origin O. Also find the direction cosines for the axis of minimum moment of inertia. y I I
b
Problem B/72
Problem B/70
c C / I
SELECTED TOPICS OF MATHEMATICS
INTRODUCTION
Appendix C contains an abbreviated summary and reminder of selected topics in basic mathematics which find frequent use in mechanics. The relationships are cited without proof. The student of mechanics will have frequent occasion to use many of these relations, and he or she wdl be handicapped if they are not well in hand. Other topics not listed will also be needed from time to time. As the reader reviews and applies mathematics, he or she should bear in mind that mechanics is an applied science descriptive of real bodies and actual motions. Therefore, the geometric and physical interpretation of the applicable mathematics should be kept clearly in mind during the development of theory and the formulation and solution of problems. C / 2
PLANE
GEOMETRY
1. When two intersecting lines are, respectively, perpendicular to two other lines, the angles formed by the two pairs are equal.
4. Circle
2. Similar triangles
5. Every triangle inscribed within a semicircle is a right triangle.
x _ h - y b h
3. Any triangle Area = }.bh
Circumference = 2nr Area — rrr2 Arc length s = rO Sector area
ÖJ + Ü2' k!2
6. Angles of a triangle 0 ! - 0 2 - 0 3 ~ 180° 0-4 — ! f i + 0 2 691
692
Appendix C
C/3
SOLID
Selected Topics of Mathematics
GEOMETRY
1. Sphere
3. Right-circular cone
Volume =
^irr3
Volume = ^nr2h
Surface area = 4rrr2
L
Lateral area = nrL
h
L -• yjr2 + h2
2. Spherical wedge
4. Any pyramid or cone Volume = ~Bh
Volume =
where B = area of base
. J- J — I
ALGEBRA
C/4
1. Quadratic equation ax2
4. Cubic equation
+ bx + c — 0
= Ax + B
-b ± yjb2 - 4ac 2a
, b2 s 4ac for real roots
Let p = A/3, q = B/2. Case I:
2. Logarithms bx — y,x = l o g 6 y
cos u = qKp , p), 0 < u < 180°
Natural logarithms
.r-L = 2 v p cos(u/3)
b = e = 2.718 282 e* = y, x = l o g c y = In y
x2 - 2 Jp cos (zi/3 + 120°)
ic3 = 2 Jp cos (w/3 + 240°)
log (afi) = log a + logf) log (a/b) — log a — logf> log(l//i) = - l o g n log a' = n l o g o log 1 = 0 log1(pc = 0.4343 In .r
Case II:
bi b
2
=
aih
~
Case III:
— p:i positive (one root real, two roots imaginary) vV
- p 3 ) , / S + ( g - Jq 2 ~ j ? } v s
q
-p
= 0 (three roots real, two roots equal)
H
=
2qlri, x2
= x3
-
-qm
For general cubic equation
a2bi
x3 + ax2 + bx + c = 0
3rd order bi =
6a
q2
jct - (
3. Determinants 2nd order
2
q2 — p'J negative (three roots real and distinct)
+a1b.2c3 + a2b3c1 + a.,b1cz -aJb2c1
-
a,2blc3
-
a1b3c2
Substitute x — '$Q — a/3 and get x03 = Ar a + B. Then proceed as above to find values of XQ from which x = 3t0 — a/3.
A r t i c l e C/6
Trigonometry
ANALYTIC GEOMETRY
C/5
3. Parabola
1. Straight line y
Ib I
_ ± _
y - a + mx
y = o—
2. Circle
4. Ellipse
X2 +yz = r2 (x-cir +
„2 xy _= a
TRIGONOMETRY
C/6
1. Definitions sin 0 — ale csc 9 — c/a cos 6 — b/c sec 0 — c/b tan 0 — alb cot 0 — b/a 2. Signs in the four quadrants (+)
X .
X
II
(+)
— t — i • (+>< - ) — •
(-)III
0/' ' IV
I
II
Ill
IV
sin &
+
+
-
-
cos 0
+
-
-
+
tan é
+
-
+
csc 0
+
+
-
-
sec 0
+
-
-
+
cot 0
+
-
+
-
-
693
694
Appendix C
Selected Topics of Mathematics
3. Miscellaneous relations
4. Law of sines
sin 2 0 + cos 2 9 = 1 1 + tan 2 0 = sec 2 B 1 + cot 2 0 = csc 2 0 a
a _ sin A b sin B
j,
sin — — yf 2(1 — cos 9) cos - = v ' ^ i l + cos 0) sin 2fi — 2 sin 6 cos 0 cos 2H — cos 2 0 — sin' ß sin (a ± b) — sin a cos b ± cos a sin b cos (a ± b) — cos a cos b + sin a sin b
VECTOR
C / 7
5. Law of cosines c a = a2 + b2
OPERATIONS
1. Notation. Vector quantities are printed in boldface type, and scalar quantities appear in lightface italic type. Thus, the vector quantity V has a scalar magnitude V. In longhand work vector quantities should always he consistently indicated by a symbol such as V or V to distinguish them from scalar quantities.
2.
Addition
Triangle addition
P + Q = it
Parallelogram addition Commutative law Associative law
3.
P + Q = R
P 1- Q = Q + P P + (Q + i i ) = (P + Q) + R
Subtraction P - Q = P + (— Q )
4, Unit vectors
i, j, k V = V r i + Vyj + V J t
where
I v| = V— J v ? T v f + v?
5. Direction cosines l, m, n are the cosines of the angles between V and the J:-, y-, «-axes. Thus, I = VJV so that and
m = VyiV V = V(/i + mj + nk) I2 + m2 + n2 — 1
„ 2ab cos C
c2 = a2 + b2 + 2ab cos D
n = Vz/V
A r t i c l e C/7 H.
Vector Operations
695
Dot or scalar product P Q = PQ cos 0 This product may be viewed as the magnitude of P multiplied by the component Q cos 0 of Q in the direction of P, or as the magnitude of Q multiplied by the component P cos 0 of P in the direction of Q. Commutative law
cos 6
P*Q = Q•P
From the definition of the dot product M = j-j
= k k -
1
i j = j i = i k = k i = j k = k j = 0
P Q = iPxi + PJ + P, k! • (Q r i + Qyj + Q z k) = PSQX + PyQy + PQ p-p = p 2 x
+
p 2 y
+
p2 z
It follows from the definition of the dot product that two vectors P and Q are perpendicular when their dot product vanishes, P Q = 0. The angle 8 between two vectors P[ and P 2 may be found from their dot product expression P, • P 2 = PlP2 cos 0, which gives COS
0 —
P 1 ; P 3 j + P1 P2 + P, P%
p ,p r.
