How much work has the motor done on the rotor when the rotor has rotated through four revolutions?

Transcription

1 Problem 9. The moment of inertia of the rotor of the medical centrifuge is I =. kg-m. The rotor starts from rest and the motor exerts a constant torque of.8 N-m on it. (a) (b) How much work has the motor done on the rotor when the rotor has rotated through four revolutions? What is the rotor s angular velocity (in rpm) when it has rotated through four revolutions? (a) (b) W = (.8 N-m)(4 rev) Iω = W, ω = 4. rad/s ( ) π rad =. N-m W =. N-m rev (. N-m) (. kg-m )ω =. N-m ω =. kg-m Problem 9. The 7.8 N slender bar is.6 m in length. It started from rest in an initial position relative to the inertial reference frame. When it is in the position shown, the velocity of the end is 6.7i + 4.7j (m/s) and the bar has a counterclockwise angular velocity of rad/s. How much work was done on the bar as it moved from its initial position to its present position? y 3 B x Work = Change in kinetic energy. To calculate the kinetic energy we will first need to find the velocity of the center of mass v G = v + ω r G/ = ( 6.7i + 4.7j)( m/s) + ( rad/s)k (.35m)(cos 3 i + sin 3 j) = ( 4.7i + 8.5j) m/s Now we can calculate the work, which is equal to the kinetic energy. W = mv + Iω ( ) = 7.8 N [( ) + (. ) ] 9.8 m/s 4.7 m/s 8 5 m/s [ ( ) ] N (.6m) ( rad/s) 9.8m/s = 86. N-m. W = 86. N-m. 56 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

2 Problem 9.3 The -kg disk is at rest when the constant N-m counterclockwise couple is applied. Determine the disk s angular velocity (in rpm) when it has rotated through four revolutions (a) by applying the equation of angular motion M = α, and (b) by applying the principle of work and energy..5 m N-m (a) First we find the angular acceleration M = Iα ( N-m) = ( kg)(.5 m) α α = 6 rad/s Next we will integrate the angular acceleration to find the angular velocity. ω dω ω θ dθ = α ω dω = α dθ ω = αθ ω = ( rev ) ( ) αθ = (6 rad/s 6 s )(4 rev) = 7 rev π rad min min ω = 7 rpm. (b) pplying the principle of work energy W = Iω ( π rad ( N-m)(4 rev) rev ) = [ ] ( kg)(.5 m) ω ( rev ) ( ) 6 s ω = 8.4 rad/s = 7 rpm π rad min ω = 7 rpm. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 56

3 Problem 9.4 The space station is initially not rotating. Its reaction control system exerts a constant couple on it until it has rotated 9, then exerts a constant couple of the same magnitude in the opposite direction so that its angular velocity has decreased to zero when it has undergone a total rotation of 8. The maneuver takes 6 hours. The station s moment of inertia about the axis of rotation is I =.5 kg-m. How much work is done in performing this maneuver? In other words, how much energy had to be expended in the form of reaction control fuel? We need to solve for the moment that causes a 9 rotation in a 3 hr time period. We will use M = Iα and the principle of work energy. M = Iα = I dω dt dω dt = M I ω = M I t W = Mθ = Mθ Iω ω = I Solving these two equations together, we find ( ) M = θi (9 π rad ) 8 (.5 kg-m ) t = [ ( )( )] 6 min 6 min = 44 N-m (3 hr) hr s The total work to accomplish the entire maneuver is then ( ) W = (44 N-m)(8 π rad ) 8 = 7 N-m. W = 7 N-m. 56 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

4 Problem 9.5 The helicopter s rotor starts from rest. Suppose that its engine exerts a constant 67 N-m couple on the rotor and aerodynamic drag is negligible. The rotor s moment of inertia is I = 54 kg-m. (a) (b) Use work and energy to determine the magnitude of the rotor s angular velocity when it has rotated through five revolutions. What average power is transferred to the rotor while it rotates through five revolutions? (a) U = ( 67 N-m)(π rad) = ( 54 N -s -m)ω ω = 3.7 rad/s (b) To find the average power we need to know the time 67 N-m = ( 54 N-s -m)α α = 3 rad/s, ω = (3 rad/s )t, θ = (3 rad/s )t π rad = (3 rad/s )t t = 4.58 s Power = U t = ( 67 N)(π rad) 4.58 s = 7 N-m/s = 5. hp Problem 9.6 The helicopter s rotor starts from rest. The moment exerted on it (in N-m) is given as a function of the angle through which it has turned in radians by M = 65 θ. The rotor s moment of inertia is I = 54 kg-m. Determine the rotor s angular velocity (in rpm) when it has turned through revolutions. We will integrate to find the work. W = Mdθ (π) = (65 θ)dθ = [65 θ θ ] (π) = N-m Using the principle of work energy we can find the angular velocity. W = W ( N-m) Iω ω = = I 54 kg-m = 37. rad/s ( )( ) rev 6 s ω = 37. rad/s = 353 rpm. π rad min ω = 353 rpm. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 563

5 Problem 9.7 During extravehicular activity, an astronaut s angular velocity is initially zero. She activates two thrusters of her maneuvering unit, exerting equal and opposite forces T = N. The moment of inertia of the astronaut and her equipment about the axis of rotation is 45 kg-m. Use the principle of work and energy to determine the angle through which she has rotated when her angular velocity reaches 5 per second.. m T T The moment that is exerted on the astronaut is M = Td = ( N)( m) = N-m. Using work energy we have ( 5 ) Mθ = (45 kg-m ) Iω θ = Iω M = 8 π rad/s =.77 rad. ( N-m) Thus θ = 44. ( 8 ) θ =.77 rad = 44. π rad Problem 9.8 The 8-kg slender bar is released from rest in the horizontal position and falls to position. m y (a) (b) (c) How much work is done by the bar s weight as it falls from position to position? How much work is done by the force exerted on the bar by the pin support as the bar falls from position to position? Use conservation of energy to determine the bar s angular velocity when it is in position. x (a) W = mgh = (8 kg)(9.8 m/s )( m) = 78.5 N-m. W = 78.5 N-m. (b) The pin force is applied to point, which does not move. Therefore W =. (c) Using conservation of energy, we have T + V = T + V + = ω = 3.84 rad/s ( ) 6gh 6(9.8 m/s 3 ml ω mgh ω = L = )( m) m 564 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

6 Problem 9.9 The -N bar is released from rest in the horizontal position and falls to position. In addition to the force exerted on it by its weight, it is subjected to a constant counterclockwise couple M = 3 N- m. Determine the bar s counterclockwise angular velocity in position. M 4 m 4 y x We will use the energy equation in the form T + V + U = T + V + + Mθ = ( ) 3 ml ω mg L sin θ M ω = 6Mθ ml + 3g L sin θ Thus ( 4 ) 6(3 N-m) 8 ω = π rad ( ) + N (4 m) 9.8 m/s ω =.93 rad/s. 3( 9.8 m/s ) sin(4 ) =.93 rad/s. 4m Problem 9. The object consists of an 35.6 N slender bar welded to a circular disk. When the object is released from rest in position, its angular velocity in position is 4.6 rad/s. What is the weight of the disk? 7 mm 559 mm 45 y x Using conservation of energy we have T + V = T + V + = T + V ( [ ] = [ [ ] W s ] [ 35.6 N W s ] [ ) + + (4.6 rad/s) m/s.559 m 9.8 m.7 m ] [ ] m ( 35.6 N) (.8 m ) sin (45 ) W (.686) sin (45 ) Solving for W we find W = 98.7 N. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 565

7 Problem 9. The object consists of an 35.6 N slender bar welded to a 53.4 N circular. disk The object is released from rest in position. Determine the x and y components of force exerted on the object by the pin support when it is in position. 7 mm 559 mm 45 y x We first determine the moment of inertia about the fixed point and the distance from to the center of mass. ( ) ( ) ( I = 35.6 N-s 53.4 N-s 53.4 N-s (.559 ) + (.7 m) m 9.8 m 9.8 m ) (.686 m) =.98 N-s -m ( 35.6N)(.8 m) + ( 53.4 N)(.686 m) d = ( 35.6N) + ( 53.4 N) =.53 m. We now write the equations of motion and the work energy equation ( ) 89 N-s F x : x = (α d sin 45 + ω d cos 45 ), 9.8 m ( ) 89 N-s F y : y ( 89 N) = ( α d cos 45 + ω d sin 45 ), 9.8 m M : ( 89 N) d cos 45 = I α, + = ( 89 N) d sin 45 + I ω. Solving we have ω = 4.7 rad/s, α=. rad/s, x =. N, y = 5. 9 N. x =. N, y = 5.9 N Problem 9. The mass of each box is 4 kg. The radius of the pulley is mm and its moment of inertia is.3 kg-m. The surfaces are smooth. If the system is released from rest, how fast are the boxes moving when the left box has moved.5 m to the right? 3 Use conservation of energy. The angular velocity of the pulley is v/r. T + V = T + V + = (4kg)v + (4kg)v + ( (.3 v ) kg-m ). m (4 kg)(9.8 m/s )(.5 m) sin 3 Solving we have v =.39 m/s. 566 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

8 Problem 9.3 The mass of each box is 4 kg. The radius of the pulley is mm and its moment of inertia is.3 kg-m. The coefficient of kinetic friction between the boxes and the surfaces is µ k =.. If the system is released from rest, how fast are the boxes moving when the left box has moved.5 m to the right? 3 v/r. Use work energy. The angular velocity of the pulley is T + V + U = T + V + (.)(8 kg)(9.8 m/s )(.5 m) = (8kg)v + ( (.3 v ) kg-m ). m (4 kg)(9.8 m/s )(.5 m) sin 3 Solving we have v =. m/s. Problem 9.4 The 4-kg bar is released from rest in the horizontal position and falls to position. The unstretched length of the spring is.4 m and the spring constant is k = N/m. What is the magnitude of the bar s angular velocity when it is in position..6 m m 6 In position the spring is stretched a distance d =.6 m.4 m=. m, while in position the spring is stretched a distance d = (.6 m) + ( m) (.6 m)( m) cos 6.4 m=. m Using conservation of energy we have T + V = T + V + kd = [ ] (4kg)( m) ω 3 (4 kg)(9.8 m/s )(.5 m) sin 6 + kd Solving we find ω = 3.33 rad/s. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 567

9 Problem 9.5 The moments of inertia of gears that can turn freely on their pin supports are I =. kg-m and I B =.6 kg-m. The gears are at rest when a constant couple M = N-m is applied to gear B. Neglecting friction, use principle of work and energy to determine the angular velocities of the gears when gear has turned revolutions. From the principle of work and energy: U = T T, where T = since the system starts from rest. The work done is 6 mm B M 9 mm θb U = Mdθ = Mθ B. Gear B rotates in a positive direction; gear rotates in a negative direction, θ = π() = π rad. The angle traveled by the gear B is θ B = r ( ).6 θ = ( π) = 48.9 rad, r B.9 from which U = Mθ B = (48.9) = N-m. The kinetic energy is T = ( r where ω B = r B ( ) ( ) I ω + I B ωb, ) ω, from which U = ( ) ( ( ).6. + (.6)) ω = N-m,.9 from which ω =± 359,39 = 599. rad/s, and ω B = ( ).6 ω = rad/s.9 Problem 9.6 The moments of inertia of gears and B are I =. kg-m and I B =.9 kg-m. Gear is connected to a torsional spring with constant k = N-m/rad. If gear B is given an initial counterclockwise angular velocity of rad/s with the torsional spring unstretched, through what maximum counterclockwise angle does gear B rotate? B mm 4 mm The counterclockwise velocity of B and clockwise velocity of are related by.ω B =.4ω. pplying conservation of energy, I ω + I Bω B = kθ : [( ) ]. (.) () +.4 (.9)() = ()θ Solving, we obtain θ =.4 rad so θ B =.4. θ =.73 rad = c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

10 Problem 9.7 The moments of inertia of three pulleys that can turn freely on their pin supports are I =. kg-m, I B =.36 kg-m, and I C =.3 kg-m. They are stationary when a constant couple M = N-m is applied to pulley. What is the angular velocity of pulley when it has turned revolutions? mm mm M mm B C mm ω C = ω B = from which ω C = ( ). ω B,. ( ). ω,. ( ). ω.. ll pulleys rotate in a positive direction: From the principle of work and energy: U = T T, where T = since the pulleys start from a stationary position. The work done is The kinetic energy is T = = ( ) ( ) ( ) I ω + I B ωb + I C ωc ( ) ( I + ( ). I B +. ( ) ). 4 I C ω., from which T =.65ω, and U = 4π =.65ω. Solve: ω = 4π = 39. rad/s.65 θ U = Mdθ = (π)() = 4π N-m. Problem 9.8 Model the arm BC as a single rigid body. Its mass is 3 kg, and the moment of inertia about its center of mass is I = 36 kg-m. Starting from rest with its center of mass m above the ground (position ), the BC is pushed upward by the hydraulic cylinders. When it is in the position shown (position ), the arm has a counterclockwise angular velocity of.4 rad/s. How much work do the hydraulic cylinders do on the arm in moving it from position to position?.8 m.4 m B.3 m.8 m.7 m C.5 m From the principle of work and energy: U = T T, where T = since the system starts from rest. The work done is U = U cylinders mg(h h ), and the kinetic energy is T = ( ) I ω, from which U cylinders mg(h h ) = ( ) I ω In position, h = m above the ground. In position, h = = 3.35 m. The distance from to the center of mass is d =.8 +. =. m, from which I = I + md = 695 kg-m. Substitute: U cylinders = 66., from which U cylinders = 563 N-m c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 569

