MEN 201 Problems For Chapter 19 with Solutions, Spring 2012

Engineering Mechanics - Dynamics Chapter 19 Problem 19-1 The rigid body (slab) has a mass m and is rotating with an angular velocity ω about an axis passing through the fixed point O. Show that the momenta of all the particles composing the body can be represented by a single vector having a magnitude mvG and acting through point P, called the center of percussion, which lies at a distance rPG = k2 G / rGO from the mass center G. Here kG is the radius of gyration of the body, computed about an axis perpendicular to the plane of motion and passing through G. Solution: Ho rGO + rPG ( )mvG = rGO mvG + IGω = Where IG mkG 2 = rGO mvG + rPGmvG rGO mvG mkG 2ω + = rPG kG 2ω vG = kG 2 vG vG rGO ⎛⎜ ⎝ ⎞⎟ ⎠ = kG 2 rGO = Q.E.D Problem 19-2 At a given instant, the body has a linear momentum L = mvG and an angular momentum HG = IGω computed about its mass center. Show that the angular momentum of the body computed about the instantaneous center of zero velocity IC can be expressed as HIC = IICω where IIC represents the body’s moment of inertia computed about the instantaneous axis of zero velocity. As shown, the IC is located at a distance rGIC away from the mass center G. Solution: HIC rGICmvG + IGω = Where vG ωrGIC = 632 Engineering Mechanics - Dynamics Chapter 19 HIC rGICmωrGIC + IGω = HIC IG mrGIC 2 + ( )ω = HIC = IICω Q.E.D. Problem 19-3 Show that if a slab is rotating about a fixed axis perpendicular to the slab and passing through its mass center G, the angular momentum is the same when computed about any other point P on the slab. Solution: Since vG = 0, the linear momentum L = mvG = 0. Hence the angular momentum about any point P is HP IGω = Since ω is a free vector , so is HP. Q.E.D. *Problem 19-4 Gear A rotates along the inside of the circular gear rack R. If A has weight W and radius of gyration kB, determine its angular momentum about point C when (a) ωR = 0, (b) ωR = ω. Given: W = 4 lbf r 0.75 ft = ωCB 30 rad s = a 1.5 ft = kB = 0.5 ft ω 20 rad s = g 32.2 ft s2 = Solution: a ( ) ωR 0 rad s = vB = aωCB ωA ωR a + r ( ) ωCB a − r = Hc W g ⎛⎜⎝ ⎞⎟⎠ vBa W g ⎛⎜⎝ ⎞⎟⎠ kB 2ωA + = Hc 6.52 slug ft2 ⋅ s = 633 Engineering Mechanics - Dynamics Chapter 19 b ( ) ωR ω = vB = aωCB ωA ωR a + r ( ) ωCB a − r = Hc W g ⎛⎜⎝ ⎞⎟⎠ vBa W g ⎛⎜⎝ ⎞⎟⎠ kB 2ωA + = Hc 8.39 slug ft2 ⋅ s = Problem 19-5 The fan blade has mass mb and a moment of inertia I0 about an axis passing through its center O. If it is subjected to moment M = A(1 − ebt) determine its angular velocity when t = t1 starting from rest. Given: mb = 2 kg A 3 N m ⋅ = t1 4 s = IO 0.18 kg m2 ⋅ = b −0.2 s 1 − = Solution: 0 0 t1 t A 1 ( − eb t ) ⌠ ⎮⌡ d + = IOω1 ω1 1 IO 0 t1 t A 1 ( − eb t ) ⌠ ⎮⌡ d = ω1 20.8 rad s = Problem 19-6 The wheel of mass mw has a radius of gyration kA. If the wheel is subjected to a moment M = bt, determine its angular velocity at time t1 starting from rest. Also, compute the reactions which the fixed pin A exerts on the wheel during the motion. Given: mw = 10 kg t1 3 s = kA = 200 mm g 9.81 m s2 = b 5 N m ⋅ s = Solution: 0 0 t1 bt t ⌠ ⎮⌡ d + mw kA 2ω1 = ω1 1 mw kA 2 0 t1 bt t ⌠ ⎮⌡ d = ω1 56.25 rad s = 0 + Ax t1 = 0 Ax = 0 Ax 0.00 = 634 Engineering Mechanics - Dynamics Chapter 19 0 + Ay t1 − mw gt1 = 0 Ay = mw g Ay 98.10 N = Problem 19-7 Disk D of weight W is subjected to counterclockwise moment M = bt. Determine the angular velocity of the disk at time t2 after the moment is applied. Due to the spring the plate P exerts constant force P on the disk. The coefficients of static and kinetic friction between the disk and the plate are μs and μk respectively. Hint: First find the time needed to start the disk rotating. Given: W = 10 lb μs 0.3 = b 10 lb ft ⋅ s = μk 0.2 = t2 = 2 s r 0.5 ft = P = 100lb g 32.2 ft s2 = Solution: When motion begins bt1 = μsP r t1 μsP