=
— ...
P
PiP2
=
&
1^2
+
mm2
+ n1l2
where l, m, n stand for the respective direction cosines of the vectors. It is also observed that twro vectors are perpendicular to each other when their direction cosines obey the relation IJ2 + m iin 2 + n ^ i 2 = 0.
Distributive law
P ( Q + R) = P Q + P R
7, Cross or vector product. The cross product P x Q of the two vectors P and Q is defined as a vector with a magnitude
PxQ
|P x Q| = PQ sin 0 and a direction specified by the right-hand rule as showrn. Reversing the vector order and using the right-hand rule give Q x P = P x Q. Distributive law
P x ( Q + R) = P x Q + P x R
From the definition of the cross product, using a right-handed Q xP = - P xQ
coordinate system, we get i x j = k j x i = - k
j x k = i k x j = -i
i x i = j x j
k x i = j i x k = -j
= k x k = 0
696
Appendix C
Selected Topics of Mathematics With the aid of these identities and the distributive law, the vector product may be written P x Q = (PJ + P J + Pzk) x (Qxi + QJ + Qzk) - (PyQ,. - r*Qy)i + && ~ PxQJi + (P&y - PyQx)k The cross product may also be expressed by the determinant i
p 8.
X
k
j
Q = P* Py Qx
%
P. Qz
Additional relations Triple scalar product (P X Q R R - i P x Q) The dot and cross may be interchanged as long as the order of the vectors is maintained. Parentheses are unnecessary since P x (Q-R,l is meaningless because a vector P cannot be crossed into a scalar Q * R . Thus, the expression may be written P x Q R - P Q x R The triple scalar product has the determinant expansion P* P x Q R - Q, Rx
Py Qy Ry
Pz Qz Rz
Triple vector product. (P x Q) x tt = - R x (P x Q) = R x (Q x P). Here we note that the parentheses must be used since an expression P x Q x R would be ambiguous because it would not identify the vector to be crossed. It may be shown that the triple vector product is equivalent to (P x Q) x R = R P Q - R - Q P or
P x (Q x R) = P - R Q - P Q R
The first term in the first expression, for example, is the dot product R • P, a scalar, multiplied by the vector Q. 9. Derivatives of vectors obey the same rales as they do for scalars. ^ = P - PA + PA + Pz dt rti x yJ z. d'Vu) = Pit + Pu dt d{ P Q) dt d i p x Q) dt
P Q + P Q
P x Q + P x Q
k
A r t i c l e C/9 10. Integration of vectors. If V is a function of it, y, and z and an element of volume is clr = dx dy dz, the integral of V over the volume may be written as the vector sum of the three integrals of its components. Thus, | V dr = i J %dr + j J Vydr + k J Vz dr
SERIES
C / 8
(Expression convergence.) (1 ¿xV =
in brackets following series
nx +
±
X5 5!
x7 7!
r 2 , x4 2! 4!
6!
xa 3!
nf*
,
2)
range
of
#+• •• •• 1x2 <1] [x2 < [x 2 < be]
e1 — e~x sinh* = — 2 e = *X++ ' + |e +
cosh
~ ^ '
indicates
• •
[x 2 < »]
e* + e~x . ac2 ^ ac4 . x s . * - _ _ = 1 + _ T+ _ + - - -
r 2 ^ „i < œ]
tf
J-Î a0 V riTTX , , • TITTX fix) = -++ ¿ » » m —+ 2 b„sin~r £ ji=1 ( n= 1 ' where
= y I 1 J
/(_r) cos - i
I
dx,
brl = y I / ( x ) sin nlfx dx I J-i t [Fourier expansion for — I < x < I
C / 9
dx' _ dx
DERIVATIVES
_ j d(uv) _ udii + dx dx
d(%) _u v V / __ dx dx dx dx u2
lim sin i r = sin dx = tan dx = dx it '0 lim cos àx = cos dx — 1 it '0 d sinx — COS X, dx d sinh x = cosh x, dx
d COS X — —sin x, dx d cosh X = sinh x, dx
d tan x _ dx
se
d tanh .r = sech 2 x dx
Derivatives
697
698
Appendix C
Selected Topics of Mathematics
INTEGRALS
C/TO
~n + 1
A
dx , — = In x
y
a + bx dx = -xrf(a + bx)3 3b
xja + bx dx =
2
x2Ja + bx dx =
dx
f ~ (36.r - 2a) J (a + feje)3
15b1
105 i>3
(8a2 - 12abx + 15¿A2) J (a + bx)3
_ 2 J a + bx
Ja + bx Ja + x :b -x
x
a + bx
dx — — J a + x Jb — x + (a + b) sin
1
/— ~ fa — b
= ^r [a + bx — a In (a + b
xdx (a + bx)n
(a + bx) ^ ' ' / n +bx _ a 2 V 2 - 71 X — 7i J b,2
dx
1 * = —= tan a + bx2 Jab x d x
a + bx2
1
,xjab —i
i , — : tanh J-ab
or
a
1
.xj-ab —-—
= ~ In (a + bx2) 2b
Jx2 ± a2 dx .=• _xjx2 ± a2 ± a2 In (x + Jx2 ± a 2 ) ]
v'a 2 —
d.r —
v;a2
— .r2 + a 2 s i n - 1 ^ j
x^a2 - x2 dx — — oi/C- .r2)3
x 2 Jet2 - x 2 dx -
x3Ja2 - x2 dx =
,/(a 2 - a:2)3 + ^ ^ a
+
|aV(a2
-
x2)3
2
- x 2 + a 2 s i n ' 1 |I
A r t i c l e C/10
N
dx = = — lu f Ja + bx + ex2 + x Je + ^ ] Je 2 Je) a + bx +cx22
dx / v2 -tdx Ja*
-
or
—— sin J-c
= In (.r + Jx2 ± a2)
• -ix = = sm 1 a x .2 -
x dx r2 a =s — v'.r - a£ r~ô — a2
v'jr
/ 9 r 5 v a ± .r
- ± Jà2 ± x2
xjx2 i a~ dx = il Jix2 T. a2)3
x2Jx* ± a2 dx — ^(x2 ± a2)3 + 4 sin x dx = - c o s x
cos Jt
4
2 j x , sin 2x cos x dx = - - — 2 4 j sin a x sin x cos x dx = —-—
sinh x dx — cosh x
cosh x dx = sinh x
tanh x dx = ln cosh x
ln x dx — x In x — x
o
xjô? ± a2 -
o
In (x + Jx? ± a 2 )
1
( ^ 'LX ) Jb2 - 4acf
Intégrais
699
700
Appendix C
e^dx
=
Selected Topics of Mathematics
—
a
xe'-x dx — — (ax — 1) a2 eax sin px dx =
eax cos px dx =
in 2 x dx =
eax (a sin px — p cos px) a2 + p2 ef* (a cos px + p sin px) a2 + p2 (
<> 2I a sin 2 x — sin 2x + a l 4 + a2
•4»cos
eax cos 2 x dx —
Ï)
Ï + sin 2x + —
a*
% (a . , - sin 2x — cos 2x a2 12 S1
e111 sin i' cos x dx =
'
3
J
sin x dx =
C O S
o
X
,
(2
•
1

+ sin x)
, = sinj: n 1 cos a x dx —— ,(2 + cos2
cos° x dx — sin « — i sin'! x + - sin' x
x sin x dx = sin x —ix cos JC
x cos x dx = cos x + x sin x
x 2 sinx dx = 2x s i n « — U 2 — 2) c o s x
x2 cos x dx — 2x cos x + (x2 — 2) sin x (dyf Pxy ~
3/2
Xdx) dx2
Radius of curvature
(IR
Pro = r2 + 2
(dr d0
3/2
rd2r
'do2
)
A r t i c l e C/11
C / I 1
NEWTON'S
INTRACTABLE
METHOD
FOR
Newton's Method for 5olving intractable Equations
SOLVING
EQUATIONS