11 Problem 9.9 The mass of the circular disk is 5 kg and its radius is R =. m. The disk is stationary when a constant clockwise couple M = N-m is applied to it, causing the disk to roll toward the right. Consider the disk when its center has moved a distance b =.4 m. (a) How much work has the couple M done on the disk? (b) How much work has been done by the friction force exerted on the disk by the surface? (c) What is the magnitude of the velocity of the center of the disk? (See ctive Example 9..) ( b (a) U = Mθ = M R (b) ) (.4 m = ( N-m). m The friction force is a workless constraint force and does no work. U =. ) = N-m. U = N-m. M b (c) Use work energy U = mv + ( ) ( v ) mr R N-m = (5kg)v + ( ) ( v ) [5 kg][. m]. m Solving we find v =.3 m/s. Problem 9. The mass of the homogeneous cylindrical disk is m = 5 kg and its radius is R =. m. The angle β = 5. The disk is stationary when a constant clockwise couple M = N-m is applied to it. What is the velocity of the center of the disk when it has moved a distance b =.4 m? (See ctive Example 9..) R M b β The angle through which the disk rolls is θ = b/r. The work done by the couple and the disk s weight is v = M ( ) b + mgb sin β. R Equating the work to the disk s kinetic energy M ( ) b + mgb sin β = R mv + Iω and using the relation ω = v/r, we obtain v = = b =.59 m/s. ( ) M mr + g sin β [ ] sin 5 (5)(.) 57 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

12 Problem 9. The mass of the stepped disk is 8 kg and its moment of inertia is.8 kg-m. If the disk is released from rest, what is its angular velocity when the center of the disk has fallen m?. m. m The work done by the disk s weight is v = mgh = (8 kg) (9.8 m/s )(m) = 76.6 N-m. We equate the work to the final kinetic energy, 76.6 = mv + Iω = (8)v + (.8)ω. Using the relation v = (.)ω and solving for ω, we obtain ω = 7.7 rad/s. Problem 9. The -kg homogenous cylindrical disk is at rest when the force F = 5 N is applied to a cord wrapped around it, causing the disk to roll. Use the principle of work and energy to determine the angular velocity of the disk when it has turned one revolution. 3 mm F From the principle of work and energy: U = T T, where T = since the disk is at rest initially. The distance traveled in one revolution by the center of the disk is s = πr =.6π m. s the cord unwinds, the force F acts through a distance of s. The work done is s U = Fds= F(.6π) = N-m. The kinetic energy is T = ( ) ( ) Iω + mv, where I = mr, and v = Rω, from which T = 3 4 mr ω = 6.75ω. U = T, = 6.75ω, from which ω = 6.7 rad/s (clockwise). c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 57

13 Problem 9.3 The5 kg homogenous cylindrical disk is given a clockwise angular velocity of rad/s with the spring unstretched. The spring constant is k = 43.8 N/m. If the disk rolls, how far will its center move to the right? k.3 m From the principle of work and energy: U = T T, where T = since the disk comes to rest at point. The work done by the spring is ks U = ks. The kinetic energy is T = Iω + mv. By inspection v = ωr, from which T = ( ) ( m R + mr ) ω = 3 4 mr ω =.75( ) = 4.7 N- m, U = T, ( 43.8) S = 4.7, from which S =.86 =.43 m. Problem 9.4 The system is released from rest. The moment of inertia of the pulley is.4 kg-m. The slanted surface is smooth. Determine the magnitude of the velocity of the N weight when it has fallen m. 5 N 6 cm N Use conservation of energy T = V = ( ) T = 5 N-s v + ( ( ) v ).4 kg-m 9.8 m. 6m V = ( N)( m) + (5 N)( m) sin T + V = T + V v =.6 m/s. 57 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

14 Problem 9.5 The system is released from rest. The moment of inertia of the pulley is.4 kg-m. The coefficient of kinetic friction between the 5-N weight and the slanted surface is µ k =.3. Determine the magnitude of the velocity of the -N weight when it has fallen m. 5 N 6 cm N Use work energy. T = V = U = (.3)(5 N) cos (m) ( ) T = 5 N-s v + ( (.4 v ) kg-m ) 9.8 m. 6m V = ( N)( m) + (5 N)( m) sin T + V + U = T + V v =.48 m/s. Problem 9.6 Each of the cart s four wheels weighs N, has a radius of 5 cm, and has moment of inertia I =. kg-m. The cart (not including its wheels) weighs 8 N. The cart is stationary when the constant horizontal force F = 4 N is applied. How fast is the cart going when it has moved m to the right? F Use work energy U = Fd = ( 4 N)( m) = 8 N-m T = T = ( 8 N-s 9.8 m ) v + 4 [ ( ) v + ( ) ] N-s v (. kg-m ) 9.8 m.5 m T + U = T v =.9 m/s. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 573

15 Problem 9.7 The total moment of inertia of car s two rear wheels and axle is I R, and the total moment of inertia of the two front wheels is I F. The radius of the tires is R, and the total mass of the car, including the wheels, is m. The car is moving at velocity v when the driver applies the brakes. If the car s brakes exert a constant retarding couple M on each wheel and the tires do not slip, determine the car s velocity as a function of the distance s from the point where the brakes are applied. s When the car rolls a distance s, the wheels roll through an angle s/r so the work done by the brakes is V = 4M(s/R). Let m F be the total mass of the two front wheels, m R the mass of the rear wheels and axle, and m c the remainder of the car s mass. When the car is moving at velocity v its total kinetic energy is T = m cv + m Rv + I R(v/R) + I F(v/R) = [m + (I R + I F )/R ]v. From the principle of work and energy, V = T T : 4M(s/R) = [m + (I R + I F )/R ]v [m + (I R + I F )/R ]v Solving for v, we get v = v 8Ms/[Rm + (I R + I F )/R]. Problem 9.8 The total moment of inertia of the car s two rear wheels and axle is.4 kg-m. The total moment of inertia of the two front wheels is. kg-m. The radius of the tires is.3 m. The mass of the car, including the wheels, is 48 kg. The car is moving at km/h. If the car s brakes exert a constant retarding couple of 65 N-m on each wheel and the tires do not slip, what distance is required for the car to come to a stop? (See Example 9..) From the solution of Problem 9.7, the car s velocity when it has moved a distance s is v = v 8Ms/[Rm + (I R + I F )/R]. Setting v = and solving for s we obtain s = [Rm + (I R + I F )/R]v /(8M). The car s initial velocity is v =,/36 = 7.8 m/s So, s = [(.3)(48) + (.4 +.)/.3](7.8) /[8(65)] = 66. m. 574 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

16 Problem 9.9 The radius of the pulley is R = mm and its moment of inertia is I =. kg-m. The mass m = 5 kg. The spring constant is k = 35 N/m. The system is released from rest with the spring unstretched. Determine how fast the mass is moving when it has fallen.5 m. R k m T =, V =, T = (5kg)v + (. kg-m ) ( v ). m V = (5 kg)(9.8 m/s )(.5 m) + (35 N/m)(.5 m) T + V = T + V v =. m/s Problem 9.3 The masses of the bar and disk are 4 kg and 9 kg, respectively. The system is released from rest with the bar horizontal. Determine the angular velocity of the bar when it is vertical if the bar and disk are welded together at. O. m.3 m The work done by the weights of the bar and disk as they fall is U = m bar g(.6 m) + m disk g(. m) = (4)(9.8)(.6) + (9)(9.8)(.) = 88.4 N-m. The bar s moment of inertia about the pinned end is I = 3 m barl = 3 (4)(.) = 6.7 kg-m, So the bar s final kinetic energy is T bar = I ω so the disk s final kinetic energy is T disk = m disk(lω) + I ω = [(9)(.) +.45]ω = 6.68ω. Equating the work to the final kinetic energy, U = T bar + T disk 88.4 = 3.36ω ω, we obtain ω = 4.33 rad/s. = 3.36ω. The moment of inertia of the disk about is I = m diskr = (9)(.3) =.45 kg-m, c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 575

17 Problem 9.3 The masses of the bar and disk are 4 kg and 9 kg, respectively. The system is released from rest with the bar horizontal. Determine the angular velocity of the bar when it is vertical if the bar and disk are connected by a smooth pin at. See the solution of Problem 9.3. In this case the disk does not rotate, so its final kinetic energy is T disk = m disk(lω) = (9)(.) ω = 6.48ω. Equating the work to the final kinetic energy, U = T bar + T disk : 88.4 = 3.36ω ω, we obtain ω = 4.38 rad/s. Problem 9.3 The 45-kg crate is pulled up the inclined surface by the winch. The coefficient of kinetic friction between the crate and the surface is µ k =.4. The moment of inertia of the drum on which the cable is being wound is I = 4 kg-m. The crate starts from rest, and the motor exerts a constant couple M = 5 N-m on the drum. Use the principle of work and energy to determine the magnitude of the velocity of the crate when it has moved m..5 m M The normal force is M N = mg cos. s the crate moves m, the drum rotates through an angle θ = ( m)/r, so the work done is mg R U = M ( ) mg sin () µ k (mg cos )(). R The final kinetic energy is µ k N N T = mv + Iω. Equating the work to the final kinetic energy and using the relation v = Rω, we solve for v, obtaining v =.384 m/s. 576 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

18 Problem 9.33 The.6 m slender bars each weigh 7.8 N, and the rectangular plate weighs 89 N.. If the system is released from rest in the position shown, what is the velocity of the plate when the bars are vertical?, 45 The work done by the weights: U = W bar h + W plate h, where h = L ( cos 45 ) =. 89 m is the change in height, from which U = 9.6 N-m. The kinetic energy is ω v ω T = ( )( Wplate g ) [ v + ( Wbar L )] ω = 5.v, 3g where ω = v has been used. L Substitute into U = T and solve: v =.93 m/s. Problem 9.34 The mass of the -m slender bar is 8 kg. torsional spring exerts a counterclockwise couple kθ on the bar, where k = 4 N-m/rad and θ is in radians. The bar is released from rest with θ = 5. What is the magnitude of the bar s angular velocity when θ = 6? k u We will use conservation of energy T = V = (8 kg)(9.8 m/s )( m) cos 5 + T = [ ] (8kg)( m) ω 3 V = (8 kg)(9.8 m/s )( m) cos 6 + ( 4 N-m rad ( 4 N-m rad )( ) 5 8 π rad )( ) 6 8 π rad T + V = T + V ω =.79 rad/s. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 577

19 Problem 9.35 The mass of the suspended object is 8 kg. The mass of the pulley is 5 kg, and its moment of inertia is.36 kg-m. If the force T = 7N is applied to the stationary system, what is the magnitude of the velocity of when it has risen. m? T mm When the mass rises. m, the end of the rope rises.4 m. T =, V =, T = (3 kg)v + ( (.36 v ) kg-m ). m V = (3 kg)(9.8 m/s )(. m), U = (7 N)(.4 m) T + V + U = T + V v =.568 m/s Problem 9.36 The mass of the left pulley is 7 kg, and its moment of inertia is.33 kg-m. The mass of the right pulley is 3 kg, and its moment of inertia is.54 kg-m. If the system is released from rest, how fast is the 8-kg mass moving when it has fallen. m? B. m.3 m 8 kg 9 kg When mass C falls a distance x, the center of pulley rises x. The potential energy is ω B v = m c gx + (m + m D )g x. RB The angular velocity of pulley B is w B = v/r B, and the angular velocity of pulley is w = v/r. The velocity of the center of pulley is w R = v/. The total kinetic energy is T = m cv + I B(v/R B ) + (m + m D )(v/) + I (v/r ). m ω v pplying conservation of energy to the initial and final positions, O = m c g(.) + (m + m D )g (.) + m cv + I B(v/R B ) R m C x v + (m + m D )(v/) + I (v/r ). m D Solving for v, we obtain v =.899 m/s. 578 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

20 Problem 9.37 The 8-kg ladder is released from rest with θ =. The wall and floor are smooth. Modeling the ladder as a slender bar, use conservation of energy to determine the angular velocity of the bar when θ = 4. θ 4 m Choose the datum at floor level. The potential energy at the initial position is V = mg ( ) L cos. t the final position, V = mg ( ) L cos 4. The instantaneous center of rotation has the coordinates (L sin θ, L cos θ), where θ = 4 at the final position. The distance of the center of rotation from the bar center of mass is L. The angular velocity ( ) about this center is ω = v, where v is the velocity of the center L of mass of the ladder. The kinetic energy of the ladder is T = ( ) ( mv + where v = )( ml ) ω = ( ) L ω has been used. From the conservation of energy, V = V + T, from which mg = mg ( ) L cos 4 + Solve: ω =.69 rad/s. ( ml 6 ( ml 6 ( ) L cos ) ω. ) ω, c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 579

21 Problem 9.38 The 8-kg slender bar is released from rest with θ = 6. The horizontal surface is smooth. What is the bar s angular velocity when θ = 3. The bar s potential energy is m V = mg l sin θ. θ No horizontal force acts on the bar, so its center of mass will fall straight down. The bar s kinetic energy is T = mv G + Iw, where I = ml. pplying conservation of energy to the initial and final states, we obtain mg l sin 6 = mg l sin 3 + mv G + Iw. () y ω G To complete the solution, we must determine v G in terms of ω. v G We write the velocity of pt. in terms of the velocity of pt. G. θ v = v G + ω r /G : i j k v i = v G j + ω l cos θ l. sin θ The j component of this equation is = v G + ω l cos θ. Setting θ = 3 in this equation and solving it together with Eq. (), we obtain ω =.57 rad/s. Problem 9.39 The mass and length of the bar are m = 4 kg and l =. m. The spring constant is k = 8 N/m. If the bar is released from rest in the position θ =, what is its angular velocity when it has fallen to θ =? k If the spring is unstretched when θ =, the stretch of the spring is S = l( cos θ). The total potential energy is θ l v = mg(l/) cos θ + kl ( cos θ). From the solution of Problem 9.37, the bar s kinetic energy is T = 6 ml ω. We apply conservation of energy. T + V = T + V : + mg(l/) cos + kl ( cos ) = 6 ml ω + mg(l/) cos + kl ( cos ). Solving for ω, we obtain ω =.84 rad/s. 58 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