r b = t1 1.50 s = At a later time we have 0 t1 t2 t bt − μk P r ( ) ⌠ ⎮ ⌡ d + W g ⎛⎜⎝ ⎞⎟⎠ r2 2 ω2 = ω2 2g W r2 t1 t2 t bt − μk P r ( ) ⌠ ⎮ ⌡ d = ω2 96.6 rad s = *Problem 19-8 The cord is wrapped around the inner core of the spool. If block B of weight WB is suspended from the cord and released from rest, determine the spool’s angular velocity when t = t1. Neglect the mass of the cord. The spool has weight WS and the radius of gyration about the axle A is kA. Solve the problem in two ways, first by considering the “system” consisting of the block and spool, and then by considering the block and spool separately. Given: WB 5 lb = 635 Engineering Mechanics - Dynamics Chapter 19 t1 3 s = WS 180 lb = kA 1.25 ft = ri 1.5ft = ro 2.75 ft = g 32.2 ft s2 = Solution: (a) System as a whole 0 + WBri t1 WS g ⎛⎜⎝ ⎞⎟⎠ kA 2ω WB g ⎛⎜⎝ ⎞⎟⎠ (riω )ri + = ω WBri t1 g WS kA 2 WBri 2 + = ω 2.48 rad s = (b) Parts separately Guesses T = 1 lb ω 1 rad s = Given 0 + T t1 ri WS g ⎛⎜⎝ ⎞⎟⎠ kA 2ω = T t1 − WBt1 WB − g ⎛⎜⎝ ⎞⎟⎠ (riω ) = T ⎛⎜⎝ω ⎞⎟⎠ Find T ω ( , ) = T = 4.81 lb ω 2.48 rad s = Problem 19-9 The disk has mass M and is originally spinning at the end of the strut with angular velocity ω. If it is then placed against the wall, for which the coefficient of kinetic friction is μk determine the time required for the motion to stop. What is the force in strut BC during this time? Given: M = 20 kg θ 60 deg = ω 60 rad s = r 150 mm = μk = 0.3 g 9.81 m s2 = 636 Engineering Mechanics - Dynamics Chapter 19 Solution: Initial Guess: FCB = 1 N t = 1 s NA 1 N = Given −M r2 2 ⎛⎜ ⎝ ⎞⎟ ⎠ ω + μk NArt 0 = FCB cos θ( )t − NAt 0 = FCB sin θ( )t − M gt + μk NAt 0 = FCB NA t ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find FCB NA , t, ( ) = NA = 96.55 N FCB = 193 N t 3.11 s = Problem 19-10 A flywheel has a mass M and radius of gyration kG about an axis of rotation passing through its mass center. If a motor supplies a clockwise torque having a magnitude M = kt, determine the flywheel’s angular veliocity at time t1. Initially the flywheel is rotating clockwise at angular velocity ω0. Given: M = 60 kg kG = 150 mm k 5 N m ⋅ s = t1 = 3 s ω0 2 rad s = Solution: M kG 2ω0 0 t1 kt t ⌠ ⎮⌡ d + M kG 2ω1 = ω1 ω0 1 M kG 2 0 t1 kt t ⌠ ⎮⌡ d + = ω1 18.7 rad s = Problem 19-11 A wire of negligible mass is wrapped around the outer surface of the disk of mass M. If the disk is released from rest, determine its angular velocity at time t. 637 Engineering Mechanics - Dynamics Chapter 19 Given: M 2 kg = t 3 s = r 80 mm = Solution: 0 + M grt 3 2M r2ω = ω 2 3 g r ⎛⎜⎝ ⎞⎟⎠ t = ω 245 rad s = *Problem 19-12 The spool has mass mS and radius of gyration kO. Block A has mass mA, and block B has mass mB. If they are released from rest, determine the time required for block A to attain speed vA. Neglect the mass of the ropes. Given: mS = 30 kg mB = 10 kg ro 0.3 m = kO = 0.25 m vA 2 m s = g 9.81 m s2 = mA = 25 kg ri 0.18 m = Solution: Guesses t = 1 s vB 1 m s = ω 1 rad s = Given vA = ωro vB ωri = 0 + mAgt ro − mBgtri mAvAro + mBvBri mS kO 2ω + = t vB ω ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find t vB , , ω ( ) = vB 1.20 m s = ω 6.67 rad s = t 0.530 s = 638 Engineering Mechanics - Dynamics Chapter 19 Problem 19-13 The man pulls the rope off the reel with a constant force P in the direction shown. If the reel has weight W and radius of gyration kG about the trunnion (pin) at A, determine the angular velocity of the reel at time t starting from rest. Neglect friction and the weight of rope that is removed. Given: P = 8 lb t = 3 s g 32.2 ft s2 = W = 250 lb θ 60 deg = kG = 0.8 ft r 1.25 ft = Solution: 0 + P rt W g ⎛⎜⎝ ⎞⎟⎠ kG 2ω = ω P rt g W kG 2 = ω 6.04 rad s = Problem 19-14 Angular motion is transmitted from a driver wheel A to the driven wheel B by friction between the wheels at C. If A always rotates at constant rate ωA and