Frequently, the application of the fundamental principles of mechanics leads to an algebraic or transcendental equation which is not solvable (or easily solvable) in closed form. In such cases, an iterative technique, such as Newton's method, can be a powerful tool for obtaining a good estimate to the root or roots of the equation. Let us place the equation to be solved in the form /(x) = 0. Part a of the accompanying figure depicts an arbitrary function fix) for values of x in the vicinity of the desired root xr. Note that xT is merely the value
of x at which the function crosses the .t-axis. Suppose that we have available (perhaps via a hand-drawn plot) a rough estimate X of this root. Provided that Xi does not closely correspond to a maximum or minimum value of the function f(x), we may obtain a better estimate of the root xr by extending the tangent to f(0 at Xj so that it intersects the x-axis at %2- From the geometiy of the figure, we may write
where / ' ( x i ) denotes the derivative of /(.r) with respect to .r evaluated at x = xi. Solving the above equation for xi results in
A')
X 1i
_
s m .r'i / fer)
The term —f&i)//'£%) is the correction to the initial root estimate Once x2 is calculated, we may repeat the process to obtain x a , and so forth. Thus, we generalize the above equation to
xk + 1 — Xj,
f m f(Xk)
701
702
Appendix C
Selected Topics of Mathematics
where xl!+1 — the (k + l)th estimate of the desired root xr xk — the ¿th estimate of the desired root xr f(xk) — the function /U') evaluated at x = x;, f'(xk) = the function derivative evaluated at x — xk
This equation is repeatedly applied until f(Xk+0 is sufficiently close to zero and = xk. The student should verify that the equation is valid for all possible sign combinations of xk, f(xk), and f'(xk). Several cautionary notes are in order:
1. Clearly, /'(x&) must not be zero or close to zero. This would mean, as restricted above, that xh exactly or approximately corresponds to a minimum or maximum o f / ( x ) . If the slope /'(.%) is zero, then the tangent to the curve never intersects the x-axis. If the slope /' (.rt ) is small, then the correction to xk may be so large that xh + L is a worse root estimate than xk. For this reason, experienced engineers usually limit the size of the correction term; that is, if the absolute value of f(xk)/f'{xk) is larger than a preselected maximum value, that maximum value is used. 2. If there are several roots of the equation /(x) = 0, we must be in the vicinity of the desired root xr in order that the algorithm actually converges to that root. Part b of the figure depicts the condition when the initial estimate Xj will result in convergence to :i'F i rather than i . 1
1
3. Oscillation from one side of the root to the other can occur if, for example, the function is antisymmetric about a root which is an inflection point. The use of one-half of the correction will usually prevent this behavior, which is depicted in part c of the accompanying figure. Example: Beginning with an initial estimate of Xj = 5, estimate the single root of the equation e1 — 10 cos x — 100 — 0. The table below summarizes the application of Newton's method to the given equation. The iterative process was terminated when the absolute value of the correction —f{x k )/f'(x k ) became less than 10~B.
k
xh
1 2 3 4 5
5.000 4.071 4.596 4.593 4.593
fW>
/(**) 000 695 498 215 209
45.576 537 7.285 610 0.292 886 0.000 527 -2(10'b)
Xk+i
138.823 96.887 89.203 88.882 88.881
916 065 650 536 956
x
'
-0.328 -0.075 -0.003 -0.000 2.25(10
305 197 283 006 IU)
A r t i c l e C/12
C / I 2
SELECTED
NUMERICAL_
TECHNIQUES
Selected Techniques for Numerical Integration
FOR
INTEGRATION
1. Area determination. Consider the problem of determining the shaded area under the curve y — f(x) from .r = a to .r = b, as depicted in part a of the figure, and suppose that analytical integration is not feasible. The function may be known in tabular form from experimental measurements or it may he known in analytical form. The function is taken to be continuous within the interval a < x < b. We may divide the area into n vertical strips, each of width Ax — (b — a)/n, and then add the areas of all strips to obtain A — Jy dx. A representative strip of area Aj is shown with darker shading in the figure. Three useful numerical approximations are cited. In each case the greater the number of strips, the more accurate becomes the approximation geometrically. As a general rule, one can begin with a relatively small number of strips and increase the number until the resulting changes in the area approximation no longer improve the accuracy obtained.
Rectangular
/ I m
(b)
A;
A1=ymAx yi*l
ym
A-jy dx^Zym Ax
Ax
I. Rectangular [Figure (6)1 The areas of the strips are taken to be rectangles, as shown by the representative strip whose height y m is chosen visually so that the small cross-hatched areas are as nearly equal as possible. Thus, we form the sum Zy,„ of the effective heights and multiply by A«. For a function known in analytical form, a value f o r y m equal to that of the function at the midpoint xt + Ax/2 may be calculated and used in the summation. II. Trapezoidal [Figure (c)] The areas of the strips are taken to be trapezoids, as shown by the representative strip. The area A, is the average
703
704
Appendix C
Selected Topics of M a t h e m a t i c s
height (y, + v, * i)/2 times i t . Adding the areas gives the area approximation as tabulated. For the example with the curvature shown, clearly the approximation will be on the low side. For the reverse curvature, the approximation will be on the high side.