22 Problem 9.4 The 4-kg slender bar is pinned to a -kg slider at and to a 4-kg homogenous cylindrical disk at B. Neglect the friction force on the slider and assume that the disk rolls. If the system is released from rest with θ = 6, what is the bar s angular velocity when θ =? (See Example 9.3.) m θ B mm Choose the datum at θ =. The instantaneous center of the bar has the coordinates (L cos θ,lsin θ) (see figure), and the distance from the center of mass of the bar is L, from which the angular velocity about the bar s instantaneous center is v = ( ) L ω, where v is the velocity of the center of mass. The velocity of the slider is θ V ω v = ωl cos θ, and the velocity of the disk is v B = ωl sin θ The potential energy of the system is V = m gl sin θ + mg ( ) L sin θ. t the datum, V =. The kinetic energy is T = ( ) ( ) ( m v + mv + ( )( mb R ) ( vb ) +, R where at the datum )( ml ) ( ) ω + m B vb v = ωl cos = ωl, v B = ωl sin =, v = ω ( ) L. From the conservation of energy: V = T. Solve: ω = 4.5 rad/s. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 58

23 Problem 9.4* The sleeve P slides on the smooth horizontal bar. The mass of each bar is 4 kg and the mass of the sleeve P is kg. If the system is released from rest with θ = 6, what is the magnitude of the velocity of the sleeve P when θ = 4?. m Q. m O θ P We have the following kinematics Q v Q = ω OQ k r Q/O = ω OQ k (. m)(cos θi + sin θj) = ω OQ (. m)( sin θi + cos θj). m. m G v P = v Q + ω PQ r P/Q θ = ω OQ (. m)( sin θi + cos θj) + ω PQ k (. m)(cos θi sin θj) O P = [(ω PQ ω OQ )(. m) sin θ]i + [(ω PQ + ω OQ )(. m) cos θ]j Point P is constrained to horizontal motion. We conclude that ω PQ = ω OQ ω, v P = ω(. m) sin θ We also need the velocity of point G v G = v Q + ω PQ r G/Q = ω(. m)( sin θi + cos θj) + ωk (.6 m)(cos θi sin θj) = [(.8 m)ω sin θ]i [(.6 m)ω cos θ]j Now use the energy methods T =, V = (4 kg)(9.8 m/s )(.6 m) sin 6 ; T = ( ) [4 kg][. m] ω + ( ) [4 kg][. m] ω 3 + (4kg)([(.8 m)ω sin 4 ] + [(.6 m)ω cos 4 ] ) + (kg)(ω[. m] sin 4 ) V = (4 kg)(9.8 m/s )(.6 m) sin 4 ; Solving T + V = T + V ω =.5 rad/s v P =.94 m/s 58 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

24 Problem 9.4* The system is in equilibrium in the position shown. The mass of the slender bar BC is 6 kg, the mass of the slender bar BD is 3 kg, and the mass of the slider at C is kg. The spring constant is k = N/m. If a constant -N downward force is applied at, what is the angular velocity of the bar BC when it has rotated from its initial position? Choose a coordinate system with the origin at D and the x axis parallel to DC. The equilibrium conditions for the bars: for bar BD, m D m B m 5 5 C k Fx = B x + D x =, Fy = B y m BD g + D y =. F ( MD = B x sin 5 B y + M ) BDg cos 5 =. For the bar BC, Fx = B x F =, F y = F + C m BC g + B y =. D x B x W BD B y B y B x W BC C F D y MC = (F B y + m BC g) cos 5 B x sin 5 =. t the initial position F =. The solution: B x = 3.87 N, D x = 3.87 N, B y =.7 N, D y = 5.5 N, F = 3.87 N, C= N. [Note: Only the value F = 3.87 N is required for the purposes of this problem.] The initial stretch of the spring is S = F k = 3.87 =.54 m. The distance D to C is cos θ, so that the final stretch of the spring is S = S + ( cos 3 cos 5 ) =.6 m. From the principle of work and energy: U = T T, where T = since the system starts from rest. The work done is U = U force + U BC + U BD + U spring. The height of the point is sin θ, so that the change in height is h = (sin 5 sin 3 ), and the work done by the applied force is h U force = F dh = ( sin 5 sin 3 ) = 53. N-m. The height of the center of mass of bar BD is sin θ, so that the work done by the weight of bar BD is h U BD = m BD gdh= m BDg (sin 3 sin 5 ) = 3.9 N-m. The height of the center of mass of bar BC is sin θ, so that the work done by the weight of bar BC is h U BC = m BC gdh= m BC g(sin 3 sin 5 ) The work done by the spring is U spring = S S ks ds = 5.66 N-m. = k (S S ) = 33.7 N-m. Collecting terms, the total work: U = 39.7 N-m. The bars form an isosceles triangle, so that the changes in angle are equal; by differentiating the changes, it follows that the angular velocities are equal. The distance D to C is x DC = cos θ, from which v C = sin θω, since D is a stationary. The kinetic energy is T = ( ) I BD ω + + ( ) m C vc = 5ω, where I BD = m BD 3 ( ), I BC = m BC ( ), v BC = ()ω, ( ) I BC ω + ( ) m BC vbc v C = sin θω. Substitute into U = T and solve: ω =.8 rad/s c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 583

25 Problem 9.43* The masses of bars B and BC are 5 kg and 3 kg, respectively. If the system is released from rest in the position shown, what are the angular velocities of the bars at the instant before the joint B hits the smooth floor? m B C m m is The work done by the weights of the bars as they fall y v = m B g(.5 m) + m BC g(.5 m) = (5 + 3)(9.8)(.5) ω B = 39.4 N-m. Consider the two bars just before B hits. B G ω BC C x The coordinates of pt B are ( 3, )m. The velocity of pt B is v B = v + w B r B/ i j k = + ω B 3 = ω B i 3ω B j. In terms of pt. C, v B = v C + ω BC r B/C i j k = v C i + ω BC = v C i ω BC j. Equating the two expressions for v B, we obtain v C = ω B i, ω BC = 3 ω B. () The velocity of pt G is v G = v C + ω BC r G/C i j k = ω B i + ω BC = ω B i ωbc j = ω B i 3ωB j. () We equate the work done to the final kinetic energy of the bars: U = [ ] 3 m B() ωb + m BC v G + [ m BC( ] ) ωbc. (3) Substituting Eqs. () and () into this equation and solving for ω B, we obtain ω B =.49 rad/s. Then, from Eq. (), ω BC = 3.5 rad/s. 584 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

26 Problem 9.44* Bar B weighs. N. Each of the sleeves and B weighs 8.9 N. The system is released from rest in the position shown. What is the magnitude of the angular velocity of the bar when sleeve B has moved 76. mm to the right? y 8.6 mm B x Find the geometry in position ( ( ) + ( ) = ( + x) x ).9.47 m.9m.38 m +. x =.63 m Now do the kinematics in position v B = v + ω r B/ (..9 v B i = v i.5.5 ) j 35 mm.6 mm Position 8.6 mm.6 mm 35 mm + ωk (.445i.4j) m = (.46v + [.4 in]ω)i + (.94v + [.445 m]ω)j Equating components we find v = (. 485 m)ω, v B = (.34 m)ω 9 4 Position Now find the velocity of the center of mass G v G = v B + ω r G/B 9x/4 B = (.34 m)ωi + ωk (.i +.7j) m x 38 m = ω(.69i.j) m Now we can do work energy T =, V = (.N ) (.4m) + ( 8.9N ) (.9m), V = (.N ) (.7m) + ( 8.9N ) (.4m) ( T = [ ].N 9.8 m/s ) (.9m) + (.47m) ] ω [ + ( ) 8.9N 9.8 m/s (.485 m) ω + ( ).N 9.8 m/s (.69 m) + (. m) [ ] ω + ( ) 8.9N (.34 m) 9.8 m/s ω Thus T + V = T + V ω =.34 rad/s c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 585

27 Problem 9.45* Each bar has a mass of 8 kg and a length of l m. The spring constant is k = N/m, and the spring is unstretched when θ =. If the system is released from rest with the bars vertical, what is the magnitude of the angular velocity of the bars when θ = 3? k θ θ The stretch of the spring is l( cos θ), so the poten- tial energy is y v = k[l( cos θ)] + mg l 3l cos θ + mg cos θ = kl ( cos θ) + mgl cos θ. ω G C θ l cos θ The velocity of pt B is i j k v B = v + ω B r B/ = + ω l sin θ lcos θ = ωl cos θi ωl sin θj. B ω θ l cos θ 3l cos θ x The velocity of pt. G is v G = v B + ω BC r G/B i j k = ωl cos θi ωl sin θj + ω l sin θ l cos θ = ω l 3l cos θi ω sin θj. From this expression, v G = ( + 8 sin θ) 4 l ω. pplying conservation of energy to the initial and final positions, mgl = kl ( cos 3 ) + mgl cos 3 + ( ) 3 ml ω + ( ) ml ω + m( + 8 sin 3 ) 4 l ω. Solving for ω, we obtain ω =.93 rad/s. 586 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

28 Problem 9.46* The system starts from rest with the crank B vertical. constant couple M exerted on the crank causes it to rotate in the clockwise direction, compressing the gas in the cylinder. Let s be the displacement (in meters) of the piston to the right relative to its initial position. The net force toward the left exerted on the piston by atmospheric pressure and the gas in the cylinder is 35/( s) N. The moment of inertia of the crank about is.3 kg-m. The mass of the connecting rod BC is.36 kg, and the center of mass of the rod is at its midpoint. The connecting rod s moment of inertia about its center of mass is.4 kg-m. The mass of the piston is 4.6 kg. If the clockwise angular velocity of the crank B is rad/s when it has rotated 9 from its initial position, what is M? (Neglect the work done by the weights of the crank and connecting rod). M 5 mm B 5 mm C s the crank rotates through an angle θ the work done by the couple is Mθ. s the piston moves to the right a distance s, the work done on the piston by the gas is y θ ω B ω BC s 35 ds s. rb/ r C/B C x Letting s = z, this is. s. 35 dz z = 35ln( s). The total work is U = Mθ + 35ln( s). From the given dimensions of the crank and connecting rod, the vector components are r B/x =.5 sin θ, () r B/y =.5 cos θ (); r C/Bx = (.5) (.5 cos θ) (3); r C/By =.5 cos θ (4). The distance s that the piston moves to the right is s = r B/x + r C/Bx (.5) (.5) (5). The velocity of B is i j k v B = v + ω r B/x r B/y = ωr B/yi ωr B/x j. The velocity of the center of mass G (the midpoint) of the connecting rod BC is v G = v B + ω BC r C/B = ωr B/y i ωr B/x j + i j k ω BC r C/By, or r C/Bx v C = ( ωr B/y ω BCr C/By ) i ( ωrb/x ω BCr C/Bx ) j (8). Let I be the moment of inertia of the crank about, I BC the moment of inertia of the connecting rod about its center of mass, m BC the mass of the connecting rod, and m p the mass of the piston. The principle of work and energy is: U = T T : Mθ + 35ln( s) = I ω + I BCω BC + m BC v G + m pv C (9). Solving this equation for M and using Equations () (8) with ω = rad/s and θ = π/ rad, we obtain M = 8. N-m. The velocity of C is v C = v B + ω BC r C/B : i j k v C i = ωr B/y i ωr B/x j + ω BC r c/bx r c/by. Equating i and j components, we can solve for v c and ω BC in terms of ω: v C = [r B/y r C/By (r B/x /r C/Bx )]ω (6); ω BC = (r B/x /r C/Bx )ω (7). c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 587

29 Problem 9.47* In Problem 9.46, if the system starts from rest with the crank B vertical and the couple M = 4 N-m, what is the clockwise angular velocity of B when it has rotated 45 from its initial position? In the solution to Problem 9.46, we substitute Equations () (8) into Equation (9), set M = 4 N-m and θ = π/4 rad and solve for ω obtaining ω = 49.6 rad/s. Problem 9.48 The moment of inertia of the disk about O is kg-m.tt =, the stationary disk is subjected to a constant 5 N-m torque. 5 N-m (a) Determine the angular impulse exerted on the disk from t = tot = 5s. (b) What is the disk s angular velocity at t = 5s? O (a) 5s ngular Impulse = Mdt = (5 N-m)(5 s) = 5 N-m-s CCW (b) 5 N-m-s = ( kg-m )ω ω =.4 rad/s counterclockwise Problem 9.49 The moment of inertia of the jet engine s rotating assembly is 4 kg-m. The assembly starts from rest. t t =, the engine s turbine exerts a couple on it that is given as a function of time by M = 65 5t N-m. (a) (b) What is the magnitude of the angular impulse exerted on the assembly from t = tot = s? What is the magnitude of the angular velocity of the assembly (in rpm) at t = s? (a) s ngular Impulse = Mdt = (65 5t) N-m dt = 5 kn-m-s (b) 5 kn-m-s = (4 kg-m )ω ω = 63 rad/s (5 rpm) 588 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

