Dynamics Homework 8 - ENGINEERING MECHANICS - Dynamics...

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Unformatted text preview: ENGINEERING MECHANICS - Dynamics 18-1* A lO—lb uniform wheel l6—in. in diameter is at rest when it is W.F. Riley & L.D. Sturges placed in contact with a moving belt as shown. The kinetic coefficient of friction between the belt and the wheel is Mk = 0.1 and the belt moves with a constant speed of 30 ft/s. Determine the number of revolutions that the wheel turns before it begins to roll without slipping on the moving belt. Solution The mass center of the wheel is fixed +T 2F = ma : N'— W = 0 Y Y N = W = 10 lb and while the wheel slips F = 0.1N = 1 1b Neither P, W, nor N'do work. The work done by the friction force F while slip occurs (and the wheel rotates through angle 6) is U = I F ds = J Fr d6 = (1)[—Eg—] I d9 = —I§— 6 lb-ft The initial kinetic energy of the wheel is zero. The wheel is in fixed—axis rotation and its final kinetic energy is given by T -inZ f ' 2 G l 10 8 2 2 IG = -§-[-3§TE-][-:3-] = 0.06901 slug°ft Therefore, the work—energy equation Ti + U = Tf gives 8 l 2 + 6 = ——— . w 0 12 2 (0 06901) The wheel begins to roll without slipping when the velocity of the contact point is the same as that of the moving belt 8 ——w 12 3o ft/s m 45 rad/s 6 = 104.81 rad = 16.68 rev ........... . . . . . .............. Ans . ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 18-5 A loo-lb rectangular plate of uniform thickness is supported by a pin and a cable as shown. If the cable at B breaks, determine for the ensuing motion a. The angular speed of the plate and the magnitude of the pin reaction at A when the mass center of the plate is directly below pin A. b. The angular speed of the plate and the magnitude of the pin reaction at A when O the plate has rotated 90 (when pin B is directly below pin A). ( c. The maximum angle Gmax through which the plate will rotate. Solution The plate rotates about a fixed axis through A. The distance between A and the mass center G is 11.18034 in., ¢ = tan_1 5/10 = 26.5050, and the mass moment of inertia of the plate about A is 2 2 2 1 100 20 10 IA"IG+md'12[32.2][[12]+[12J] 2 O . [A] = 3.59443 slug.ft2 32.2 12 The initial kinetic energy of the plate is zero and the final kinetic energy of the plate is 2 2 T = L I m = %(3.59443)w A" a. As the plate rotates so that the mass center is directly below A, the mass center drops (moves in the direction of the weight force) a distance of 11.18034 - 5 = 6.18034 in. Therefore, the work—energy ‘ + = T ' equation Ti U f gives 0 + lOO[—§Ll§9§fi-] = 1.797220.)2 12 w = 5.35 rad/s ......... . . . . . . . . . . . . . ... . . . . . . . . . . . . . .. Ans. (Problem 18—5 continues ...) ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges (Problem 18-5 - cont.) Then, the equations of motion for the plate 100 11.18034 2 + = — = —— —— . T 2F AD 100 [ 32.2 ][ 12 ](5 35322) 100 11.18034 9 = = —————— —————————— + 2F At [ 32.2 ][ 12 ] +9 EMA = o = IAa give 2 a = O rad/s An = 182.9180 lb A = 182.9 lb . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 0 b. As the plate rotates 90 , the mass center drops 10 - 5 = 5 in. and the work- energy equation gives 5 2 + — = . w 0 100[ 12 J 1 79722 w = 4.81 rad/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Now the equations of motion 100 11.18034 2 + = — = --——-— —-——-—— . 2 BF AH 100 cos ¢ [ 32.2 ][ 12 ](4 81498) . _ 100 11.18034 +2 2F — At + 100 Sln ¢ — [—EETE—J L————I§————) 5 +5 EMA — 12 (100) — 3.59443a give 2 = 11.592 rad/s An = 156.525 1b At = —11.180 lb = 156.9 lb . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. c. The maximum rotation of the plate corresponds to a final kinetic energy of zero. Therefore, the net work done by the weight force must be zero - the weight must end up at the same level as it started. Then 11.18034 sin(9 + ¢) = 11.18034 sin ¢ 6 + ¢ = 1800 - ¢ 6 = 1800 - 2¢ = 126.870 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. '——————-—-—-———-—-————————_—_——— l ? ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges t———————————————————————————————————————————————————_—————________________ 18-13 A 16—lb bowling ball is placed on a 28° inclined surface and released from rest. Assume that the ball is a uniform sphere 8.6 in. in diameter. If the static coefficient of friction between the ball and the surface is 0.25, 3. Verify that the ball will begin to roll without slipping. b. Determine the speed v and the angular velocity w of the ball after it has rolled 20 ft down the incline. c. Compare the speed of part b with the speed of a 16—lb particle that slides (without friction) the same distance down the incline. Solution a. From the equations of motion +5 2F = ma : N - 16 cos 280 = 0 H II . ° _ 16 4.3 +F ZFt - mat. 16 Sin 28 F — [—§§T§—)[ 12 )a 4 3 2 16 4 3 2 c+ 2M6 = Ia“: F[ 12 J = 5 [ 32.2 ][ 12 ] a Therefore 2 N = 14.12716 lb a = 30.13352 rad/s C F = 2.14616 lb F/N = 0.15192 < us = 0.25 and the ball will roll without slipping. b. Neither the normal force N nor the friction force F do work. The weight W = mg has a potential; the zero of gravitational potential energy is set at the initial position. The kinetic energy is given by l 2 1 2 1 2 l 2 2 2 7 2 = ——— + ——— w = ——— + ——— ——— = T 2 mVG 2 I6 2 va 2 [ 5 mr )(vg/r) lo va and the work-energy equation Ti + vi + U = Tf + Vf gives 2 o O + O + O = 7va/10 + mg(—20 sin 28 ) VG = 20.8 ft/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... Ans. w = vG/r = 55.4 rad/s . . . . . ........... . . . . . . . . . . ... . . . . .. Ans. c. VFor a particle sliding without friction, work—energy gives 2 o O + O + O = mVP/2 + mg(-20 sin 28 ) V = 24.6 ft/s .................... . . . . . . . . ............ Ans. P ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges ———_———_—_—_—_—_——— 18—21* Two 16—in. diameter disks and an 8—in. diameter cylinder are fastened together to form a spool weighing 50 lb that has a radius of gyration of 5 in. with respect to an axis through the mass center of the spool. The spool is initially at rest when a constant force N of magnitude 50 lb is applied to the spool through a cord wrapped around the 8-in. diameter cylinder. If the spool rolls without slipping on the horizontal surface, determine the velocity VG of the mass center of the spool and the angular velocity w of the spool after it has rolled 10 ft to the right. Solution Neither the normal force An, the frictional force Af, nor the weight W do any work on the spool. As the center of the spool moves 10 ft to the right, the point where the 50 lb force is applied moves 10(4/8) ft to the right and the 50 lb force does work on the spool 4 U = so 10 — = 250 lb°ft [ [8]] Am The spool starts from rest (I: = O) and the kinetic energy at the final position is l = —~— + Tf 2 G 2 G l 50 8 2 + 1 50 5 2 wz 2 32.2 12 2 32.2 12 2 0.47986 0 lb’ft II and the work—energy equation Ti + U = Tf gives 2 0 + 250 = 0.47986 w w = 22.8 rad/s ... . . . . . . . . . . .... . . . . . . . . . . . . . . . . . . . . . .. Ans. V = (8/12)w = 15.22 ft/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. G ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 18—39* A l6—lb ladder 12 ft long slides in a smooth corner as shown. The ladder is at rest with 9 = 00 when the lower end is disturbed slightly. Determine the angle 9, the angular velocity w, and the velocity 38 of end B when end A loses contact with the vertical wall. Solution Neither normal force A nor B /‘ does work. The weight W = mg has a a potential; the zero of gravitational potential energy is at the level of B. The moment of inertia of the rod is 1 16 2 Is ‘ 12 [32.2 ]‘12) 2 = 5.9627 slug°ft Using the relative velocity equation to relate the velocities of A and 3 gives v j = VB 1 + 12w(-cos 9 1 — sin 6 3) A =—2m'6ft VA 1 Sin /s E5 VB = 12w cos 9 ft/s Then the velocity of the mass center G is G; _'=~+_° = w 6"+w— 6"—'6" VG VB VG/B 12 cos 1 6 ( cos 1 Sin J) 69 2 2 2 2 2 = + = 3 w VG VGX VGy 6 (ft/s) 0‘ and the work-energy equation Ti + vi + U = Tf + V? gives f3 2 2 o + 16(6) + 0 %[%](36m ) + %(5.9627)w -+ 16(6 cos 9) 96 = 11.92550.)2 + 96 cos 9 2 w H 8.05(l - cos 9) (rad/s)2 Using the relative acceleration equation to relate the accelerations of A and B gives A A A A 2 A A = a 1 + 12a(-cos 9 1 — sin 6 J) + 12w (sin 9 1 — cos 6 J) a A J 8 (Problem 18—39 continues ...) ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges (Problem 18-39 - cont.) 2 2 a3 = 12a cos 9 — 12w sin 9 ft/s and then the acceleration of the mass center G is g a q 2 A = = 9 — ' 3G 38 + saw (120: cos 12w sm 6) 1 + 6a(—cos 9 T - sin 6 j) 2 A A + 6w (sin 9 1 - cos 6 J) 2 2 6a cos 6 — 60 sin 6 ft/s D) II 2 2 -6a sin 9 — 60 cos 9 ft/s m I! Therefore the equations of motion are 16 2 _) Z = 3 = -—-———- _ ' + I; maGX A [ 32.2 )(Gd cos 6 6w Sin 6) 16 2 +T 2 = : _ = ._.._____._. _ ' 6 _ w Ey maGy B 16 [ 32-2 )( 6a Sin 6 cos 9) C+ EMG = Igor: B(6 sin 6) — A(6 cos 6) = 5.96270: End A loses contact with the vertical wall when A = O at which time 2 a = w tan 9 = 8.05(1 - cos 9) tan 9 (a) and a = 1.06263 sin 9 2 . 1.0626(16 - 2.9814a sin 9 - 2.9814w cos 9) sin 9 2 17.002 sin 9 — 3.1680G sin 6 - 25.503(1 - cos 9) sin 9 cos 9 (b) Substituting Eq. a into Eq. b and solving for 9 gives 6 = 48.190 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. w = 1.638 rad/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. VB = 13.10 ft/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. (Solving with symbols instead of numbers would show that 9 = cos—1(2/3) independent of W, C, or 9. Also, note that when 6 = 48.190 that end B is still in contact with and sliding on the floor a = 3.00 rad/s2 B = 4.00 lb > O & L.D. ENGINEERING MECHANICS - Dynamics W.F. Riley 18-47 The 5—lb uniform rod is 3 ft long. rod move in frictionless guides and weigh 2 1b each. the slider at A has a modulus of 5 = 15 lb/ft and is stretched 2 ft when the rod is vertical. If the system is released from rest when the rod is vertical, determine the angular velocity w of the rod and the velocity 68 of end B a. When the stretch in the spring is zero. b. When the bar is horizontal. Solution Neither normal force A nor B does work; the weights and the spring force £5 all have potentials. The zero of gravitational potential energy is at the level of A. The moment of inertia of the rod is 1 5 2 Is ' 12 [ 32.2 ](3) 2 0.11646 slug'ft Using the relative velocity equation I <1 gives VB J VA 1 3 (Sin 1 cos J) 00 VA = —3w sin 9 ft/s fix VB = —30 cos 9 ft/s to relate the velocities of A and B on Then the velocity of the mass center G is = —3w sin 9 i + 1.5w(sin 9 i - cos 6 3) <1 V=V+ G A 2 2 2 2 2 = + = . (1.) VG VGX VGy 2 25 (ft/s) G/A (Problem 18-47 continues Sturges The sliders at the ends of the A spring attached to ...) ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges (Problem 18—47 — cont.) 0 Initially, 9 = O , VA = VB = w = 0, Ti = 0, and 2 vi 2(3) + 5(1.5) + o + —%— (15)(2) = 43.5 lb°ft The final kinetic and potential energy are l 5 2 l 2 Tf — 2 [ 32.2 )(2.250 ) + 2 (0.11646)w 1 2 2 _ 2 l 2 2 2 + 2 [ 32.2 ](9w Sin 9) + 2 [ 32.2 ](9w cos 9) 2 0.51242w lb'ft Vf = 2(3 sin 9) + 5(1.5 sin 9) + 2(0) + -%- (15)(2 - 3 cos 9)2 13.5 sin 9 + 7.5(2 - 3 cos 9)2 lb°ft Then the work-energy equation Ti + Vi + U = Tf + V? gives 2 2 o + 43.5 + o = 0.51242w + 13.5 sin e + 7.5(2 - 3 cos 6) a. The stretch in the spring is zero when 2 - 3 cos 6 = 0 6 = 48.190 m = 8.08 rad/s D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. v3 = —3(8.08) cos 48.190 = —16.16 ft/s = 16.16 ft/s ¢ . . . . . . . . . . . . . . . . . . . . . . . .. Ans. b. The bar is horizontal when 9 = O , w = 8.38 rad/s D . . . . . . . . . . . . . . . . . . . . . . . . . . ..' . . . . . . . . . . .. Ans. 0 VB = -3(8.38) cos 0 = -25.1 ft/s = 25.1 ft/s ¢ . . . . . . . . . . . . . . . . . . . . . . . .. Ans. ...
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Dynamics Homework 8 - ENGINEERING MECHANICS - Dynamics...

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