Trapezoidal
A A =fydx={^+y = 1+yi y-i
(c)
+
.-
+y|
Í 1 + j ) Ax
Xi+1
Ax
Parabolic AA = — ( y ¡ + 4y¡ + 1 + y ! +
AA
y i+-¿
A =, jy dx = y ( y u + 4 y : + 2y,2 + 4y:i + 2y4 +
id)
Ax
- + 2yn,2 + 4yrl.1+yn)Ax
Ax
III. Parabolic [Figure (d)l The area between the chord and the curve (neglected in the trapezoidal solution) may be accounted for by approximating the function by a parabola passing through the points defined by three successive values o f y . This area may be calculated from the geometry of the parabola and added to the trapezoidal area of the pair of strips to give the area AA of the pair as cited. Adding all of the AA's produces the tabulation shown, which is known as Simpson's rule. To use Simpson's rule, the number n of strips must be even. Example: Determine the area under the curve y = -t' -Jl + x2 from x = 0 to x = : 2. (An integrable function is chosen here so that the three approximations can be compared with the exact value, which is A =
fix
+x?dx = ¿ ( 1 + x2)ml =
NUMBER OF SUBINTERVALS 4 10 50 too 1000 2500
- 1) = 3 . 3 9 3 4 4 7 ) .
AREA APPROXIMATIONS RECTANGULAR 3.3B1 3.388 3.393 3.393 3.393 3.393
704 399 245 396 446 447
TRAPEZOIDAL 3.456 3.403 3.393 3.393 3.393 3.393
731 536 850 547 448 447
PARABOLIC 3.392 3.393 3.393 3.393 3.393 3.393
214 420 447 447 447 447
A r t i c l e C/12
Selected Techniques for Numerical Integration
Note that the worst approximation error is less tiran 2 percent, even with only four strips.
2. Integration of first-order ordinary differential equations. The application of the fundamental principles of mechanics frequently results in differential relationships. Let us consider the first-order form dyidt — fit), where the function f(t) may not be readily integrable or may be known only in tabular form. We may numerically integrate by means of a simple slope-projection technique, knowrn as Euler integration, which is illustrated in the figure.
/
Beginning at tlt at which the value yj is known, we project the slope over a horizontal subinterval or step it2 — f j ) and see that y2 = yi + f itft2 — ty. At t2, the process may be repeated beginning aty 3 , and so forth until the desired value of t is reached. Hence, the general expression is +
f(h)(tk+i-tk)
I f y versus t were linear, i.e., if / ( t ) were constant, the method would be exact, and there would be no need for a numerical approach in that case. Changes in the slope over the subinterval introduce error. For the case shown in the figure, the estimate y2 is clearly less than the true value of the function y(i) at t2. More accurate integration techniques (such as Runge-Kutta methods) take into account changes in the slope over the subinterval and thus provide better results. As with the area-determ¡nation techniques, experience is helpful in the selection of a subinterval or step size when dealing with analytical functions. As a rough rule, one begins with a relatively large step size and then steadily decreases the step size until the corresponding changes in the integrated result are much smaller than the desired accuracy. A step size which is too small, however, can result in increased error due to a very large number of computer operations. This type of error is generally known as 'round-off error,' while the error which results from a large step size is known as algorithm error.
705
706
Appendix C
Selected Topics of Mathematics Example: For the differential equation clyjdt = 51 with the initial condition y = 2 when t — 0, determine the value of y for t = 4. Application of the Euler integration technique yields the following results:
NUMBER OF SUBINTERVALS 10 100 500 1000
STEP SIZE
y at f = 4
PERCENT ERROR
0.4 0.04 0.008 0.004
38 41.6 41.92 41.96
9.5 0.95 0.19 0.10
This simple example may be integrated analytically. The result is y = 42 (exactly).
USEFUL TABLES
TABLE D/I
PHYSICAL PROPERTIES
Density (kg/m3) and specific weight (lb/ft3)
Air* Aluminum Concrete (av.) Copper Earth (wet, av.) (dry, av.) Glass Gold Ice Iron (cast)
kg/m 3
lb/ft 3
1.2062 2 690 2 400 8 910 1 760 1280 2 590 19 300 900 7 210
0.07530 168 150 556 110 80 162 1205 56 450
Lead Mercury Oil (av.) Steel Titanium Water (fresh) (salt) Wood (soft pine) (hard oak)
kg/m 3
lb/ft 3
11370 13 570 900 7 830 3 080 1000 1030 480 800
710 847 56 489 192 62.4 64 30 50