30 Problem 9.5 n astronaut fires a thruster of his maneuvering unit, exerting a force T = ( + t) N, where t is in seconds. The combined mass of the astronaut and his equipment is kg, and the moment of inertia about their center of mass is 45 kg-m. Modeling the astronaut and his equipment as a rigid body, use the principle of angular impulse and momentum to determine how long it takes for his angular velocity to reach. rad/s. T 3 mm From the principle of impulse and angular momentum, t t Mdt= Iω Iω, where ω =, since the astronaut is initially stationary. The normal distance from the thrust line to the center of mass is R =.3 m, from which t ( + t)(r) dt = Iω. ( ).6 t + t = 45(.). Rearrange: t + bt + c =, where b =, c = 5. Solve: t = b ± b c = 3or 5. Since the negative solution has no meaning here, t = 3s c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 589

31 Problem 9.5 The combined mass of the astronaut and his equipment is kg, and the moment of inertia about their center of mass is 45 kg-m. The maneuvering unit exerts an impulsive force T of.-s duration, giving him a counterclockwise angular velocity of rpm. (a) (b) What is the average magnitude of the impulsive force? What is the magnitude of the resulting change in the velocity of the astronaut s center of mass? T 3 mm (a) From the principle of moment impulse and angular momentum, t t Mdt= Iω Iω, where ω = since the astronaut is initially stationary. The angular velocity is ω = π =.47 rad/s 6. from which TRdt = Iω, from which T(.3)(.) = 45(ω ), T = 45ω.6 = N (b) From the principle of impulse and linear momentum t t Fdt= m(v v ) where v = since the astronaut is initially stationary.. Tdt= mv, from which v = T(.) m =.9 m/s 59 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

32 Problem 9.5 flywheel attached to an electric motor is initially at rest. t t =, the motor exerts a couple M = e.t N-m on the flywheel. The moment of inertia of the flywheel is kg-m. (a) What is the flywheel s angular velocity at t = s? (b) What maximum angular velocity will the flywheel attain? (a) From the principle of moment impulse and angular momentum, t t Mdt= Iω Iω, where ω =, since the motor starts from rest. e.t dt = [ e.t ]. = [ e ], = 64. N-m-s. From which ω = 64.4 = 6 rad/s. (b) n inspection of the angular impulse function shows that the angular velocity of the flywheel is an increasing monotone function of the time, so that the greatest value occurs as t. [ ω max = lim e.t ] rad/s t ()(.) Problem 9.53 main landing gear wheel of a Boeing 777 has a radius of.6 m and its moment of inertia is 4 kg-m. fter the plane lands at 75 m/s, the skid marks of the wheel s tire is measured and determined to be 8 m in length. Determine the average friction force exerted on the wheel by the runway. ssume that the airplane s velocity is constant during the time the tire skids (slips) on the runway. The tire skids for t = 8 m =.4 s. 75 m/s When the skidding stops the tire is turning at the rate ω = rad/s Frt = Iω F = Iω rt = (4 kg-m )( rad/s) (.6 m)(.4 s) 75 m/s.6 m = F = 9.5 kn c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 59

33 Problem 9.54 The force a club exerts on a.45-kg golf ball is shown. The ball is 4 mm in diameter and can be modeled as a homogeneous sphere. The club is in contact with the ball for.6 s, and the magnitude of the velocity of the ball s center of mass after the ball is hit is 36 m/s. What is the magnitude of the ball s angular velocity after it is hit?.5 mm F Linear momentum: Ft = mv F = mv t ngular momentum Ftd = Iω ω = Ftd I = mvd I = (.45 kg)(36 m/s)(.5 m) /5 (.45 kg)(. m) Solving ω = 5 rad/s Problem 9.55 Disk initially has a counterclockwise angular velocity ω = 5 rad/s. Disks B and C are initially stationary. t t =, disk is moved into contact with disk B. Determine the angular velocities of the three disks when they have stopped slipping relative to each other. The masses of the disks are m = 4kg,m B = 6 kg, and m C = 9 kg. (See ctive Example 9.4.) v. m.4 m B.3 m C The FBDs Given: F F ω = 5 rad/s m = 4kg, r =. m, I = / m r =.8 kg-m B C m B = 6 kg, r B =.4 m, I B = / m B r B =.8 kg-m m C = 9kg, r C =.3 m, I C = / m C r C =.45 kg-m Impulse Momentum equations: F tr = I ω I ω, F tr B F tr B = I B ω B, F tr C = I C ω C Constraints when it no longer slips: ω r = ω B r B = ω C r C We cannot solve for the slipping time, however, treating F t and F t as two unknowns we have ω = 6.9 rad/s counterclockwise ω B = 3.45 rad/s clockwise ω C = 4.6 rad/s counterclockwise F F 59 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

34 Problem 9.56 In Example 9.5, suppose that in a second test at a higher velocity the angular velocity of the pole following the impact is ω =.8 rad/s, the horizontal velocity of its center of mass is v = 7.3 m/s, and the duration of the impact is t =.9 s. Determine the magnitude of the average force the car exerts on the pole in shearing off the supporting bolts. Do so by applying the principle of angular impulse and momentum in the form given by Eq. (9.3). Using the data from Example 9.5 we write the linear and angular impulse momentum equations for the pole. F is the force of the car and S is the shear force in the bolts (F S) t = mv ( ) L F t h S t L = ml ω Putting in the numbers we have (F S)(.9 s) = (7 kg)(7.3 m/s) F(.9 s)(.5 m) S(.9 s)(43 m) = (7 kg)(6 m) ω Solving we find F = 33 N, S = 46 N. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 593

35 Problem 9.57 The force exerted on the cue ball by the cue is horizontal. Determine the value of h for which the ball rolls without slipping. (ssume that the frictional force exerted on the ball by the table is negligible.) R h momentum, From the principle of moment impulse and angular F t t (h R)F dt = I(ω ω ), h R where ω = since the ball is initially stationary. From the principle of impulse and linear momentum t t Fdt= m(v v ) where v = since the ball is initially stationary. Since the ball rolls, v = Rω, from which the two equations: (h R)F(t t ) = Iω, Substitute: (h R)mRω = mr ω. 5 Solve: h = ( 7 5 ) R F(t t ) = mrω. The ball is a homogenous sphere, from which I = 5 mr. 594 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

36 Problem 9.58 In Example 9.6, we neglected the moments of inertia of the two masses m about the axes through their centers of mass in calculating the total angular momentum of the person, platform, and masses. Suppose that the moment of inertia of each mass about the vertical axis through its center of mass is I M =. kg-m. If the person s angular velocity with her arms extended to r =.6 misω = revolution per second, what is her angular velocity ω when she pulls the masses inward to r =. m? Compare your result to the answer obtained in Example 9.6. r v r Using the numbers from Example 9.6, we conserve angular momentum H O = (I P + mr + I M )ω ( = (.4 kg-m + [4 kg][.6 m] + [. kg-m ]) rev ) s H O = (I P + mr + I M )ω = (.4 kg-m + [4 kg][. m] + [. kg-m ])ω r r H O = H O ω = 4.55 rev s. If we include the moments of inertia of the weights then ω = 4.55 rev s. Without the moments of inertia (Example 9.6) we found ω = 4.56 rev s. v Problem 9.59 Two gravity research satellites (m = 5 kg, I = 35 kg-m ; m B = 5 kg, I B = 6 kg-m ) are tethered by a cable. The satellites and cable rotate with angular velocity ω =.5 rpm. Ground controllers order satellite to slowly unreel 6 m of additional cable. What is the angular velocity afterward? ω B mass is The initial distance from to the common center of m x = ()(5) + ()(5) = m. The final distance from to the common center of mass is x = ()(5) + (8)(5) = 3m. The total angular momentum about the center of mass is conserved. x m (x ω ) + I ω + ( x )m B [( x )ω ] + I B ω = xm (x ω ) + I ω + (8 x)m B [(8 x)ω] + I B ω; or ()(5)()ω + 35ω + ()(5)()ω + 6ω = (3)(5)(3)ω+ 35ω + (5)(5)(5)ω+ 6ω. we obtain ω =.459ω =.5 rpm. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 595

37 Problem 9.6 The -kg bar rotates in the horizontal plane about the smooth pin. The 6-kg collar slides on the smooth bar. ssume that the moment of inertia of the collar about its center of mass is negligible; that is, treat the collar as a particle. t the instant shown, the angular velocity of the bar is ω = 6 rpm and the distance from the pin to the collar is r =.8 m. Determine the bar s angular velocity when r =.4 m. v k ngular momentum is conserved [ ] 3 (kg)(3 m) + (6 kg)(.8 m) (6 rpm) [ ] = 3 (kg)(3 m) + (6 kg)(.4 m) ω r 3 m Solving we find ω = 37.6 rpm Problem 9.6 The -kg bar rotates in the horizontal plane about the smooth pin. The 6-kg collar slides on the smooth bar. The moment of inertia of the collar about its center of mass is. kg-m. t the instant shown, the angular velocity of the bar is ω = 6 rpm and the distance from the pin to the collar is r =.8 m. Determine the bar s angular velocity when r =.4 m and compare your answer to that of Problem 9.6. ngular momentum is conserved [ ] 3 (kg)(3 m) + (6 kg)(.8 m) + (. kg-m ) (6 rpm) [ ] = 3 (kg)(3 m) + (6 kg)(.4 m) + (. kg-m ) ω Solving we find ω = 37.7 rpm Problem 9.6* The -kg bar rotates in the horizontal plane about the smooth pin. The 6-kg collar slides on the smooth bar. The moment of inertia of the collar about its center of mass is. kg-m. The spring is unstretched when r =, and the spring constant is k = N-m. t the instant shown, the angular velocity of the bar is ω = rad/s, the distance from the pin to the collar is r =.8 m, and the radial velocity of the collar is zero. Determine the radial velocity of the collar when r =.4 m. ngular momentum is conserved [ ] 3 (kg)(3 m) + (6 kg)(.8 m) + (. kg-m ) ( rad/s) [ ] = 3 (kg)(3 m) + (6 kg)(.4 m) + (. kg-m ) ω Thus ω =.58 rad/s Energy is conserved T = [ ] 3 (kg)(3 m) + (6 kg)(.8 m) + (. kg-m ) ( rad/s) V = ( N/m)(.8 m) T = [ ] 3 (kg)(3 m) + (6 kg)(.4 m) + (. kg-m ) ω + (6kg)v r V = ( N/m)(.4 m) Solving we find that v r =.46 m/s. 596 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

38 Problem 9.63 The circular bar is welded to the vertical shafts, which can rotate freely in bearings at and B. Let I be the moment of inertia of the circular bar and shafts about the vertical axis. The circular bar has an initial angular velocity ω, and the mass m is released in the position shown with no velocity relative to the bar. Determine the angular velocity of the circular bar as a function of the angle β between the vertical and the position of the mass. Neglect the moment of inertia of the mass about its center of mass; that is, treat the mass as a particle. β R m ngular momentum about the vertical axis is conserved: B ω Iω + (R sin β )m(r sin β )ω = Iω+ (R sin β)m(r sin β)ω. Solving for ω,ω = ( ) I + mr sin β I + mr sin ω. β Problem 9.64 The -N bar is released from rest in the 45 position shown. It falls and the end of the bar strikes the horizontal surface at P. The coefficient of restitution of the impact is e =.6. When the bar rebounds, through what angle relative to the horizontal will it rotate? 3 m 45 We solve the problem in three phases. We start with a work energy analysis to find out how fast the bar is rotating just before the collision P T + V = T + V, [ ( ) ] + ( N)(.5 m) sin 45 = N-s (3m) ω m +, ω =.63 rad/s. Next we do an impact analysis to take it through the collision eω L = ω 3 L.6(.63 rad/s)(3 m) = ω 3 (3m) ω 3 =.58 rad/s. Finally we do another work energy analysis to find the rebound angle. T 3 + V 3 = T 4 + V 4, [ ( ) ] N (3 m) ω3 + = + ( )(.5 ) sin θ, m/s N m θ = 4.7. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 597

39 Problem 9.65 The -N bar is released from rest in the 45 position shown. It falls and the end of the bar strikes the horizontal surface at P. The bar rebounds to a position relative to the horizontal. If the duration of the impact is. s, what is the magnitude of the average vertical force the horizontal surface exerted on the bar at P? 3 m 45 P We solve the problem in three phases. We start with a work energy analysis to find out how fast the bar is rotating just before the collision T + V = T + V, [ ( ) ] + ( N)(.5 m) sin 45 = N-s (3m) ω m +, ω =.63 rad/s. Next we do another work energy analysis to find out how fast the bar is rotating just after the collision. T 3 + V 3 = T 4 + V 4, [ ( ) ] N-s (3m) ω3 3 + = + ( N)(.5 m) sin, 9.8 m ω 3 =.3 rad/s. Now we can use the angular impulse momentum equation about the pivot point to find the force. Iω W t L + F tl = Iω 3, ( ) N-s (3m) (.63 rad/s) ( N)(. s)(.5 m) m ( ) + F(. s)(3 m) = N-s 3 (3m) ( m rad/s) Solving we find F= 46.6 N. 598 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