the coefficient of kinetic friction between the wheels is μk, determine the time required for B to reach a constant angular velocity once the wheels make contact with a normal force FN. What is the final angular velocity of wheel B? Wheel B has mass mB and radius of gyration about its axis of rotation kG. Given: ωA 16 rad s = mB = 90 kg a = 40 mm c 4 mm = μk = 0.2 kG = 120 mm b = 50 mm FN 50 N = Solution: Guesses t = 1 s ωB 1 rad s = Given μk FN a + b ( )t mBkG 2ωB = ωB a + b ( ) ωA a 2 ⎛⎜⎝ ⎞⎟⎠ = t ⎛⎜ωB ⎝ ⎞⎟ ⎠ Find t ωB ( , ) = ωB 3.56 rad s = t 5.12 s = 639 Engineering Mechanics - Dynamics Chapter 19 Problem 19-15 The slender rod of mass M rests on a smooth floor. If it is kicked so as to receive a horizontal impulse I at point A as shown, determine its angular velocity and the speed of its mass center. Given: M 4 kg = l1 2 m = l2 1.75 m = I 8 N s = θ 60 deg = Solution: Guesses v 1 m s = ω 1 rad s = Given I sin θ( ) l2 l1 − 2 ⎛⎜⎝ ⎞⎟⎠ 1 12M l1 2ω = I M v = ω ⎛⎜⎝ v ⎞⎟⎠ Find ω v ( , ) = ω 3.90 rad s = v 2.00 m s = *Problem 19-16 A cord of negligible mass is wrapped around the outer surface of the cylinder of weight W and its end is subjected to a constant horizontal force P. If the cylinder rolls without slipping at A, determine its angular velocity in time t starting from rest. Neglect the thickness of the cord. Given: W 50 lb = P 2 lb = t 4 s = r 0.6 ft = 640 Engineering Mechanics - Dynamics Chapter 19 Solution: 0 P t 2r ( ) + 1 2 W g ⎛⎜⎝ ⎞⎟⎠ r2 ⎡⎢⎣ ⎤⎥⎦ ω W g ⎛⎜⎝ ⎞⎟⎠ (rω )r + = ω 4P t g = 3rW ω 11.4 rad s = Problem 19-17 The drum has mass M, radius r, and radius of gyration kO. If the coefficients of static and kinetic friction at A are μs and μk respectively, determine the drum's angular velocity at time t after it is released from rest. Given: Given: M = 70 kg μs = 0.4 θ 30 deg = r = 300mm μk = 0.3 g 9.81 m s2 = kO = 125 mm t 2 s = Solution: Assume no slip Guesses Ff = 1 N FN 1 N = ω 1 rad s = v 1 m s = Fmax 1 N = Given 0 + Ff rt M kO 2ω = v = ωr Fmax μsFN = M gsin θ( )t − Ff t = M v FN t − M gcos θ( )t 0 = Ff Fmax FN ω v ⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠ Find Ff Fmax , , FN , ω , v ( ) = Ff Fmax FN ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ 51 238 595 ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ N = v 8.36 m s = ω 27.9 rad s = Since Ff = 51 N < Fmax = 238 N then our no-slip assumption is good. Problem 19-18 The double pulley consists of two wheels which are attached to one another and turn at the same rate. The pulley has mass mp and radius of gyration kO. If the block at A has mass mA, determine 641 Engineering Mechanics - Dynamics Chapter 19 the speed of the block at time t after a constant force F is applied to the rope wrapped around the inner hub of the pulley. The block is originally at rest. Neglect the mass of the rope. Units Used: kN 103 N = Given: mp = 15 kg F 2 kN = kO = 110 mm ri 75 mm = mA = 40 kg ro 200 mm = t 3 s = Solution: Guess vA 1 m s = ω 1 rad s = Given −F ri t + mAgro t −mp kO 2ω − mAvAro = vA ωro = vA ⎛⎜ ω ⎝ ⎞⎟ ⎠ Find vA ω , ( ) = ω 120.44 rad s = vA 24.1 m s = Problem 19-19 The spool has weight W and radius of gyration kO. A cord is wrapped around its inner hub and the end subjected to a horizontal force P. Determine the spool’s angular velocity at time t starting from rest. Assume the spool rolls without slipping. Given: W = 30 lb t 4 s = kO = 0.45 ft ri 0.3 ft = P = 5 lb ro 0.9 ft = Solution: Guesses ω 1 rad s = vO 1 ft s = Given −P ro − ri ( )t W − g ⎛⎜⎝ ⎞⎟⎠ vOro W g ⎛⎜⎝ ⎞⎟⎠ kO 2ω − = vO ωro = vO ⎛⎜ ω ⎝ ⎞⎟ ⎠ Find vO ω , ( ) = vO 3.49 m s = ω 12.7 rad s = *Problem 19-20 The two gears A and B have weights WA, WB and radii of gyration kA and kB respectively. If a motor transmits a couple moment to gear B of M = M0 ( 1 − e-bt ), determine the angular velocity of gear A 642 Engineering Mechanics - Dynamics Chapter 19 at time t, starting from rest. Given: WA = 15 lb WB 10 lb = rA = 0.8 ft rB 0.5 ft = kA = 0.5 ft kB 0.35 ft = M0 2 lb ft ⋅ = b 0.5 s 1 − = t 5 s = Solution: Guesses ωA 1 rad s = ωB 1 rad s = ImpF 1 lb s⋅ = Given 0 t t M0 1 e b − t ( − ) ⌠ ⎮⌡ d − ImpFrB WB g ⎛⎜⎝ ⎞⎟⎠ kB 2ωB = ImpFrA WA g ⎛⎜⎝ ⎞⎟⎠ kA 2ωA = ωArA ωBrB = ωA ωB ImpF ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find ωA ωB , , ImpF ( ) = ImpF = 6.89 lbs ωB 75.7 rad s = ωA 47.3 rad s = Problem 19-21 Spool B is at rest and spool A is rotating at ω when the slack in the cord connecting them is taken up. Determine the angular velocity of each spool immediately after the cord is jerked tight by the spinning of spool A. The weights and radii of gyration of A and B are WA, kA, and WB, kA, respectively. 643 Engineering Mechanics - Dynamics Chapter 19 Given: WA = 30 lb WB 15 lb = kA = 0.8 ft kB 0.6 ft = rA = 1.2 ft rB 0.4 ft = ω 6 rad s = Solution: Guesses ωA 1 rad s = ωB 1 rad s = Imp 1 lb s⋅ = Given WA g ⎛⎜⎝ ⎞⎟⎠ kA 2ω − Imp rA WA g ⎛⎜⎝ ⎞⎟⎠ kA 2ωA = Imp rB WB g ⎛⎜⎝ ⎞⎟⎠ kB 2ωB = ωArA ωBrB = ωA ωB Imp ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find ωA ωB , , Imp ( ) = Imp 2.14 lb s⋅ = ωA ωB ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1.70 ⎛⎜⎝5.10 ⎞⎟⎠ rad s = Problem 19-22 Disk A of mass mA is mounted on arm BC, which has a negligible mass. If a torque of M = M0eat is applied to the arm at C, determine the angular velocity of BC at time t starting from rest. Solve the problem assuming that (a) the disk is set in a smooth bearing at B so that it rotates with curvilinear translation, (b) the disk is fixed to the shaft BC, and (c) the disk is given an initial freely spinning angular velocity ωDk prior to application of the torque. 644 Engineering Mechanics - Dynamics Chapter 19 mA = 4 kg M0 5 N m ⋅ = ωD −80 rad s = r = 60 mm a 0.5 s 1 − = b = 250 mm t 2 s = Solution: Guess ωBC 1 rad s = a ( ) Given 0 t t M0 ea t ⌠ ⎮⌡ d mAωBC b2 = ωBC Find ωBC ( ) = ωBC 68.7 rad s = b ( ) Given 0 t t M0 ea t ⌠ ⎮⌡ d mAωBC b2 mA r2 2 ⎛⎜ ⎝ ⎞⎟ ⎠ ωBC + = ωBC Find ωBC ( ) = ωBC 66.8 rad s = c ( ) Given −mA r2 2 ⎛⎜ ⎝ ⎞⎟ ⎠ ωD 0 t t M0 ea t ⌠ ⎮⌡ d + mAωBC b2 mA r2 2 ⎛⎜ ⎝ ⎞⎟ ⎠ ωD − = ωBC Find ωBC ( ) = ωBC 68.7 rad s = Problem 19-23 The inner hub of the wheel rests on the horizontal track. If it does not slip at A, determine the speed of the block of weight Wb at time t after the block is released from rest. The wheel has weight Ww and radius of gyration kG. Neglect the mass of the pulley and cord. Given: Wb = 10 lb ri 1 ft = t = 2 s ro 2 ft = Ww = 30 lb g 32.2 ft s2 = kG 1.30 ft = Solution: Guesses vG 1 ft s = vB 1 ft s = T = 1 lb ω 1 rad s = 645 Given: Engineering Mechanics - Dynamics Chapter 19 Given T t − Wb t Wb − g ⎛⎜⎝ ⎞⎟⎠ vB = T ro + ri ( )t Ww g ⎛⎜⎝ ⎞⎟⎠ vGri Ww g ⎛⎜⎝ ⎞⎟⎠ kG 2ω + = vG = ωri vB ω ri + ro ( ) = vG vB ω T ⎛⎜ ⎜ ⎜ ⎜ ⎝ ⎞⎟ ⎟ ⎟ ⎟ ⎠ Find vG vB , , ω , T ( ) = T = 4.73 lb ω 11.3 rad s = vG 11.3 ft s = vB 34.0 ft s = *Problem 19-24 If the hoop has a weight W and radius r and is thrown onto a rough surface with a velocity vG parallel to the surface, determine the amount of backspin, ω0, it must be given so that it stops spinning at the same instant that its forward velocity is zero. It is not necessary to know the coefficient of kinetic friction at A for the calculation. Solution: W g ⎛⎜⎝ ⎞⎟⎠ vGr W g ⎛⎜⎝ ⎞⎟⎠ r2ω0 − 0 = ω0 vG r = Problem 19-25 The rectangular plate of weight W is at rest on a smooth horizontal floor. If it is given the horizontal impulses shown, determine its angular velocity and the velocity of the mass center. 