* At 20°C (68°F) and atmospheric pressure
Coefficients of friction (The coefficients in the following table represent typical values under normal working conditions. Actual coefficients for a given situation will depend on the exact nature of the contacting surfaces. A variation of 25 to 100 percent or more from these values could be expected in an actual application, depending 011 prevailing conditions of cleanliness, surface finish, pressure, lubricat ion, and velocity. ) TYPICAL VALUES OF COEFFICIENT OF FRICTION CONTACTING SURFACE Steel on steel(dry) Steel on steel(greasy) Teflon on steel Steel on babbitt (dry) Steel on babbitt (greasy) Brass on steel (dry) Brake lining on cast iron Rubber tires on smooth pavement (dry) Wire rope on iron pulley (dry) Hemp rope on metal Metal on ice
STATIC, 0.6 0.1 0.04 0.4 0.1 0.5 0.4 0.9 0.2 0.3
KINETIC, p.* 0.4 0.05 0.04 0.3 0.07 0.4 0.3 0.8 0.15 0.2 0.02 707
708
Appendix 0
TABLE D/2
Useful Tables
SOLAK SYSTEM CONSTANTS Universal gravitational constant Mass of Earth Period of Earth's rotation (1 sidereal day) Angular velocity of Earth Mean angular velocity of Earth-Sun line Mean velocity of Earth's center about Sun
BODY
MEAN DISTANCE TO SUN km (mi)
ECCENTRICITY OF ORBIT e
PERIOD OF ORBIT solar days
Sun Moon
384 398* (238 854)* Mercury 57.3 x 10' (35.6 X 10e) Venus 108 X 10e (67.2 x 10e) Earth 149.6 x 10e (92.96 X 10e) Mars 227.9 X 10e (141.6 x 10e)
0.055
27.32
0.206
87.97
0.0068
224.70
0.0167
365.26
0.093
686.98
G = B, 673(11}- n ) rrrVikg-s2) = 3.439(10' s ) ft4/(lbf-s4) m, = 5,976! 1 0 % kg = 4.095(10a:i) Ibf-s2/ft = 23 h 56 min 4 s =•-•23.9344 h Ù> --= 0.7292(10 4) rad/s m' = 0.1991(10 e) rad/s = 107 200 km/h 66,610 mi/h
MEAN DIAMETER km (mi) 1 392 000 (865 000) 3 476 (2 160) 5 000 [3 100) 12 400 (7 700) 12 7421 (7 918)f 6 788 (4 218)
MASS RELATIVE TO EARTH 333 000 0.0123 0.054 0.815 1.000 0.107
SURFACE GRAVITATIONAL ACCELERATION m/s2 (ft/s2)
ESCAPE VELOCITY km/s (mi's)
274 (898) 1.62 (5.32) 3.47 (11.4) 8.44 (27.7) 9.821* (32.22J* 3.73 (12.3)
616 (383) 2.37 (1.47) 4.17 (2.59) 10.24 (6.36) 11.18 (6.95) 5.03 (3.13)
* Mean distance to Earth (center-to-center) t Diameter of sphere of equal volume, based on a spheroidal Earth with a polar diameter of 12 714 km (7900 mi) and an equatorial diameter of 12 756 km (7926 mi) For nonrotating spherical Earth, equivalent to absolute value at sea level and latitude 37.5°
U s e f u l Tables
T A B L E D/3
PROPERTIES OF PLANE FIGURES FIGURE
AREA MOMENTS OF INERTIA
CENTROID
— _ T sinß a —
<
—
3
II
Quarter and Semicircular Arcs
—
y
f Circular Area 1
—
M)¡
—
2
i=r=Z!i
y1
Semicircular / Area / 1
x
. ^ • t P X +—r r | y l Í
18 X
'

Area of Circular Sector
/
z
=
lie
9m
-n-P-4 — 8
r4 1 = — ta - •• sin 2 a) 4 2
M
7 _ - c
9J
L = — * 16
?
1
x
y 1 1
8
t - i n t z 4
/ =
Quarter- Circular Area
v
—
—X
— _ 2 r sin a 3 a
r4 1 /,, = — ta + ¿sin 2a) * 4 2 I, = i r 4 « 2
709
710
Appendix 0
T A B L E D/3
Useful Tables
PROPERTIES OF PLANE FIGURES
FIGURE
Continued
AREA MOMENTS OF INERTIA
CENTROID
Rectangular Area
í
_
1
bh3
t =
M c T
J
•x0
*
12
—b—
12
Ir
- _ a+b ~ 3
=
12
X
Í
Triangular Area
Èftf 36
=
1
7
- M!
* r.
Area of Elliptical Quadrant
ï - i s 3?r
1
16
16 I - L
t=
L = 4
Area A- — x
/ C
y
1Ü
h =
Ï - I
9.W
= U-±-)a3b U8 9tt!
I
^
+
a¿ 3 21
5
4 =
2abs 7
ív-
2aaí> 15
8 b n
US
j _ a*b ¥' 5
Parabolic Area
Area A = ~~ 3
r = /-E-í-U»
ZS&'foî 16
Subparabolie Area y = kx'2 =
—
J, = lab
l5
21 I
7 I
Useful Tables
T A B L E D/4 (m
PROPERTIES OF HOMOGENEOUS SOLIDS
mass of body shown)
711
712
Appendix 0
T A B L E D/4
(m
Useful Tables
P R O P E R T I E S OF HOMOGENEOUS SOLIDS
mass of body shown)
Continued
Useful Tables
T A B L E D/4
PROPERTIES OF HOMOGENEOUS SOLIDS
Cm = m a s s of b o d y s h o w n )
Continued
713
714
Appendix 0
T A B L E D/4
t.m
Useful Tables
P R O P E R T I E S OF HOMOGENEOUS SOLIDS
Continued
mass of body shown) MASS CENTER
BODY
MASS MOMENTS OF INERTIA I xx ~ fyy
„ 1
İL
Vl
Half Cone
y
h
-
r x = ~ IT 3h
j
9 3 9 x y z - 'T + + —2 = 1 , ' /!' C
3
=
230
mr2 + | mh2
=
20™'*
=
iomr2
+
iOmh'2
JÜgt = l ^ i ö 2 + C 2 ) = imto 2 + c'2) 3c
Ia =
+ b2)
Ixx = ~m(b2 + g®c2) y
b2
*

y
g •
z
= im (a2 + He2) 2 + JIM1 I ** , = 5mA b i
z c
a
a2
Semiellipsoid
/ v v = g/ÜO2 + -
Elliptic Paraboloid
2c T
Ja = gm (a2 + b'2) Ixx = sm(b2 + |c2)
^
—
^
c -—
-
lyy =
1?