40 Problem 9.66 The 4-kg bar is released from rest in the horizontal position above the fixed projection at. The distance b =.35 m. The impact of the bar with the projection is plastic; that is, the coefficient of restitution of the impact is e =. What is the bar s angular velocity immediately after the impact? m. m b First we do work energy to find the velocity of the bar just before impact. T + V = T + V + mgh = mv + v = gh = (9.8 m/s )(. m) =.98 m/s ngular momentum is conserved about the impact point. fter the impact, the bar pivots about the fixed point. ( L ) mv b = m + b ω 3 ω 3 = bv (.35 m)(.98 m/s) L = + b ( m) + (.35 m) ω 3 = 3.37 rad/s. Problem 9.67 The 4-kg bar is released from rest in the horizontal position above the fixed projection at. The coefficient of restitution of the impact is e =.6. What value of the distance b would cause the velocity of the bar s center of mass to be zero immediately after the impact? What is the bar s angular velocity immediately after the impact? m. m b First we do work energy to find the velocity of the bar just before impact. T + V = T + V + mgh = mv + v = gh = (9.8 m/s )(. m) =.98 m/s ngular momentum is conserved about point. The coefficient of restitution is used to relate the velocities of the impact point before and after the collision. mv b = mv 3 b + ml ω 3,ev = ω 3 b v 3,v 3 =, Solving we have e.6 b = L = (m) =.4 m, ω 3 = v b (.98 m/s)(.4 m) L = ( m) = 5.3 rad/s. b =.4 m, ω 3 = 5.3 rad/s. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 599

41 Problem 9.68 The mass of the ship is 544, kg, and the moment of inertia of the vessel about its center of mass is 4 8 kg-m. Wind causes the ship to drift sideways at. m/s and strike the stationary piling at P. The coefficient of restitution of the impact is e =.. What is the ship s angular velocity after the impact? 6 m 45 m P ngular momentum about P is conserved 45 m 45 m (45)(mv) = (45)(mv ) + Iω. () The coefficient of restitution is e = v P v, () ω = v Before impact P ω v fter impact P v P where v P is the vertical component of the velocity of P after the impact. The velocities v P and v are related by v = v P + (45)ω. (3) Solving Eqs. () (3), we obtain ω =.96 rad/s, v =.68 m/s, and v P =. m/s. Problem 9.69 In Problem 9.68, if the duration of the ship s impact with the piling is s, what is the magnitude of the average force exerted on the ship by the impact? See the solution of Problem Let F p be the average force exerted on the ship by the piling. We apply linear impulse and momentum. F p t = mv mv: F p () = (544,)(.68.). Solving, we obtain F p = 74 N. 6 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

42 Problem 9.7 In ctive Example 9.7, suppose that the ball weighs N, the bar B weighs 6 N, and the length of the bar is m. The ball is translating at v = 3 m/s before the impact and strikes the bar at h=.6 m. What is the angular velocity of the bar after the impact if the ball adheres to the bar? C h v ngular momentum is conserved about point C. ( ) m v h = 3 m BL + m h ω m v h ω = = 3 m BL + m h ω =. 3 rad/s. ( N)( 3 m/s)(.6m) 3 (6N)( m ) + ( N)(.6m) =. 3 rad/s. Problem 9.7 The -kg sphere is moving toward the right at 4 m/s when it strikes the end of the 5-kg slender bar B. Immediately after the impact, the sphere is moving toward the right at m/s. What is the angular velocity of the bar after the impact? O B m 4 m/s ngular momentum is conserved about point O. m v L = m v L + 3 m BL ω ω = 3m (v v ) m B L ω = 3( kg)(4 m/s m/s) (5 kg)( m) ω = 3.6 m/s counterclockwise. = 3.6 m/s. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 6

43 Problem 9.7 The -kg sphere is moving toward the right at 4 m/s when it strikes the end of the 5-kg slender bar B. The coefficient of restitution is e =.4. The duration of the impact is. seconds. Determine the magnitude of the average horizontal force exerted on the bar by the pin support as a result of the impact. O B m System angular momentum is conserved about point O. The coefficient of restitution is used to relate the relative velocities before and after the impact. We also use the linear impulse momentum equation for the ball and for the bar. m v L = m v L + 3 m BL ω, ev = ω L v, 4 m/s m v F t = m v, Solving we find R = ( + e)m m B v (3m + m B ) t R =.7 kn. F t + R t = m B ω L. =.4( kg)(5 kg)(4 m/s) ( kg)(. s) = 7 N Problem 9.73 The -kg sphere is moving toward the right at m/s when it strikes the unconstrained 4-kg slender bar B. What is the angular velocity of the bar after the impact if the sphere adheres to the bar? B m The coefficient of restitution is e =. ngular momentum for the system is conserved about the center of mass of the bar. The coefficient of restitution is used to relate the relative velocities before and after the impact. Linear momentum is conserved for the system. ( ) ( ) L L m v h = m v h + m BL ω, m/s.5 m ev = = v B + ω ( L h ) v, m v = m v + m B v B. Solving we find 6(L h)m v ω = h m hlm + L (4m + m B ) = 6( m [.5 m])( kg)( m/s) (.5 m) (kg) (.5 m)( m)( kg) + ( m) ( kg) ω = 8 rad/s. 6 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

44 Problem 9.74 The -kg sphere is moving to the right at m/s when it strikes the unconstrained 4-kg slender bar B. The coefficient of restitution of the impact is e =.6. What are the velocity of the sphere and the angular velocity of the bar after the impact? B m m/s.5 m ngular momentum for the system is conserved about the center of mass of the bar. The coefficient of restitution is used to relate the relative velocities before and after the impact. Linear momentum is conserved for the system. ( ) ( ) L L m v h = m v h + m BL ω, ev = v B + ω ( L h ) v, m v = m v + m B v B. Solving we find 6( + e)(l h)m v ω = h m hlm + L (4m + m B ), = 6(.6)( m [.5 m])( kg)( m/s) (.5 m) (kg) (.5 m)( m)( kg) + ( m) ( kg), v = (h[h L]m + L [4m em B ])v h m hlm + L (4m + m B ), = ([.5 m][.75 m]( kg) + [ m] [5.6 kg])( m/s) (.5 m) (kg) (.5 m)( m)( kg) + ( m) ( kg). ω =.8 rad/s counterclockwise, v =.47 m/s to the right. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 63

45 Problem 9.75 The.4 N ball is translating with velocity v = 4.4 m/s perpendicular to the bat just before impact. The player is swinging the 8.6 N bat with angular velocity ω = 6π rad/s before the impact. Point C is the bat s instantaneous center both before and after the impact. The distances b = mm and y = 66.4 mm. The bat s moment of inertia about its center of mass is I B =.45 kg-m. The coefficient of resti- tution is e =.6, and the duration of the impact is.8 s. Determine the magnitude of the velocity of the ball after the impact and the average force x exerted on the bat by the player during the impact if (a) d =, (b) d = 76. mm, and (c) d = 3 mm. d ω y By definition, the coefficient of restitution is y x () e = v P v v v P. The angular momentum about is conserved: b () m v (d + y b) + m B v B (y b) I B ω C = m v (d + y b) + m Bv B (y b) I Bω. From kinematics, the velocities about the instantaneous center: v v P v v P (3) v P = ω(y + d), v B v B (4) v B = ωy, ω ω (5) v P = ω (y + d), x (6) v B = ωy. Since ω, y, and d are known, v B and v P are determined from (3) and (4), and these six equations in six unknowns reduce to four equations in four unknowns. v P, v B, v, and ω. Further reductions may be made by substituting (5) and (6) into () and (); however here the remaining four unknowns were solved by iteration for values of d =, d =.76m, d =.3 m. The reaction at is determined from the principle of angular impulse-momentum applied about the point of impact: Only the values in the first two columns are required for the problem; the other values are included for checking purposes. Note: The reaction reverses between d =.76 m and d =.3 m, which means that the point of zero reaction occurs in this interval. (7) t t x (d + y b) dt = (dm B v B + I Bω ) (dm B v B + I B ω), where t t =.8 s. Using (4) and (5), the reaction is x = (I B dm B y) (d + y b)(t t ) (ω ω), where the unknown, ω, is determined from the solution of the first six equations. The values are tabulated: d, m v, x,n v P, B,, rad/s c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

46 Problem 9.76 In Problem 9.75, show that the force x is zero if d = I B /(m B y), where m B is the mass of the bat. From the solution to Problem 9.75, the reaction is x = (I B dm B y) (d + y b)(t t ) (ω ω). Since (ω ω) =, the condition for zero reaction is I B dm B y =, from which d = I B. ym B Problem N slender bar of length l = m is released from rest in the horizontal position at a height h = m above a peg (Fig. a). small hook at the end of the bar engages the peg, and the bar swings from the peg (Fig. b). What it the bar s angular velocity immediately after it engages the peg? l h (a) (b) impact. Work energy is used to find the velocity just before T + V = T + V + mgh = mv + v = gh ngular momentum is conserved about the peg. l mv = 3 ml ω 3, ω 3 = 3v l = 3 gh l ω 3 = 4.7 rad/s. = 3 ( 9.8 m/s )( m) = 4.7 rad/s. ( m) c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 65

47 Problem N slender bar of length l = m is released from rest in the horizontal position at a height h = m above a peg (Fig. a). small hook at the end of the bar engages the peg, and the bar swings from the peg (Fig. b). l h (a) (b) Through what maximum angle does the bar rotate relative to its position when it engages the peg? t the instant when the bar has reached the angle determined in part (a), compare its gravitational potential energy to the gravitational potential energy the bar had when it was released from rest. How much energy has been lost? (a) (b) impact. Work energy is used to find the velocity just before T + V = T + V, + mgh = mv + v = gh ngular momentum is conserved about the peg. l mv = 3 ml ω 3,ω 3 = 3v = 3 gh l l Work energy is used to find the angle through which the bar rotates T 3 + V 3 = T 4 + V 4, ( ) 3 ml ω (h 3 mgh = mg + l ) sin θ θ = sin ( lω 3 3g ) ( = sin 3h ) ( = sin 3 ) = 9. l 4 The energy that is lost is the difference in potential energies [ V V 4 = mg (h + l )] ( sin θ = ( N) m+ m ) sin[9 ] (a) θ = 9,(b).5 N- m. =.5 N-m. Problem 9.79 The 4.6 kg disk rolls at velocity v = 3.5 m/s toward a 5.4 mm step. The wheel remains in contact with the step and does not slip while rolling up onto it. What is the wheel s velocity once it is on the step? 457. mm ft/s 5.4 mm We apply conservation of angular momentum about to analyze the impact with the step. ( v ) (R h)mv + I = Rmv + I R ( v R ). () v v v 3 Then we apply work and energy to the climb onto the step. [ mgh = mv 3 + ( I v3 ) ] [ R mv + ( I v ) ]. () R R O h Solving Eqs. () and () with v = obtain v 3 =.9 m/s. 3.5 m/s and I = mr, we 66 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

48 Problem 9.8 The 4.6 kg disk rolls toward a 5.4 mm step. The wheel remains in contact with the step and does not slip while rolling up onto it. What is the minimum velocity v the disk must have in order to climb up onto the step? See the solution of Problem 9.79 solving Eqs. () and () with v 3 =, we obtain v =.8 m/s. Problem 9.8 The length of the bar is m and its mass is kg. Just before the bar hits the floor, its angular velocity is ω = and its center of mass is moving downward at 4 m/s. If the end of the bar adheres to the floor, what is the bar s angular velocity after the impact? v 6 Given: ω =, L = m, m = kg, v G = 4 m/s, sticks to floor ngular momentum about the contact point mv G L cos 6 = 3 ml ω ω = 3 rad/s counterclockwise Problem 9.8 The length of the bar is m and its mass is kg. Just before the bar hits the smooth floor, its angular velocity is ω = and its center of mass is moving downward at 4 m/s. If the coefficient of restitution of the impact is e =.4, what is the bar s angular velocity after the impact? Given ω =, L = m, m = kg, v G = 4 m/s, e =.4, smooth floor. ngular momentum about the contact point L mv G cos 6 = ml ω + mv L G cos 6 Coefficient of restitution ev G = ω ( ) L cos 6 v G Solving we find ω = 9.6 rad/s counterclockwise c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 67

49 Problem 9.83 The length of the bar is m and its mass is kg. Just before the bar hits the smooth floor, it has angular velocity ω and its center of mass is moving downward at 4 m/s. The coefficient of restitution of the impact is e =.4. What value of ω would cause the bar to have no angular velocity after the impact? Given L = m, m = kg, v G = 4 m/s, e =.4, ω = smooth floor. ngular momentum about the contact point L mv G cos 6 + ml ω = mv L G cos 6 Coefficient of restitution e (v G ω L ) cos 6 = v G Solving we find ω = 4 rad/s ω = 4 rad/s clockwise Problem 9.84 During her parallel-bars routine, the velocity of the 4 N gymnast s center of mass is.i 3.5j (m/s) and her angular velocity is zero just before she grasps the bar at. In the position shown, her moment of inertia about her center of mass is.44 kg-m. Ifshe stiffens her shoulders and legs so that she can be modeled as a rigid body, what is the velocity of her center of mass and her angular velocity just after she grasps the bar? y x ( 3., 558.8) mm Let v and ω be her velocity and angular velocity after she grasps the bar. The angle θ =. and r =.59 m. Conservation of angular momentum about is y k (r mv) = rmv + Iω : i j k k x y m m = rmv + Iω, mx.my = rmv + Iω. r θ x Solving this equation together with the relation v = rω, we obtain ω = 3.5 rad/s, v =.87 m/s. Her velocity is ω v = v cos θi v sin θj (x, y) v =.76i.64j ( m/s). 68 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