646 Engineering Mechanics - Dynamics Chapter 19 Given: W = 10 lb d 0.5 ft = I1 = 20 lb s⋅ θ 60 deg = I2 = 5 lb s⋅ e 3 = a = 0.5 ft f 4 = b = 2 ft g 32.2 ft s2 = c 1 ft = Solution: Guesses vx 1 ft s = vy 1 ft s = ω 1 rad s = Given f e2 f 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ I1 + I2 cos θ( ) W g ⎛⎜⎝ ⎞⎟⎠ vx = e e2 f 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ I1 − I2 sin θ( ) W g ⎛⎜⎝ ⎞⎟⎠ vy = f e2 f 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ I1 a I2 sin θ( ) b − d ( ) − 1 12 W g ⎛⎜⎝ ⎞⎟⎠ b2 + c2 ( )ω W g ⎛⎜⎝ ⎞⎟⎠ vx c 2 ⎛⎜⎝ ⎞⎟⎠ + W g ⎛⎜⎝ ⎞⎟⎠ vy b 2 ⎛⎜⎝ ⎞⎟⎠ + = vx vy ω ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find vx vy , , ω ( ) = vG vx vy ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = ω −119 rad s = vG 59.6 ⎛⎜⎝24.7 ⎞⎟⎠ ft s = vG 64.5 ft s = Problem 19-26 The ball of mass m and radius r rolls along an inclined plane for which the coefficient of static friction is μ. If the ball is released from rest, determine the maximum angle θ for the incline so that it rolls without slipping at A. 647 Engineering Mechanics - Dynamics Chapter 19 Solution: Nt − mgcos θ( )t = 0 N mgcos θ( ) = μNrt 2 = 5mr2ω t 2ωr 5μgcos θ( ) = mgsin θ( )t − μNt = mrω t rω g sin θ( ) − μ cos θ( ) ( ) = Thus 2ωr 5μgcos θ( ) rω g sin θ( ) − μ cos θ( ) ( ) = θ atan 7μ 2 ⎛⎜⎝ ⎞⎟⎠ = Problem 19-27 The spool has weight Ws and radius of gyration kO. If the block B has weight Wb and a force P is applied to the cord, determine the speed of the block at time t starting from rest. Neglect the mass of the cord. Given: Ws = 75lb t 5s = 648 Engineering Mechanics - Dynamics Chapter 19 kO = 1.2ft ro 2ft = Wb = 60lb ri 0.75ft = P 25lb = Solution: Guess vB 1 ft s = ω 1 rad s = Given −P ro t + Wb ri t Wb − g ⎛⎜⎝ ⎞⎟⎠ vBri Ws g ⎛⎜⎝ ⎞⎟⎠ kO 2ω − = vB ωri = vB ⎛⎜ ω ⎝ ⎞⎟ ⎠ Find vB ω , ( ) = ω 5.68 rad s = vB 4.26 ft s = *Problem 19-28 The slender rod has a mass m and is suspended at its end A by a cord. If the rod receives a horizontal blow giving it an impulse I at its bottom B, determine the location y of the point P about which the rod appears to rotate during the impact. Solution: FΔt mvcm = mΔt Iω = I y I'ω = I y m l2 12 y l ⎛⎜⎝ − 2 ⎞⎟⎠ 2 + ⎡⎢ ⎣ ⎤⎥ ⎦ ω = 1 ( ) I mvcm = I m l 2ω = 2 ( ) Divide Eqs (1) by (2) yl 2 l2 12 y l ⎛⎜⎝ − 2 ⎞⎟⎠ 2 + ⎡⎢ ⎣ ⎤⎥ ⎦ = y 3 4 33 + 12 ⎛⎜⎝ ⎞⎟⎠ l = 649 Engineering Mechanics - Dynamics Chapter 19 Problem 19-29 A thin rod having mass M is balanced vertically as shown. Determine the height h at which it can be struck with a horizontal force F and not slip on the floor. This requires that the frictional force at A be essentially zero. Given: M = 4 kg L 0.8 m = Solution: F t Mω L 2 = F t h M L2 3 ω = Thus Mω L 2 h M L2 3 ω = h 2 = 3L h 0.53 m = Problem 19-30 The square plate has a mass M and is suspended at its corner A by a cord. If it receives a horizontal impulse I at corner B, determine the location y' of the point P about which the plate appears to rotate during the impact. Solution: I a 2 1 = 6M a2ω ω 6 2 I M a = I = M vG vG I M = y' vG ω = y' 2 6 a = y a 2 − y' = y 2 3 a = Problem 19-31 Determine the height h of the bumper of the pool table, so that when the pool ball of mass m strikes it, no frictional force will be developed between the ball and the table at A. Assume the bumper exerts only a horizontal force on the ball. 650 Engineering Mechanics - Dynamics Chapter 19 Solution: FΔt = MΔv FΔt h 7 = 5M r2Δω Δv rΔω = Thus M rΔωh 7 = 5M r2Δω h 7 5r = *Problem 19-32 The double pulley consists of two wheels which are attached to one another and turn at the same rate. The pulley has a mass MC and a radius of gyration kO. If the block at A has a mass MA and the container at B has a mass MB, including its contents, determine the speed of the container at time t after it is released from rest. Given: MC = 15 kg kO 110 mm = MA = 40 kg r1 200 mm = MB = 85 kg r2 75 mm = t 3 s = Solution: Guess vA 1 m s = vB 1 m s = ω 1 rad s = Given MAgtr1 − MBgt r2 MAvAr1 + MBvBr2 MC kO 2ω + = vA = ωr1 vB ωr2 = vA vB ω ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find vA vB , , ω ( ) = ω 21.2 rad s = vA 4.23 m s = vB 1.59 