z
c
1 a
/ X
Kec t angular Tetrahedron
r b
/ t , = j^mlö 2 + e 2 )
a
= ^mla 2 + c 2 )
b
= JH m(a 2 + b2)
4 c z=— 4 y =
4r = 80' i ( & 2 + In =
g3Qm(a2
7,, =
+ c2) + f>2)
z
j Half Torus
y
, =2
+ |c2)
, ' -if - -
Ä
a 2 + 4Ü2 * ' 2-rrR
% = 'rv
=
2' ! j R 2 + 8'°'
/„ = mi? 2 -r |ma2
INDEX
Absolute measurements, 5, 121 Absolute motion, 22,91, 344, 357, 381 Absolute system of units, 7, 122 Acceleration: absolute, 92, 121, 381, 398 angular, 334, 336, 531 average, 23, 41 constant, 25, 334 Coriolis, 121, 398 cylindrical components of, 81 function of displacement, 26 function of time, 25 function of velocity, 26 graphical determination of, 24 due to gravity, 9, 10, 122 instantaneous, 23, 42 normal components of, 56 polar components of, 69 rectangular components of, 43, 81 relative to rotating axes, 397, 543 relative to translating axes, 92, 248, 381, 542 spherical components of, 82 tangential component of, 50 vector representation of, 42 from work-energy principle, 489 Acceleration-displacement diagr am, 24 Acceleration-time diagram, 24 Accelerometer, 626 Action and reaction, principle of, 6 Active-force diagram, 103, 420, 473 Addition of vectors, 5, 694 Amplitude ratio, 622, 623 Amplitude of vibration, 604 Angular acceleration, 334, 336, 531 Angular displacement, 333
Angular impulse, 210, 499 Angular momentum: applied to fluid streams, 295 conservation of, 212, 234, 282, 501 of a particle, 209, 250 relative, 250, 2S0 of a rigid body, 421, 499, 554 of a system, 278 units of, 209 vector representation of, 209 Angular motion: of a line, 333 vector representation of, 335, 528 Angular velocity, 333, 334, 528, 531 absolute, 357, 381 of the eart h, 121, 708 vector representation of, 335, 530, 531 Apogee velocity, 238 Ar ea moments of inertia, 661 Associative law, 694 Astronomical frame of reference, 4, 120, 248 Axes: rotating, 395, 543 translating, 91, 248, 356, 542 Balancing in rotation, 568 Base units, 7, 122 Bodies, interconnected, 276, 426, 489, 500 Body, rigid, 5, 273, 332, 419, 528 Body centrode, 372 Body cone. 531, 582 Cajori, F., 4 Center: of curvature, 56 715
716
Index
Center: (Continued) of mass, motion of, 275 of percussion, 441 Central-force motion, 234 Centrifugal force, 249 Centrode, body and space, 372 Centroids, table of, 709 Circular frequency, natural, 604 Circular motion, 57, 70 Coefficient: of friction, 707 of restitution, 222, 502 of viscous damping, 605 Commutative law, 694, 695 Complementary solution, 621 Computer-oriented problems, 14, 115, 268, 416, 521, 658, 689 Cone, body and space, 531, 582 Conservation: of energy, 180, 281,645 of momentum, 195, 212, 222, 234, 282, 501, 581 Conservative force, 180 Conservative system, 281, 474, 645 Constant of gravitation, 8, 708 Constrained motion, 22, 101, 125, 425, 455 Constraint, equations of, 101, 490 Coordinates: cartesian, 22 choice of, 22, 43, 91, 109, 124, 125, 664 cylindrical, 81 normal and tangential, 55 polar, 68 rectangular, 43, 81 rotating, 395, 543 spherical, 82 transformation of, 83, 110 translating, 91, 248, 356, 542 Coriolis, G„ 4, 398 Coriolis acceleration, 121, 398 Couple: gyroscopic, 577 resultant, 489, 567 work of, 471 Critical frequency, 622 Cross or vector product, 209, 335, 395, 695 Curvature: center of, 56 radius of, 55, 700 Curvilinear motion: in cylindrical coordinates, 81 in normal and tangential coordinates, 55, 140 of a particle, 40, 81, 140 in polar coordinates, 68, 140 in rectangular coordinates, 43, 81, 140 in spherical coordinates, 82 Curvilinear translation, 332, 429, 528 D'Alembert, J„ 4, 249 D'Alembert's principle, 248 Damped forced vibration, 622
Damped free vibration, 605 Damping: coefficient, 605 critical, 606 ratio, 606 viscous or fluid, 605 Dashpot, 605 Degrees of freedom, 101, 125, 602 Densities, table of, 707 Derivative: table of, 697 transformation of, 397, 544 of a vector, 41, 096 Descartes, R., 22 Diagram: acceleration-displacement, 24 acceleration-time, 24 active-force, 163, 420, 473 displacement-time, 24 force-displacement, 158, 160 force-time, 195 free-body, 14, 125, 421, 473, 602 impulse-momentum, 194, 499 kinetic, 421, 422, 423, 424, 429, 441, 454 velocity-displacement, 24 velocity-time, 24 Dimensions, homogeneity of, 11 Direction cosines, 694 Discrete or lumped-parameter model, 601 Displacement: angular, 333 in curvilinear motion, 40 graphical determination of, 24 linear, 22 virtual, 490 Displacement meter, 620 Displacement-time diagram, 24 Distance, 40 Distributed-parameter system, 601 Distributive law, 695 Dot or scalar product, 157, 158, 162, 695 Dynamical energy, conservation of, 180, 281 Dynamic balance in rotation, 568 Dynamic equilibrium, 249 Dynamics, 3 Earth, angular velocity of, 121, 708 Earth satellites, equations of motion for, 234 Efficiency, 164 Einstein, A„ 4, 122 Elastic impact, 223 Elastic potential energy, 178, 473 Electric circuit analogy, 626 Energy: conservation of, 180, 281, 645 kinetic, 162, 275, 472, 557 potential, 177, 473, 644 in satellite motion, 237 in vibration, 644 Equations of constraint, 101, 490
Index Equations of motion: for fixed-axis rotation, 441 for particles, 127, 140 for plane motion, 422, 454 in polar' coordinates, 140 for rectilinear and curvilinear translation, 429 for a rigid body, 421, 566 for rotation about a point, 529 for a system of particles, 275 Equilibrium, dynamic, 249 Euler, L„ 4, 567 Euier s equations, 567 Fluid damping, 605 Fluid streams, momentum equations for, 294, 295 Foot, 7 Force: centrifugal, 249 concept of, 5 conservative, ISO external, 275 inertia, 249 internal, 274 gravitational, 8,11, 123 resultant, 6, 126, 162, 193, 275, 421, 489, 567 units of, 7 work of, 157, 471 Force-displacement diagram, 158, 160 Forced vibration, 620, 634 damped. 622 equation for, 621 frequency ratio for, 622 magnification factor for, 622, 623 resonant frequency of, 622 steady-state, 622, 623 undamped, 621 Force field, conservative, 180 Force-time diagram, 195 Forcing functions, 620 Formulation of problems, 12 Frame of reference, 6, 91, 248, 250, 395, 542, 543 Free-body diagram, 14, 125, 421, 473, 602 Freedom, degrees of, 101, 125, 602 Free vibration: damped. 605 energy solution for, 644 equations for. 602, 605 undamped, 602 vector representation of, 604 Frequency: critical, 622 damped. 608 natural and circular, 604 Frequency ratio, 622 Friction: coefficients of, 707 work of, 474 Galileo, 3 Gradient, 181