50 Problem 9.85 The -kg homogenous rectangular plate is released from rest (Fig. a) and falls mm before coming to the end of the string attached at the corner (Fig. b). ssuming that the vertical component of the velocity of is zero just after the plate reaches the end of the string, determine the angular velocity of the plate and the magnitude of the velocity of the corner B at that instant. 3 mm 5 mm (a) B mm ( b) B We use work and energy to determine the plate s downward velocity just before the string becomes taut. y mg(.) = mv. Solving, v =.98 m/s. The plate s moment of inertia is ω I = ()[(.3) + (.5) ] =.567 kg-m. ngular momentum about is conserved: G v Just before G v Just after B x.5(mv) =.5(mv ) + Iω. () The velocity of just after is v = v G + ω r /G = v i j k j + ω.5.5. The j component of v is zero: v +.5ω =. () Solving Eqs. () and (), we obtain v =.36 m/s and ω = 5.45 rad/s. The velocity of B is v B = v G + ω r B/G = v i j k j + ω.5.5 v B =.88i.76j (m/s). c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 69

51 Problem 9.86* Two bars and B are each m in length, and each has a mass of 4 kg. In Fig. (a), bar has no angular velocity and is moving to the right at m/s, and bar B is stationary. If the bars bond together on impact, (Fig. b), what is their angular velocity ω after the impact? B B m/s ω ' (a) (b) Linear momentum is conserved: mv = mv + mv B. () ngular momentum about any point is conserved. bout P, mv ( l ) = mv ( ) l mv B ( ) l + Iω, () v B where I = ml. The velocities are related by P v = v B + lω. (3) Solving Eqs. () (3), we obtain ω =.375 rad/s. v ω v Before fter 6 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

52 Problem 9.87* In Problem 9.86, if the bars do not bond together on impact and the coefficient of restitution is e =.8, what are the angular velocities of the bars after the impact? Linear momentum is conserved: mv = mv + mv B. () ngular momentum of each bar about the point of contact is conserved: mv ( l ) ( ) l = mv + Iω, () = mv B Coefficient of restitution: ( ) l + Iω B. (3) v v ' P ω B ' v B ' v B ' P v ' e = v BP v P v. (4) The velocities are related by v = v P + ( l v BP = v B + ( l ) ω, (5) ) ω B. (6) ω ' Before fter Solving Eqs. () (6), we obtain ω = ω B =.675 rad/s. Problem 9.88* Two bars and B are each m in length, and each has a mass of 4 kg. In Fig. (a), bar has no angular velocity and is moving to the right at m/s, and B is stationary. If the bars bond together on impact (Fig. b), what is their angular velocity ω after the impact? B B ω' m/s Linear momentum is conserved: (a) (b) x DIR: m v = m v x + m Bv Bx, () y DIR: = m v y + m Bv By. () v' By v' Bx Total angular momentum is conserved. Calculating it about, v' y = l m v y + I ω l m Bv Bx + I Bω. (3) v O v' x O ω' We also have the kinematic relation Before impact fter impact v B = v + ω r B/ : i j k v Bx i + v By j = v x i + v y j + ω l l, v Bx = v x l ω, (4) v By = v y + l ω. (5) Solving Eqs. () (5), we obtain ω =.3 rad/s. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 6

53 Problem 9.89* The horizontal velocity of the landing airplane is 5 m/s, its vertical velocity (rate of descent) is m/s, and its angular velocity is zero. The mass of the airplane is Mg, and the moment of inertia about its center of mass is 5 kg-m. When the rear wheels touch the runway, they remain in contact with it. Neglecting the horizontal force exerted on the wheels by the runway, determine the airplane s angular velocity just after it touches down. (See Example 9.8.).3 m.8 m Use a reference frame that moves to the left with the airplane s horizontal velocity. Before touchdown, the velocity of the center of mass is v G = j (m/s). Because there is no horizontal force, v Gx = (see Fig.), and it is given that v Py =, where P is the point of contact. v' Gy v' Gx ngular momentum about P is conserved: m( m/s)(.3 m) = mv Gy (.3 m) + Iω. () y ω' v' Px The velocities are related by v G = v P + ω r G/P : v Gy j = v Px i + i j k ω.3.8. () The j component of this equation is x v' Py v Gy =.3ω. (3) Solving Eqs. () and (3), we obtain ω =.7 rad/s. Problem 9.9* Determine the angular velocity of the airplane in Problem 9.89 just after it touches down if its wheels don t stay in contact with the runway and the coefficient of restitution of the impact is e =.4. (See Example 9.8.) See the solution of Problem In this case v Py is not zero but is determined by e = v Py v Py :.4 = v Py ( ) We see that v Py =.8 m/s. From Eqs. () and () of the solution of Problem 9.89,.6m =.3mv Gy + Iω, v Gy = v Py.3ω. Solving these two equations, we obtain ω =.997 rad/s. 6 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

54 Problem 9.9* While attempting to drive on an icy street for the first time, a student skids his 6- kg car () into the university president s unoccupied 7-kg Rolls-Royce Corniche (B). The point of impact is P. ssume that the impacting surfaces are smooth and parallel to the y axis, and the coefficient of restitution of the impact is e =.5. The moments of inertia of the cars about their centers of mass are I = 4 kg-m and I B = 76 kg-m. Determine the angular velocities of the cars and the velocities of their centers of mass after the collision. (See Example 9.9.).6 m 5 km/hr.7 m y P.6 m 3. m x B v = 5 =.39 m/s 36 Car s initial velocity is The force of the impact is parallel to the x-axis so the cars are moving in the x direction after the collision. Linear momentum is conserved: m v = m v + m Bv B () The angular momentum of each car about the point of impact is conserved. The x-components of the velocities at P are related by the coefficient of restitution: e = (v B.6ω B ) (v +.6ω ) v (4). Solving Equations () (4), we obtain v =.74 m/s,v B =.567 m/s ω =.383 rad/s,ω B =. rad/s..6m v =.6m v + I ω () =.6m B v B + I Bω B (3). The velocity of car at the point of impact after the collision is v P = v + ω r P/ = v i + i j k ω.7.6 v P = (v +.6ω )i +.7ω j. The corresponding equation for car B is v BP = v B + ω B r P/B = v B i + i j k ω B 3..6, or v BP = (v B.6ω B )i 3.ω B j Problem 9.9* The student in Problem 9.9 claimed he was moving at 5 km/h prior to the collision, but police estimate that the center of mass of the Rolls- Royce was moving at.7 m/s after the collision. What was the student s actual speed? (See Example 9.9.) Setting v B =.7 m/s in Equations () (4) of the solution of Problem 9.9 and treating v as an unknown, we obtain v = 4.7 m/s = 5. km/h. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 63

55 Problem 9.93 Each slender bar is 48 cm long and weighs N. Bar is released in the horizontal position shown. The bars are smooth, and the coefficient of restitution of their impact is e =.8. Determine the angle through which B swings afterward. 8 cm Choose a coordinate system with the x axis parallel to bar in the initial position, and the y axis positive upward. The strategy is (a) from the principle of work and energy, determine the angular velocity of bar the instant before impact with B; (b) from the definition of the coefficient of restitution, determine the value of e in terms of the angular velocities of the two bars at the instant after impact; (c) from the principle of angular impulse-momentum, determine the relation between the angular velocities of the two bars after impact; (d) from the principle of work and energy, determine the angle through which bar B swings. The angular velocity of bar before impact: The angle of swing of bar is ( ) 8 β = sin = β ω h ω B ' B Denote the point of impact by P. The point of impact is h = L cos β = 3.5 cm ω' h The change in height of the center of mass of bar is L cos β = h. v' p β v' BP From the principle of work and energy, U = T T, where T =, since the bar is released from rest. The work done by the weight of bar is h U = W dh= Wh. The kinetic energy the bar is T = ( ) I ω, β F F h Wh 3g cos β from which ω = = = 4.43 rad/s, I L where I = ml 3. The angular velocities at impact: By definition, the coefficient of restitution is e = v BPx v P x v P x v BPx. Bar B is at rest initially, from which v BPx =,v BPx = v BP, and v P x = v P cos β,v P x = v P cos β, from which e = v BP v P cos β. v P cos β 64 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

56 The angular velocities are related by v BP = hω B,v P = Lω,v P = Lω, from which e = hω B (L cos β)ω (L cos β)ω = ω B ω ω, from which () ω B ω = eω The force reactions at P: From the principle of angular impulsemomentum, t Fhdt = Fh(t t ) = I B (ω B ), and t t t F(Lcos β)dt = F(Lcos β)(t t ) = I (ω ω ). Divide the second equation by the first: L cos β h from which = = ω ω ω B, () ω B + ω = ω. Solve () and (): ω ω B ( e) = ω, ( + e) = ω. The principle of work and energy: From the principle of work and energy, U = T T, where T =, since the bar comes to rest after rotating through an angle γ. The work done by the weight of bar B as its center of mass rotates through the angle γ is L cos γ U = L W B dh = W The kinetic energy is T = ( ) ( ) I B ω B = I B ( + e) ω 8 = I B( + e) (3g cos β) 8L Substitute into U = T to obtain cos γ = ( + e) cos β 4 ( ) L ( cos γ). = WL( + e) cos β. 8 =.34, from which γ = 7 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 65

57 Problem 9.94* The pollo CSM () approaches the Soyuz Space Station (B). The mass of the pollo is m = 8 Mg, and the moment of inertia about the axis through the center of mass parallel to the z axis is I = 4 Mg-m. The mass of the Soyuz is m B = 6.6 Mg, and the moment of inertia about the axis through its center of mass parallel to the z axis is I B = 7 Mg-m. The Soyuz is stationary relative to the reference frame shown and the CSM approaches with velocity v =.i +.5j (m/s) and no angular velocity. What is the angular velocity of the attached spacecraft after docking? () 7.3 m y 4.3 m (B) x The docking port is at the origin on the Soyuz, and the configuration the instant after contact is that the centers of mass of both spacecraft are aligned with the x axis. Denote the docking point of contact by P.(P is a point on each spacecraft, and by assumption, lies on the x-axis.) The linear momentum is conserved: m v G = m v G + m Bv GB, ω' v' Gy v' Gx.6 m v' GBy v' GBx from which m (.) = m v Gx + m Bv GBx, and () m (.5) = m v Gy + m Bv GBy. Denote the vectors from P to the centers of mass by r P/G = 7.3i (m), and r P/GB =+4.3i (m). The angular momentum about the origin is conserved: r P/G m v G = r P/G m v G + I ω + r P/GB m B v GB. Denote the vector distance from the center of mass of the pollo to the center of mass of the Soyuz by r B/ =.6i (m). From kinematics, the instant after contact: v GB = v G + ω r B/. Reduce: i j k v GB = v G + ω.6 = v Gx i + (v Gy +.6ω )j, from which v GBx = v Gx, i j k r P/G m v G = 7.3 m v Gx m v Gy = ( 7.3m v Gy )k. i j k r P/GB m B v GB = 4.3 m B v Gx m (v Gy +.6ω ) = 4.3m B (v Gy +.6ω )k. Collect terms and substitute into conservation of angular momentum expression, and reduce:.365m = ( 7.3m + 4.3m B )v Gy From () and (), + (49.9m B + I + I B )ω. v Gy =.5m.6m B ω m + m B. These two equations in two unknowns can be further reduced by substitution and algebraic reduction, but here they have been solved by iteration: ω =.3359 rad/s, v Gy =.474 m/s, from which v GBy = v Gy +.6ω =.87 m/s () v GBy = v Gy +.6ω r P/G m v G i j k = 7.3 = (.365m )k,.m.5m 66 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

58 Problem 9.95 The moment of inertia of the pulley is. kg-m. The system is released from rest. Use the principle of work and energy to determine the velocity of the -kg cylinder when it has fallen m. 5 mm Choose a coordinate system with the y axis positive upward. Denote m L = 5kg,m R = kg, h R = m,r =.5 m. From the principle of work and energy, U = T T where T = since the system is released from rest. The work done by the left hand weight is U L = hl m L gdh= m L gh L. T L T L 5 kg kg T R T R The work done by the right hand weight is m L g m R g hr U L = m R gdh= m R gh R. Since the pulley is one-to-one, h L = h R, from which U = U L + U R = (m L m R )gh R. The kinetic energy is T = ( ) ( ) ( ) I P ω + m L vl + m R vr. Since the pulley is one-to-one, v L = v R. From kinematics ω = v R R, from which T = ( )( ) IP R + m L + m R vr. Substitute and solve: v R = (m L m R )gh R ( ) =.6...=.3 m/s IP R + m L + m R c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 67

59 Problem 9.96 The moment of inertia of the pulley is. kg-m. The system is released from rest. Use momentum principles to determine the velocity of the -kg cylinder s after the system is released. Use the coordinate system and notations of Problem From the principle of linear impulse-momentum for the left hand weight: t t (T L m L g) dt = m L (v L v L ) = m L v L, since v L =, from which () T L (t t ) = m L g(t t ) + m L v. For the right hand weight, t t (T R m R g) dt = m R v R, from which () T R (t t ) = m R g(t t ) + m R v R. from which (3) (T L T R )(t t ) = I P R ω. Substitute () and () into (3): (m L m R )g(t t ) + m L v L m R v R = I P R ω. Since the pulley is one to one, v L = v R. From kinematics: ω = v R R, from which v R = (m R m L )g(t t ) I P R + m L + m R =.5 m/s From the principle of angular impulse-momentum for the pulley: t t (T L T R )R dt = I P ω, Problem 9.97 rm BC has a mass of kg, and the moment of inertia about its center of mass is 3 kg-m. Point B is stationary. rm BC is initially aligned with the (horizontal) x axis with zero angular velocity, and a constant couple M applied at B causes the arm to rotate upward. When it is in the position shown, its counterclockwise angular velocity is rad/s. Determine M. y 3 mm M 4 B C x ssume that the arm BC is initially stationary. Denote R =.3 m. From the principle of work and energy, U = T T, where T =. The work done is θ R sin θ U = Mdθ + mg dh = Mθ mgr sin θ. The angle is ( π ) θ = 4 =.698 rad. 8 The kinetic energy is T = ( ) ( ) mv + I BC ω. From kinematics, v = Rω, from which T = ( ) (mr + I BC )ω. Substitute into U = T and solve: M = (mr + I BC )ω + mgr sin θ θ = 44. N-m 68 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