m s = Problem 19-33 The crate has a mass Mc. Determine the constant speed v0 it acquires as it moves down the conveyor. The rollers each have radius r, mass M, and are spaced distance d apart. Note that friction causes each roller to rotate when the crate comes in contact with it. 651 Engineering Mechanics - Dynamics Chapter 19 Solution: Assume each roller is brought to the condition of roll without slipping. In time t, the number of rollers affected is v0t/d. Mc gsin θ( )t − F t 0 = F Mc gsin θ( ) = F t r 1 ⎛⎜⎝2M r2 ⎞⎟⎠ v0 r v0 ⎛⎜⎝ d t ⎞⎟⎠ = v0 2gsin θ( )d MC M = Problem 19-34 Two wheels A and B have masses mA and mB and radii of gyration about their central vertical axes of kA and kB respectively. If they are freely rotating in the same direction at ωA and ωΒ about the same vertical axis, determine their common angular velocity after they are brought into contact and slipping between them stops. Solution: mAkA 2ωA mBkB 2ωB + mAkA 2 mBkB 2 + ( )ω = ω mAkA 2ωA mBkB 2ωB + mAkA 2 mBkB 2 + = Problem 19-35 The Hubble Space Telescope is powered by two solar panels as shown. The body of the telescope has a mass M1 and radii of gyration kx and ky, whereas the solar panels can be considered as thin plates, each having a mass M2. Due to an internal drive, the panels are given an angular velocity of ω0j, measured relative to the telescope. Determine the angular velocity of the telescope due to the rotation of the panels. Prior to rotating the panels, the telescope was originally traveling at vG = (vxi + vyj + vzk). Neglect its orbital rotation. 652 Engineering Mechanics - Dynamics Chapter 19 Units Used: Mg 103kg = Given: M1 = 11 Mg ω0 0.6 rad s = vx −400 m s = M2 54 kg = a = 1.5 m vy 250 m s = kx 1.64 m = ky = 3.85 m b = 6 m vz 175 m s = Solution: Angular momentum is conserved. Guess ωT 1 rad s = Given 0 2 1 ⎛⎜⎝12M2 b2 ⎞⎟⎠ ω0 − ωT ( ) M1 ky 2 ( )ωT − = ωT Find ωT ( ) = ωT 0.00 rad s = *Problem 19-36 The platform swing consists of a flat plate of weight Wp suspended by four rods of negligible weight. When the swing is at rest, the man of weight Wm jumps off the platform when his center of gravity G is at distance a from the pin at A. This is done with a horizontal velocity v, measured relative to the swing at the level of G. Determine the angular velocity he imparts to the swing just after jumping off. Given: Wp = 200lb a 10ft = Wm = 150lb b 11ft = v 5 ft s = c 4ft = Solution: Guess ω 1 rad s = Given 0 Wm − g v ( − ωa )a Wp g c2 12 + b2 ⎛⎜ ⎝ ⎞⎟ ⎠ ω + = ω Find ω ( ) = ω 0.190 rad s = 653 Engineering Mechanics - Dynamics Chapter 19 Problem 19-37 Each of the two slender rods and the disk have the same mass m. Also, the length of each rod is equal to the diameter d of the disk. If the assembly is rotating with an angular velocity ω1 when the rods are directed outward, determine the angular velocity of the assembly if by internal means the rods are brought to an upright vertical position. Solution: H1 H2 = 1 2m d 2 ⎛⎜⎝ ⎞⎟⎠ 2 2 1 12md2 + + 2md2 ⎡⎢ ⎣ ⎤⎥ ⎦ ω1 1 2m d 2 ⎛⎜⎝ ⎞⎟⎠ 2 2m d 2 ⎛⎜⎝ ⎞⎟⎠ 2 + ⎡⎢ ⎣ ⎤⎥ ⎦ ω2 = ω2 11 3 ω1 = Problem 19-38 The rod has a length L and mass m. A smooth collar having a negligible size and one-fourth the mass of the rod is placed on the rod at its midpoint. If the rod is freely rotating with angular velocity ω about its end and the collar is released, determine the rod’s angular velocity just before the collar flies off the rod. Also, what is the speed of the collar as it leaves the rod? Solution: H1 H2 = 1 3mL2ω m 4 ⎛⎜⎝ ⎞⎟⎠ L 2 ⎛⎜⎝ ⎞⎟⎠ ω L 2 ⎛⎜⎝ ⎞⎟⎠ + 1 3mL2ω' m 4 ⎛⎜⎝ ⎞⎟⎠ Lω' L + = ω' 19 28ω = T1 + V1 T2 + V2 = 1 2 1 ⎛⎜⎝3mL2 ⎞⎟⎠ ω2 1 2 m 4 ⎛⎜⎝ ⎞⎟⎠ L ⎛⎜⎝2 ω ⎞⎟⎠ 2 + 1 2 m 4 ⎛⎜⎝ ⎞⎟⎠ v'2 1 2 m 4 ⎛⎜⎝ ⎞⎟⎠ (Lω' )2 + 1 2 1 ⎛⎜⎝3mL2 ⎞⎟⎠ ω'2 + = v'2 57 112L2 ω2 = v'' = 0.985ωL v'' 57 112L2 ω2 L 19 ⎛⎜⎝28ω ⎞⎟⎠ ⎡⎢⎣ ⎤⎥⎦ 2 + = v'' 95 98 ωL = 654 Engineering Mechanics - Dynamics Chapter 19 Problem 19-39 A man has a moment of inertia Iz about the z axis. He is originally at rest and standing on a small platform which can turn freely. If he is handed a wheel which is rotating at angular velocity ω and has a moment of inertia I about its spinning axis, determine his angular velocity if (a) he holds the wheel upright as shown, (b) turns the wheel out θ = 90°, and (c) turns the wheel downward, θ = 180°. Solution: a ( ) 0 + Iω IzωM + Iω = ωM 0 = b ( ) 0 + Iω IzωM + 0 = ωM I Iz ω = c ( ) 0 + Iω IzωM − Iω = ωM 2I Iz ω = * Problem 19-40 The space satellite has mass mss and moment of inertia Iz , excluding the four solar panels A,B,C, and D. Each solar panel has mass mp and can be approximated as a thin plate.If the satellite is originally spinning about the z axis at aconstant rate ωz, when θ = 90°, determine the rate of spin if all the panels are raised and reach the upward position, θ = 0°, at the same instsnt. Given: mss = 125 kg a = 0.2 m ωz 0.5 rad s = Iz 0.940 kg m2 ⋅ = b 0.75 m = msp = 20 kg c 0.2 m = Solution: Guess ωz2 1 rad s = Given Iz 4 msp 12 b2 + c2 ( ) msp a b ⎛⎜⎝ + 2 ⎞⎟⎠ 2 + ⎡⎢ ⎣ ⎤⎥ ⎦ ⎡⎢ + ⎣ ⎤⎥ ⎦ ωz Iz 4 msp 12 c2 + msp a2 ⎛⎜⎝ ⎞⎟⎠ ⎡⎢⎣ + ⎤⎥⎦ ωz2 = ωz2 Find ωz2 ( ) = ωz2 3.56 rad s = 655 Engineering Mechanics - Dynamics Chapter 19 Problem 19-41 Rod ACB of mass mr supports the two disks each of mass md at its ends. If both disks are given a clockwise angular velocity ωA1 = ωΒ1 = ω0 while the rod is held stationary and then released, determine the angular velocity of the rod after both disks have stopped spinning relative to the rod due to frictional resistance at the pins A and B. Motion is in the horizontal plane. Neglect friction at pin C. Given: mr 2 kg = md 4 kg = ω0 5 rad s = a 0.75 m = r 0.15 m = Solution: 2 1 ⎛⎜⎝2md ⎞⎟⎠ r2ω0 2 1 ⎛⎜⎝2md ⎞⎟⎠ r2 + 2md a2 mr 12 2a ( )2 + ⎡⎢⎣ ⎤⎥⎦ ω2 = ω2 md r2 md r2 + 2md a2 mr 3 ⎛⎜⎝ ⎞⎟⎠ a2 + ω0 = ω2 0.0906 rad s = Problem 19-42 Disk A has a weight WA. An inextensible cable is attached to the weight W and wrapped around the disk.The weight is dropped distance h before the slack is taken up. If the impact is perfectly elastic, i.e., e = 1, determine the angular velocity of the disk just after impact. Given: WA = 20 lb h 2 ft = W = 10 lb r 0.5 ft = Solution: v1 2gh = Guess ω2 1 rad s = v2 1 ft s = 656 Engineering Mechanics - Dynamics Chapter 19 Given W g ⎛⎜⎝ ⎞⎟⎠ v1 r W g ⎛⎜⎝ ⎞⎟⎠ v2 r WA g ⎛⎜⎝ ⎞⎟⎠ r2 2 ω2 + = v2 ω2 r = v2 ω2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find v2 ω2 , ( ) = v2 5.67 ft s = ω2 11.3 rad s = Problem 19-43 A thin disk of mass m has an angular velocity ω1 while rotating on a smooth surface. Determine its new angular velocity just after the hook at its edge strikes the peg P and the disk starts to rotate about P without rebounding. Solution: H1 H2 = 1 ⎛⎜⎝2mr2 ⎞⎟⎠ ω1 1 2mr2 + mr2 ⎛⎜⎝ ⎞⎟⎠ ω2 = ω2 1 3ω1 = *Problem 19-44 The pendulum consists of a slender rod AB of weight Wr and a wooden block of weight Wb. A projectile of weight Wp is fired into the center of the block with velocity v. If the pendulum is initially at rest, and the projectile embeds itself into the block, determine the angular velocity of the pendulum just after the impact. Given: Wr = 5 lb Wp = 0.2 lb v 1000 ft s = g 32.2 ft s2 = Wb = 10 lb a = 2 ft b 1 ft = Solution: Wp g ⎛⎜⎝ ⎞⎟⎠ v a b ⎛⎜⎝ + 2 ⎞⎟⎠ Wr g ⎛⎜⎝ ⎞⎟⎠ a2 3 ⎛⎜ ⎝ ⎞⎟ ⎠ Wb g ⎛⎜⎝ ⎞⎟⎠ b2 6 ⎛⎜ ⎝ ⎞⎟ ⎠ + Wb g ⎛⎜⎝ ⎞⎟⎠ a b ⎛⎜⎝ + 2 ⎞⎟⎠ 2 + Wp g ⎛⎜⎝ ⎞⎟⎠ a b ⎛⎜⎝ + 2 ⎞⎟⎠ 2 + ⎡⎢ ⎣ ⎤⎥ ⎦ ω2 = ω2 Wp v a b ⎛⎜⎝ + 2 ⎞⎟⎠ Wr a2 3 Wb b2 6 + Wb a b ⎛⎜⎝ + 2 ⎞⎟⎠ 2 + Wp a b ⎛⎜⎝ + 2 ⎞⎟⎠ 2 + = ω2 6.94 rad s = 657 Engineering Mechanics - Dynamics Chapter 19 Problem 19-45 The pendulum consists of a slender rod AB of mass M1 and a disk of mass M2. It is released from rest without rotating. When it falls a distance d, the end A strikes the hook S, which provides a permanent