Graphical representation, 14, 24, 158, 334, 358, 382, 604 Gravitation: constant of, 8, 708 law of, 8 Gravitational force, 8, 11, 123 Gravitational potential energy, 177, 473 Gravitational system of units, 7, 122 Gravity: acceleration due to, 9, 10, 122 International Formula for, 10 Gyration, radius of, 665 Gyroscope, 575 Gyroscopic couple, 577 Gyroscopic motion, equation of, 576 Harmonic motion, simple, 29, 603 Hertz (unit), 604 Hodograph. 42 Horsepower, 164 Huygens, C., 3 Imbalance, rotational, 568 Impact, 221, 502 classical theory of, 223 direct central, 221 elastic, 223 energy loss in, 223 inelastic or plastic, 223 oblique, 223 Impulse: angular, 210, 499 linear, 194, 498 Impulse-momentum diagram, 194, 499 Impulse-momentum equation. 194, 210, 250 Inertia, 5, 120 area moments of, see Moments of inertia of area mass moments of, see Moments of inertia of mass principal axes of, 556, 681 products of, 555, 680 Inertia force, 249 Inertial sytem, 4, 91, 92, 120, 121 Inertia tensor or matrix, 556. 681 Instantaneous axis of rotation, 371. 530 Instantaneous center of zero velocity, 371 Integrals, table of selected, 698 Integration, numerical techniques for, 703, 705 of vectors, 697 Interconnected bodies, 276, 426, 489, 500 International Gravity Formula, 10 International System of units, 6 Joule (unit), 158 Kepler, J., 234 Kepler's laws of motion, 234, 236 Kilogram, 7, 122 Kinematics, 3, 21, 331, 528 of angular' motion, 334, 528 of curvilinear motion, 40 of rectilinear motion, 22
717
718
Index
Kinematics: {Continued) of relative motion, 22, 91, 248, 356, 381, 395, 542 of rigid bodies, 331, 528 Kinetic diagram, 421, 422, 423, 424. 429, 441. 454 Kinetic energy: of a particle, 162 of plane motion, 472 of rotation. 472, 557 of space motion, 558 of a system of particles, 276, 557 of translation, 472 units of, 162 Kinetic friction, coefficient of, 707 Kinetics, 3, 21, 119, 273, 419, 554 of particles, 119 of rigid bodies, in plane motion, 419, 568 in rotation, 441, 554 in space motion, 554 Kinetic system of units, 122 Lagrange, J. L., 4 Lagrange's equations, 427 Laplace, P., 4 Law: associative, 694 commutative. 694, 695 of conservation of dynamical energy, 180, 281, 645 distributive, 695 of gr avitation, 8 Laws of motion: Kepler's, 234, 236 Newton's, 6, 119, 248, 274, 498 Light, speed of, 4, 122 Line, angular motion of, 333 Linear displacement, 22 Linear' impulse, 194, 498 Linear' momentum: applied to fluid streams, 294 conservation of, 195, 222, 282, 501 moment of, 209 of a particle, 193 relative, 250 of a rigid body, 498 of a system, 277 Logar ithmic decrement, 608 Lumped-parameter or discrete model, 601 Magnification factor, 622, 623 Mass, 5, 119 steady flow of. 293 unit of, 7, 122 variable, 308 Mass center, motion of, 275 Mass flow, equations of motion for, 294, 295 Mass moments of inertia, see Moments of inertia of mass Mathematical model. 12 Mathematics, selected topics in, 691 Matrix, inertia, 556, 681 Measurements: absolute, 5, 121 relative, 91, 248, 250, 356, 381, 395, 542
Mechanics, 3 Meter, 7 Metric units, 6, 122 Moment center, choice of, 424, 429, 454 Moment equation of motion, 210, 278, 279, 280, 281, 422, 424, 425, 566 Moment of linear momentum, 209 Moments of inertia of area, 661, 666, 709 Moment s of inertia of mass, 422, 554, 663 choice of element of integration for, 664 for composite bodies, 667 about any prescribed axis, 681 principal axes for, 556, 681 radius of gyration for, 665 table of, 711 transfer of axes for, 665 Momentum: angular, 209, 250, 278, 295. 499, 554 conservation of, 195, 212, 222, 234, 282, 501, 581 equations for mass How, 294, 295 linear, 193, 277, 498 moment of, 209 rate of change of, 6, 193, 210, 277, 278, 279, 280, 281, 498, 499, 500, 566 vector representation of, 193, 209, 498, 499, 557 Motion: absolute, 22, 91, 344, 357, 381 angular, 333, 335, 528, 529, 531. 532 central-force, 234 circular, 57, 70 constrained, 22, 101, 125, 425, 455 curvilinear, 40, 81, 140 in cylindrical coordinates, 81 general space, 542 graphical representation of, 14, 24, 158, 334, 358, 382, 604 gyroscopic, 575 of mass center, 275 Newton's laws of, 6, 119, 248, 274, 498 in normal and tangential coordinates, 55 parallel-plane, 529, 568 plane, 22, 40, 332, 419, 454 planetary and satellite, 234 in polar' coordinates, 68 in rectangular coordinates, 43, 81 rectilinear, 22, 126 relative, 22, 91, 248, 356, 381, 395, 542 rotational, 333,441, 528, 529 simple harmonic, 29, 603 in spherical coordinates, 82 of a system of particles, 273 unconstrained, 22, 125, 455 Natural frequency, 604 Newton, Isaac, 3, 4 Newton (unit), 7 Newtonian frame of reference, 250, 274, 278 Newtonian mechanics, 122, 274 Newton's laws, 6, 119, 248, 274, 498 Newton's method, 701 Notation for vectors, 5, 41. 694
Index