60 Problem 9.98 The cart is stationary when a constant force F is applied to it. What will the velocity of the cart be when it has rolled a distance b? The mass of the body of the cart is m C, and each of the four wheels has mass m, radius R, and moment of inertia I. F From the principle of work and energy, U = T T, where T =. The work done is b U = Fdx= Fb. The kinetic energy is T = ( ) m C v + 4 ( ) mv + 4 From kinematics, v = Rω, from which ( mc T = + m + I ) R v. Substitute into U = T and solve: v = Fb ( ) I m C + 4m + 4 R ( ) Iω. Problem 9.99 Each pulley has moment of inertia I =.3 kg-m, and the mass of the belt is. kg. If a constant couple M = 4 N-m is applied to the bottom pulley, what will its angular velocity be when it has turned revolutions? mm ssume that the system is initially stationary. From the principle of work U = T T, where T =. The work done is M θ U = Mdθ = Mθ, where the angle is θ = (π) = 6.83 rad. The kinetic energy is T = ( ) ( ) m belt v + I pulley ω. From kinematics, v = Rω, where R =. m, from which T = ( ) (R m belt + I pulley )ω. Substitute into U = T and solve: Mθ ω = (R = 5.7 rad/s m belt + I pulley ) c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 69

61 Problem 9. The ring gear is fixed. The mass and moment of inertia of the sun gear are ms = 3 kg and IS = 596 kg-m. The mass and moment of inertia of each planet gear are mp = kg and IP = 88. kg-m. couple M = 84 N-m is applied to the sun gear. Use work and energy to determine the angular velocity of the sun gear after it has turned revolutions mm mm Ring gear M 58 mm Planet gears (3) Sun gear Denote the radius of planetary gear, RP =.78 m, the radius of sun gear RS =.58 m, and angular velocities of the sun gear and planet gear by ωs, ωp. ssume that the system starts from rest. From the principle of work and energy U= T T, where T =. The work done is vp U= R θ Mdθ = Mθ, ωp vs where the angle is ωs θ = ()(π ) = π = 68.3 rad. The kinetic energy is T = IS ωs + 3 mp vp + IP ωp. The velocity of the outer radius of the sun gear is vs = RS ωs. The velocity of the center of mass of the planet gears is the average velocity of the velocity of the sun gear contact and the ring gear contact, vp = vs vs + vr =, since vr =. The angular velocity of the planet gears is ωp = vp. RP Collect terms: ωp = RS RP vp = RS ωs. ωs, Substitute into U = T and solve: Mθ ωs = =.5 rad/s RS 3 IS + mp RS + IP 4 RP 6 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

62 Problem 9. The moments of inertia of gears and B are I =.9 kg-m, and I B =.36 kg-m. Gear is connected to a torsional spring with constant k =. 7 N-m/rad. If the spring is unstretched and the surface supporting the. N weight is removed, what is the velocity of the weight when it has fallen 76. mm? Denote W =. N, s =.76m is the distance the weight falls, r B =.54 m, r =.5m, r B =.76m, are the radii of the gears and pulley. Choose a coordinate system with y positive upward. From the conservation of energy T + V = const. Choose the datum at the initial position, such that V =, T =, from which T + V = at any position. The gear B rotates in a negative direction and the gear rotates in a positive direction. By inspection, 5.4 mm 76. mm 54 mm B. N θ B = s r B =. 76 = rad,. 76 ( ) rb θ = θ B =.667 rad, r v = r B ω B =. 76ω B, ( ) rb ω = ω B =.667ω B. r The moment exerted by the spring is negative, from which the potential energy in the spring is The kinetic energy of the system is T = ( ) ( ) ( )( ) W I ω + I B ωb + v. g Substitute: T = 7. 3v, from which V spring + V weight v =. Solve v =. 76 m/s downward. V spring = θ Mdθ = θ kθdθ = kθ =. 377 N-m. The force due to the weight is negative, from which the potential energy of the weight is s V weight = ( W)dy = Ws =. 695 N-m. Problem 9. Consider the system in Problem 9.. (a) What maximum distance does the. N weight fall when the supporting surface is removed? (b) What maximum velocity does the weight achieve? Use the solution to Problem 9.: V spring + V weight + 7.3v =. V spring = θ Mdθ = θ s V weight = ( W)dy = Ws, from which 64. 5s. s v =. (a) kθ dθ = kθ = 64.5 s, The maximum travel occurs when v =, from which. s max = =. 344 m (where the other solution s max = is meaningless here). (b) The maximum velocity occurs at dv ds = = s =, from which s v max =. 7 m. This is indeed a maximum, since d (v ) ( ) ds = >, 7.3 and v max =.33 m/s (downward). c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 6

63 Problem 9.3 Each of the go-cart s front wheels weighs. N and has a moment of inertia of.36 kgm. The two rear wheels and rear axle form a single rigid body weighing 78 N and having a moment of inertia of.36 kg-m. The total weight of the rider and go-cart, including its wheels, is 67 N. The go-cart starts from rest, its engine exerts a constant torque of.3 N-m on the rear axle, and its wheels do not slip. Neglecting friction and aerodynamic drag, how fast is the go-cart moving when it has traveled 5. m?. 5.4 mm.6 mm From the principle of work and energy: U = T T, where T =, since the go-cart starts from rest. Denote the rear and front wheels by the subscripts and B, respectively. The radius of the rear wheels r =.5 m. The radius of the front wheels is r B =. m. The rear wheels rotate through an angle θ = = rad, from which the work done is 54 mm U = θ Mdθ = Mθ =.3() N-m. 38 mm 5.4 mm.6 mm B The kinetic energy is T = ( ) ( ) ( ) W g v + I ω + I B ωb 46.4 mm 54 mm (for two front wheels). The angular velocities are related to the gocart velocity by ω = v = v, ω B = v = 3v, from which T = r r B. 3v N-m. Substitute into U = T and solve: v = 5.89 m/s. Problem 9.4 Determine the maximum power and the average power transmitted to the go-cart in Problem 9.3 by its engine. The maximum power is P max = Mω max, where ω max = v max R. From which P max = Mv max. Under constant torque, R the acceleration of the go-cart is constant, from which the maximum velocity is the greatest value of the velocity, which will occur at the end of the travel. From the solution to Problem 9.3, v max = 9.3 ft/s, from which P max = Mv max r =.3( 5.89) = N-m/s.. 5 The average power is P ave = U. From the solution to Problem 9.3, t U = 3 N-m. Under constant acceleration, v = at, and s = at, from which s = s vt, and t = v = 3.48 = 5.77 s, from which P ave = U t 3 = = 39 N- m/s c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

64 Problem 9.5 The system starts from rest with the 4-kg slender bar horizontal. The mass of the suspended cylinder is kg. What is the angular velocity of the bar when it is in the position shown? 3 m 45 m From the principle of work and energy: U = T T, where T = since the system starts from rest. The change in height of the cylindrical weight is found as follows: By inspection, the distance between the end of the bar and the pulley when the bar is in the horizontal position is d = + 3 = 3.6 m. The law of cosines is used to determine the distance between the end of the bar and the pulley in the position shown: d 3 β 45 d = + 3 ()(3) cos 45 =.5 m, from which h = d d =.48 m. The work done by the cylindrical weight is h U cylinder = m c gds = m c gh = 45.3 N-m. The work done by the weight of the bar is U bar = cos 45 m b gdh= m b g cos 45 = 7.75 N-m, from which U = U cylinder + U bar = 7.5 N-m. From the sketch, (which shows the final position) the component of velocity normal to the bar is v sin β, from which ω = v sin β. From the law of sines: ( ) 3 sin β = sin 45 = d The kinetic energy is T = ( ) ( ) m c v + Iωb, ( ) from which T =.73ω, where v = ω and I = m( ) sin β 3 has been used. Substitute into U = T and solve: ω =.74 rad/s. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 63

65 Problem 9.6 The.-kg slender bar and.-kg cylindrical disk are released from rest with the bar horizontal. The disk rolls on the curved surface. What is the angular velocity of the bar when it is vertical? 4 mm mm From the principle of work and energy, U = T T, where T =. Denote L =. m, R =.4 m, the angular velocity of the bar by ω B, the velocity of the disk center by v D, and the angular velocity of the disk by ω D. The work done is L/ ω B L L U = m B gdh+ m D gdh= ( ) L m B g + Lm D g. From kinematics, v D = Lω B, and ω D = v D. The kinetic energy is R ( ) ( ) ( ) T = I B ωb + m D vd + I D ωd. L/ v D ω D Substitute the kinematic relations to obtain T = ( ) ( ( ) ) L I B + m D L + I D ωb R, where I B = m BL, 3 I D = m DR, from which T = ( )( mb 3 + 3m ) D L ωb. Substitute into U = T and solve: 6g(mB + m D ) ω B = =. rad/s. (m B + 9m D )L 64 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

66 Problem 9.7 slender bar of mass m is released from rest in the vertical position and allowed to fall. Neglecting friction and assuming that it remains in contact with the floor and wall, determine the bar s angular velocity as a function of θ. l The strategy is θ (a) to use the kinematic relations to determine the relation between the velocity of the center of mass and the angular velocity about the instantaneous center, and (b) the principle of work and energy to obtain the angular velocity of the bar. The kinematics: Denote the angular velocity of the bar about the instantaneous center by ω. The coordinates of the instantaneous center of rotation of the bar are (L sin θ,lcos θ). The coordinates of the center of mass of the bar are L C ( L sin θ, L ) cos θ. The vector distance from the instantaneous center to the center of mass is G θ ω r G/C = L (i sin θ + j cos θ). The velocity of the center of mass is i j k v G = ω r G/C = ω L sin θ L cos θ = ωl (i cos θ j sin θ), from which v G = ωl. The principle of work and energy: U = T T where T =. The work done by the weight of the bar is U = mg ( ) L ( cos θ). The kinetic energy is T = ( ) ( ) mvg + I B ω. Substitute v G = ωl and I B = ml Substitute into U = T and solve: to obtain T = ml ω. 6 3g( cos θ) ω =. L c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 65

67 Problem 9.8 The 4-kg slender bar is pinned to -kg sliders at and B. If friction is negligible and the system starts from rest in the position shown, what is the bar s angular velocity when the slider at has fallen.5 m?. m Choose a coordinate system with the origin at the initial position of and the y axis positive upward. The strategy is (a) (b) (c) to determine the distance that B has fallen and the center of mass of the bar has fallen when falls.5 m, use the coordinates of, B, and the center of mass of the bar and the constraints on the motion of and B to determine the kinematic relations, and use the principle of work and energy to determine the angular velocity of the bar. The displacement of B: Denote the length of the bar by L =. +.5 =.3 m. Denote the horizontal and vertical displacements of B when falls.5 m by d x and d y, which are in the ratio d y = tan 45 =, from which d x = d y = d. The vertical d x distance between and B is reduced by the distance.5 m and increased by the distance d y, and the horizontal distance between and B is increased by the distance d x, from which L = (..5 + d y ) + (.5 + d x ). Substitute d x = d y = d and L =.3 m and reduce to obtain d + bd + c =, where b =.6, and c =.475. Solve: d, = b ± b c =.338 m, or =.54 m, from which only the positive root is meaningful. The final position coordinates: The coordinates of the initial position of the center of mass of the bar are ( L (x G,y G ) = sin θ, L ) cos θ = (.5,.6) (m), ( ).5 where θ = sin =.6 L is the angle of the bar relative to the vertical. The coordinates of the final position of the center of mass of the bar are ( L (x G,y G ) = sin θ,.5 L ) cos θ = (.469,.7), ( ).5 + d where θ = sin = L The vertical distance that the center of mass falls is h = y G y G =.469 m. The coordinates of the final positions of and B are, respectively (x,y ) = (,.5), and (x B,y B ) = (.5 + d, (. + d)) = (.838,.54). The vector distance from to B is v 45 θ.5 m ω B v B 45 r B/ = (x B x )i + (y B y )j =.838i.4j (m). 66 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

68 Check: r B/ =L =.3 m.check. The vector distance from to the center of mass is r G/ = (x G x )i + (y G y )j =.469i.569j (m). Check: r G/ = L =.65 m. check. The kinematic relations: From kinematics, v B = v + ω r B/. The slider is constrained to move vertically, and the slider B moves at a45 angle, from which v = v j (m/s), and v B = (v B cos 45 )i (v B sin 45 )j. i j k v B = v + ω = i(.4ω) + j( v +.838ω), ( ).4 from which () v B = cos 45 ω =.434ω, and () v = v B sin ω =.88ω. The velocity of the center of mass of the bar is i j k v G = v + ω r G/ = v j + ω = (.569ω)i + ( v +.469ω)j, v G =.569ωi.4ωj (m/s), from which (3) v G =.58ω The principle of work and energy: From the principle of work and energy, U = T T, where T =. The work done is d.5 h U = m B gdh+ m gdh+ m bar gdh, U = m B g(.338) + m g(.5) + m bar g(.469) = 3.93 N-m. The kinetic energy is T = ( ) ( ) ( ) ( ) m B vb + m v + m bar vg + I bar ω, substitute I bar = m barl, and (), () and (3) to obtain T =.3ω. Substitute into U = T and solve: ω = 3.93 =.77 rad/s.3 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 67