connection. Determine the angular velocity of the pendulum after it has rotated 90°. Treat the pendulum’s weight during impact as a nonimpulsive force. Given: M1 = 2kg r 0.2m = M2 = 5kg l 0.5m = d 0.3m = Solution: v1 2gd = IA M1 l2 3 M2 r2 2 + M2 l ( + r )2 + = Guesses ω2 1 rad s = ω3 1 rad s = Given M1 v1 l 2 M2 v1 l ( + r ) + = IAω2 1 2IAω2 2 1 2IAω3 2 M1 g l 2 − M2 g l ( + r ) − = ω2 ω3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find ω2 ω3 , ( ) = ω2 3.57 rad s = ω3 6.46 rad s = Problem 19-46 A horizontal circular platform has a weight W1 and a radius of gyration kz about the z axis passing through its center O. The platform is free to rotate about the z axis and is initially at rest. A man having a weight W2 begins to run along the edge in a circular path of radius r. If he has a speed v and maintains this speed relative to the platform, determine the angular velocity of the platform. Neglect friction. 658 Engineering Mechanics - Dynamics Chapter 19 Given: W1 = 300 lb v 4 ft s = W2 150 lb = kz = 8 ft r 10 ft = Solution: M vr Iω = ω W2 v r W1 kz 2 = W2 g vr W1 g kz 2ω = ω 0.312 rad s = Problem 19-47 The square plate has a weight W and is rotating on the smooth surface with a constant angular velocity ω0. Determine the new angular velocity of the plate just after its corner strikes the peg P and the plate starts to rotate about P without rebounding. Solution: W g ⎛⎜⎝ ⎞⎟⎠ a2 6 ⎛⎜ ⎝ ⎞⎟ ⎠ ω0 W g ⎛⎜⎝ ⎞⎟⎠ 2a2 3 ⎛⎜ ⎝ ⎞⎟ ⎠ ω = ω 1 4ω0 = *Problem 19-48 Two children A and B, each having a mass M1, sit at the edge of the merry-go-round which is rotating with angular velocity ω. Excluding the children, the merry-go-round has a mass M2 and a radius of gyration kz. Determine the angular velocity of the merry-go-round if A jumps off horizontally in the −n direction with a speed v, measured with respect to the merry-go-round.What is the merry-go-round’s angular velocity if B then jumps off horizontally in the +t direction with a speed v, measured with respect to the merry-go-round? Neglect friction and the size of each child. 659 Engineering Mechanics - Dynamics Chapter 19 Given: M1 = 30 kg kz 0.6 m = M2 = 180 kg a 0.75 m = ω 2 rad s = v 2 m s = Solution: a ( ) Guess ω2 1 rad s = Given M2 kz 2 + 2M1 a2 ( )ω M2 kz 2 + M1 a2 ( )ω2 = ω2 Find ω2 ( ) = ω2 2.41 rad s = b ( ) Guess ω3 1 rad s = Given M2 kz 2 + M1 a2 ( )ω2 M2 kz 2ω3 M1 v ( + ω3 a )a + = ω3 Find ω3 ( ) = ω3 1.86 rad s = Problem 19-49 A bullet of mass mb having velocity v is fired into the edge of the disk of mass md as shown. Determine the angular velocity of the disk of mass md just after the bullet becomes embedded in it. Also, calculate how far θ the disk will swing until it stops. The disk is originally at rest. Given: mb = 7 gm md = 5 kg φ 30 deg = v 800 m s = r 0.2 m = Solution: Guesses ω 1 rad s = θ 10 deg = Given mb vcos φ( )r 3 = 2md r2ω −md gr 1 2 3 ⎛⎜⎝2md r2 ⎞⎟⎠ ω2 + −md grcos θ( ) = ω ⎛⎜⎝ θ ⎞⎟⎠ Find ω θ ( , ) = ω 3.23 rad s = θ 32.8 deg = 660 Engineering Mechanics - Dynamics Chapter 19 Problem 19-50 The two disks each have weight W. If they are released from rest when θ = θ1 , determine the maximum angle θ2 after they collide and rebound from each other. The coefficient of restitution is e. When θ = 0° the disks hang so that they just touch one another. Given: g 32.2 ft s2 = W 10 lb = θ1 30 deg = e 0.75 = r 1 ft = Solution: Guesses ω1 1 rad s = ω2 1 rad s = θ2 10deg = Given −W rcos θ1 ( ) 1 2 3 2 W g r2 ⎛⎜⎝ ⎞⎟⎠ ω1 2 − W r = erω1 rω2 = −W r 1 2 3 2 W g r2 ⎛⎜⎝ ⎞⎟⎠ ω2 2 + −W rcos θ2 ( ) = ω1 ω2 θ2 ⎛⎜ ⎜ ⎜⎝ ⎞⎟ ⎟ ⎟⎠ Find ω1 ω2 , , θ2 ( ) = ω1 ω2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.40 ⎛⎜⎝1.80 ⎞⎟⎠ rad s = θ2 22.4 deg = Problem 19-51 The rod AB of weight W is released from rest in the vertical position. If the coefficient of restitution between the floor and the cushion at B is e, determi