Numerical integration, 703, 705 Nutation, 578 Oblique central impact, 223 Orbit, elliptical, 236 Osculating plane, 22 Parallel-axis theorems, for mass moments of inertia, 665 Parallel-plane motion, 529, 568 Particles, 5, 21 curvilinear motion of, 40, 81, 140 equations of motion of, 127, 140 kinematics of, 21 kinetics of, 119 motion of system of, 273 Particle vibration, 602 Particular solution, 621, 623 Path variables, 22, 56 Percussion, center of, 441 Perigee velocity, 238 Period: of orbital motion, 236, 239 of vibration, 604, 608 Phase angle, 623 Plane motion, 22, 40, 332, 419, 454 curvilinear, 40 equations of motion for, 422, 454 general, 332, 454 kinematics of, 40, 91, 332 kinetic energy of, 472 kinetics of, 419, 568 Planetary motion: Kepler's laws of, 234, 236 period of, 236 Poinsot, L., 4 Polar moment of inertia, 666 Position vector, 41 Potential energy, 177, 473, 644 Potential function, 180 Pound force, 7. 122. 124 Pound mass, 8, 123, 124 Power, 163, 474 Precession; defined, 531, 575 direct and retrograde, 582 steady, 575, 577, 580 velocity of, 575 with zero moment, 581 Primary inertial system, 4, 91, 120 Principal axes of inertia, 556, 681 Principia, 4 Principle: of action and reaction 6 of conservation of momentum, 195, 212, 222, 234, 282, 501, 581 D'Alembert's, 249 of motion of mass center, 275 Products of inertia, 555, 680 Products of vectors, 157, 209, 695 Projectile motion, 44 Propulsion, rocket, 310
719
Radius: of curvature, 55, 700 of gyration, 665 Rectilinear motion of a particle, 22, 126 Rectilinear translation, 332, 428, 528 Reference frame, 6, 91, 248, 250, 395, 542, 543 Relative acceleration, rotating axes, 397, 543 translating axes, 92, 248, 381, 542 Relative angular momentum, 250, 280 Relative linear momentum, 250 Relative motion, 22, 91, 248, 356, 381, 395, 542 Relative velocity: rotating axes, 396, 543 translating axes, 92, 356, 542 Relativity, theory of, 122 Resonance, 622 Restitution, coefficient of, 222, 502 Resultant: couple, 489, 567 force, 6, 126, 162, 193, 275,421,489, 567 Right-hand rule, 209, 695 Rigid bodies: kinematics of, 331, 528 kinetics of, 419, 568 Rigid body, 5, 273, 332, 419, 528 Rigid-body motion, general moment equations for, 421, 424, 425 Rigid-body vibration, 634 Rocket propulsion, 310 Rotating axes, 395, 543 Rotation: equations of motion for, 425, 441 finite, 529 fixed-axis, 332, 335, 441, 528 fixed-point, 529 infinitesimal, 530 instantaneous axis of, 371, 530 kinematics of, 333, 335, 531 kinetic energy of, 472, 557 of a line, 333 of a rigid body, 332, 441, 528, 529 Rotational imbalance, 568 Satellite, motion of, 234 Scalar', 5 Scalar- or dot product, 157, 158, 162, 695 Second, 7 Series, selected expansions, 697 Simple har monic motion, 29, 603 SI units, 6, 7 Slug, 7, 122, 123 Solar system constants, 708 Solution, method of, 12 Space, 4 Space centrode, 372 Space cone, 531, 582 Space motion, general, 542 Speed, 41 Spin axis. 575 Spin velocity, 575 Spring: constant or stiffness of, 159, 602, 645
720
Index
Spring: {Continued) potential energy of, 178 work done by, 159 Standard conditions, 10, 122 Static friction, coefficient of, 707 Steady mass flow, force and moment equations for, 294, 295 Steady-state vibration, 622, 623 Subtraction of vectors, 694 System: conservative, 281, 474. 645 of interconnected bodies, 276, 426, 489, 500 of particles: angular momentum of, 278 equation of motion for, 275 kinetic energy of, 276, 557 linear' momentum of, 277 of units, 6, 8, 122 Table: of ar ea moments of inert ia, 709 ofcentroids. 709 of coefficients of friction, 707 of densities, 707 of derivatives, 697 of integrals, 698 of mass centers, 711 of mass moments of inertia, 711 of mathematical relations, 691 of solar-system constants, 708 of units, 7 Tensor, inertia, 556, 681 Theory of relativity, 122 Thrust, rocket, 310 Time, 5, 7, 122 Time derivative, transformation of, 397, 544 Transfer of axes: for moments of inertia, 665 for products of inertia, 680 Transformation of derivative, 397, 544 Transient solution, 622, 623 Translating axes, 91, 248, 356, 542 Translation, rectilinear and curvilinear, 332, 428, 528 Triple scalar product, 558, 696 Triple vector product, 690 Two-body problem: perturbed, 239 restricted, 239 Unconstrained motion, 22, 125, 455 Units, 6, 8, 122 kinetic system of, 122 Unit vectors, 43, 55, 68, 81, 694 derivative of, 55, 68, 395, 543 U.S. customary units, 6, 8, 122 Variable mass, force equation of, 309 Vectors, 5, 694 addition of, 5, 694 cross or vector product of, 209, 335, 395, 695
derivative of, 41, 696 dot or scalar product of, 157, 158, 162, 695 integration of, 697 notation for, 5, 41, 694 subtraction of, 694 triple scalar product of, 696 triple vector product of, 696 unit, 43, 55, 68, 81,694 Velocity: absolute, 91 angular-, 333, 334, 528, 531 average, 23, 40 cylindrical components of, 82 defined, 23,41 graphical determination of, 24 instantaneous. 23, 41 instantaneous axis or center of, 371 in planetary motion, 238 polar- components of, 68 rectangular components of, 43, 81 relative to rotating axes, 396, 542 relative to translating axes, 92, 356 spherical components of, 82 tangential component of, 55 vector representation of, 41 Velocity-displacement diagram. 24 Velocity-time diagram, 24 Vibration: amplitude of, 604 damped. 605 energy in. 644 forced, 620, 634 free, 602 frequency of, 604, 607 over- and underdamped, 606. 607 period of. 604, 608 reduction of, 623 simple harmonic. 603 steady-state, 622, 623 transient, 622, 023 work-energy solution for, 644 Virtual displacement, 490 Virtual work, 157, 490 Viscous damping coefficient, 605 Watt, 164 Weight, 7, 8, 11, 123 Work, 157, 471 of a constant force, 159 of acouple, 471 an exact differential, 180 examples of, 158 of a force, 157, 471 of friction, 474 graphical representation of, 158 of a spring force, 159 units of, 158 virtual, 157. 490 of weight, 160 Work-energy equation, 163, 179, 250. 276, 473, 568
Conversion Charts Between SI and U.S. Customary Units
Pressure or Stress
Conversion Charts Between SI and U.S. Customary Units (cont) N -m 1000-
I It-ft -700
Mg/m 5 10 —
ibm/ftJ
hp
200-
-260
—-600
900600-
kw
-600
- 500
180-
-240
160-
-220 -200
700-
140-
500
-ISO
-400
600-
120-
-lfiO
-400 500 400300-
-300 -300
-140
100-
-120 M1-
-200
-200
200-
-100
00-
-80 -60
40-
-40
-100
100-
-100
20-21)
0Moment or Torque
Density
•0 Power