69 Problem 9.9 homogeneous hemisphere of mass m is released from rest in the position shown. If it rolls on the horizontal surface, what is its angular velocity when its flat surface is horizontal? R 3R 8 The hemisphere s moment of inertia about O is 5 mr, so its moment of inertia about G is I = ( ) 3 5 mr 8 R m = 83 3 mr. The work done is mg (R 58 ) R = 3 8 mgr. O 3R 8 G ω O G 5R 8 Work and energy is 3 8 mgr = mv + Iω = m [( 5R 8 ) ] ω + ( ) 83 3 mr ω. Solving for ω, we obtain ω = 5g 3R. Problem 9. The homogeneous hemisphere of mass m is released from rest in the position shown. It rolls on the horizontal surface. What normal force is exerted on the hemisphere by the horizontal surface at the instant the flat surface of the hemisphere is horizontal? G is See the solution of Problem 9.9. The acceleration of a G = a O + α r G/O ω r G/O i j k = a O i + α h ω ( hj), y x G O mg N 3R 8 = h so a Gy = ω h. Therefore N mg = ma Gy = mω h. Using the result from the solution of Problem 9.9, N = mg + m ( )( ) 5g 3R =.433 mg. 3R 8 68 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

70 Problem 9. The slender bar rotates freely in the horizontal plane about a vertical shaft at O. The bar weighs 89 N and its length is.83 m. The slider weighs 8.9N. If the bar s angular velocity is ω = rad/s and the radial component of the velocity of is zero when r =.3m, what is the angular velocity of the bar when r =. m? (The moment of inertia of about its center of mass is negligible; that is, treat as a particle.) O r ω From the definition of angular momentum, only the radial position of the slider need be taken into account in applying the principle of the conservation of angular momentum; that is, the radial velocity of the slider at r =. m does not change the angular momentum of the bar. From the conservation of angular momentum: I bar ω + r ( W g ) ω = I bar ω + r ( W g ) ω. Substitute numerical values: ( ) I bar + r W = W ( ) bar g 3g L + r W =. kg-m g ( ) I bar + r W = W ( ) bar g 3g L + r W =.5 kg-m, g (. ) from which ω = ω = 8.9 rad/s.5 Problem 9. satellite is deployed with angular velocity ω = rad/s (Fig. a). Two internally stored antennas that span the diameter of the satellite are then extended, and the satellite s angular velocity decreases to ω (Fig. b). By modeling the satellite as a 5-kg sphere of.-m radius and each antenna as a -kg slender bar, determine ω. (a) ω. m ω'.4 m.4 m (b) ssume (I) in configuration (a) the antennas are folded inward, each lying on a line passing so near the center of the satellite that the distance from the line to the center can be neglected; (II) when extended, the antennas are entirely external to the satellite. Denote R =. m,l = R =.4 m. The moment of inertia of the antennas about the center of mass of the satellite in configuration (a) is ( ml ) I ant-folded = = 9.6 kg-m. The angular momentum is conserved, I sphere ω + I ant-folded ω = I sphere ω + I ant-ext ω, where I ant-folded,i ant-ext are for both antennas, from which 97.6ω = 4.8ω. Solve ω =.7 rad/s The moment of inertia of the antennas about the center of mass of the satellite in configuration (b) is ( ml ) ( I ant-ext = + R + L ) m = 4.8 kg-m. The moment of inertia of the satellite is I sphere = 5 m spherer = 88 kg-m. c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 69

71 Problem 9.3 n engineer decides to control the angular velocity of a satellite by deploying small masses attached to cables. If the angular velocity of the satellite in configuration (a) is 4 rpm, determine the distance d in configuration (b) that will cause the angular velocity to be rpm. The moment of inertia of the satellite is I = 5 kg-m and each mass is kg. (ssume that the cables and masses rotate with the same angular velocity as the satellite. Neglect the masses of the cables and the mass moments of inertia of the masses about their centers of mass.) 4 rpm m m (a) d rpm (b) d From the conservation of angular momentum, 4 rpm (I + m( ))ω = (I + md )ω. Solve: ( ) (I + 8m) ω I ω d = m = 9.8 m (a) m m d rpm (b) d Problem 9.4 The homogenous cylindrical disk of mass m rolls on the horizontal surface with angular velocity ω. If the disk does not slip or leave the slanted surface when it comes into contact with it, what is the angular velocity ω of the disk immediately afterward? β ω R The velocity of the center of mass of the disk is parallel to the surface before and after contact. The angular momentum about the point of contact is conserved mvr cos β + Iω = mv R + Iω. From kinematics, v = Rω and v = Rω. Substitute into the angular moment condition to obtain: v v (mr cos β + I)ω = (mr + I)ω. Solve: ω = ( cos β + ) ω 3 63 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

72 Problem 9.5 The 44.5 N slender bar falls from rest in the vertical position and hits the smooth projection at B. The coefficient of restitution of the impact is e =.6, the duration of the impact is. s, and b =.3 m. Determine the average force exerted on the bar at B as a result of the impact..9 m Choose a coordinate system with the origin at and the x axis parallel to the plane surface, and y positive upward. The strategy is to (a) (b) (c) use the principle of work and energy to determine the velocity before impact, the coefficient of restitution to determine the velocity after impact, and the principle of angular impulse-momentum to determine the average force of impact. From the principle of work and energy, U = T T, where T =. The center of mass of the bar falls a distance h = L. The work done ( ) L by the weight of the bar is U = mg. The kinetic energy is T = ( ) Iω, where I = ml 3. Substitute into U = T and solve: 3g ω =, where the negative sign on the square root is chosen to L be consistent with the choice of coordinates. By definition, the coefficient of restitution is e = v B v, where v,v are the velocities v v B of the bar at a distance b from before and after impact. Since the projection B is stationary before and after the impact, v B = v B =, from which v = ev. From kinematics, v = bω, and v = bω, from which ω = eω. The principle of angular impulse-momentum about the point is b B where F B is the force exerted on the bar by the projection at B, from which bf B (t t ) = ml ( + e)ω. 3 Solve: F B = ml ( + e) 3g 3b(t t ) L = N. t t bf B dt = (Iω Iω) = ml 3 (ω ω), Problem 9.6 The 44.5 N bar falls from rest in the vertical position and hits the smooth projection at B. The coefficient of restitution of the impact is e =.6 and the duration of the impact is. s. Determine the distance b for which the average force exerted on the bar by the support as a result of the impact is zero. From the principle of linear impulse-momentum, If the reaction at is zero, then t t Fdt= m(v G v G ), where v G, v G are the velocities of the center of mass of the bar before and after impact, and F = F + F B ( are ) the forces exerted ( ) on the L L bar at and B. From kinematics, v G = ω, v G = ω. From the solution to Problem 9.5, ω = eω, from which ml( + e) F + F B = (t t ) ω. ml( + e) F B = (t t ) ω. From the solution to Problem 9.5, F B = ml ( + e) 3b(t t ) ω. Substitute and solve: b = 3 L=.6m c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 63

73 Problem 9.7 The -kg sphere is moving at m/s when it strikes the end of the -kg stationary slender bar B. If the velocity of the sphere after the impact is.8 m/s to the right, what is the coefficient of restitution? B m m/s 4 mm Denote the distance of the point of impact P from the end of the bar by d =.4 m. The linear momentum is conserved: () m v = m v + m Bv CM, where v CM is the velocity of the center of mass of the bar after impact, and v,v are the velocities of the sphere before and after impact. By definition, the coefficient of restitution is e = v p v, where v p,v p are the velocities of the bar v v p at the point of impact. The point P is stationary before impact, from which () v p v = ev. From kinematics, v v v P ω (3) v CM = v P ( L d ) ω. Substitute () into (3) to obtain ( ) L (4) v CM = v + ev d ω. Substitute (4) into () to obtain (5) (m em B )v = (m + m B )v ( L d ) m B ω. The angular momentum of the bar about the point of impact is conserved: ( ) L (6) = d m B v CM + I CMω. Substitute (3) into (6) to obtain, (7) ( ) ( ( ) ) L L d m B (v + ev ) = I CM + d m B ω, where I CM = m BL. Solve the two equations (5) and (7) for the two unknowns to obtain: ω =.8 rad/s and e = c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

74 Problem 9.8 The slender bar is released from rest in the position shown in Fig (a). and falls a distance h =.3 mm. When the bar hits the floor, its tip is supported by a depression and remains on the floor (Fig. b). The length of the bar is.3 m and its weight is. N. What is angular velocity ω of the bar just after it hits the floor? h 45 ω (a) (b) Choose a coordinate system with the x axis parallel to the surface, with the y axis positive upward. The strategy is to use the principle of work and energy to determine the velocity just before impact, and the conservation of angular momentum to determine the angular velocity after impact. From the principle of work and energy, U = T T, where T = since the bar is released from rest. The work done is U = mgh, where h = energy is T = ( ).3 m. The kinetic mv. Substitute into U = T T and solve: v = gh =.44 m/s. The conservation of angular momentum about the point of impact is (r mv) = (r mv ) + I B ω. t the instant before the impact, the perpendicular distance ( from ) the point of impact to L the center of mass velocity vector is cos 45. fter impact, the ( ) L center of mass moves in an arc of radius about the point of impact so that the perpendicular ( ) distance from the point of impact to L the velocity vector is. From the definition of the cross product, for motion in the x, y plane, ( ) L (r mv) k = cos 45 mv, ( ) L (r mv ) k = mv, and I B ω k = I B ω. From kinematics, v = L ω. Substitute to obtain: ω = mv(cos 45 ) ( ml + I ) = 3v B L = 3 gh = 8.5 rad/s L L c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 633

75 Problem 9.9 The slender bar is released from rest with θ = 45 and falls a distance h = m onto the smooth floor. The length of the bar is m and its mass is kg. If the coefficient of restitution of the impact is e =.4, what is the angular velocity of the bar just after it hits the floor? Choose a coordinate system with the x axis parallel to the plane surface and y positive upward. The strategy is to h θ (a) (b) use the principle of work and energy to obtain the velocity the instant before impact, use the definition of the coefficient of restitution to find the velocity just after impact, v Gy ω v Gy v Gx v px (c) get the angular velocity-velocity relations from kinematics and v py (d) use the principle of the conservation of angular momentum about the point of impact to determine the angular velocity. From the principle of work and energy, U = T T, where T =. The center of mass of the bar also falls a distance h before impact. The work done is U = h mg dh = mgh. The kinetic energy is T = ( ) mv Gy. Substitute into U = T and solve: v Gy = gh, where the negative sign on the square root is chosen to be consistent with the choice of coordinates. Denote the point of impact on the bar by P. From the definition of the coefficient of restitution, e = v Py v By v By v Py. Since the floor is stationary before and after impact, v B = v B =, and () v Py = ev Py = ev Gy. From kinematics: v P = v G + ω r P/G, ( ) L where r P/G = (i cos θ j sin θ), from which ( ) i j k L () v Py = v Gy + ω cos θ sin θ = v Gy j + ( L ) (iω sin θ + jω cos θ). Substitute () into () to obtain (3) v Gy = ev Gy ( ) L ω cos θ. The angular momentum is conserved about the point of impact: (4) ( ) ( ) L L cos θmv Gy = cos θmv Gy + I Gω. Substitute (3) into (4) to obtain L m cos θ( + e)v G = Solve: ( I G + ω ml( + e) cos θ = ( ( ) L cos θ v Gy, I G + m) ( ) L m cos θ) ω. ω 6( + e) cos θ = ( + 3 cos θ) ( gh) =.5 rad/s, where I G = ml has been used. Problem 9. The slender bar is released from rest and falls a distance h = m onto the smooth floor. The length of the bar is m and its mass is kg. The coefficient of restitution of the impact is e =.4. Determine the angle θ for which the angular velocity of the bar after it hits the floor is a maximum. What is the maximum angular velocity? From the solution to Problem 9.9, ω = 6( + e) cos θ gh. ( + 3 cos θ) Take the derivative: dω dθ 6( + e) sin θ 6( + e)(6) cos θ sin θ = = gh + gh ( + 3 cos θ) ( + 3 cos θ) from which 3 cos θ max =, cos θ max = 3, θ max = 54.74, and ω max =.7 rad/s. =, 634 c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior

76 Problem 9. nonrotating slender bar moving with velocity v strikes a stationary slender bar B. Each bar has mass m and length l. If the bars adhere when they collide, what is their angular velocity after the impact? B l From the conservation of linear momentum, mv = mv + mv B. From the conservation of angular momentum about the mass center of ω = I B ω + I Bω B + (r mv B ) i j k = (I B ω + I Bω B )k + L, v B v v B = (I B ω + I Bω B )k m ( L ) v B )k. Since the bars adhere, ω = ω B, from which I Bω = m From kinematics i j k v B = v + ω r B = v i + ω L = (v L ) ω i, from which v B = v ( L ) ω. ( ) L v B. Substitute into the expression for conservation of linear momentum to obtain ( ) L v ω v B =. Substitute into the expression for the conservation of angular momentum to obtain: ω = ( ml ) v 4I B + ml 4 I B = ml, from which ω = 6v 7L c 8 Pearson Education South sia Pte Ltd. ll rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 635