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336_PartUniversity Physics Solution

# 336_PartUniversity Physics Solution - Dynamics of...

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Dynamics of Rotational Motion 10-31 Thus 1 2 L L = gives ( ) 2 2 1 1 1 rod rod rod 8 3 16 m vL m L m L ω = + 19 1 8 48 v L ω = 6 19 / v L ω = (b) 2 2 1 1 1 rod 2 8 K mv m v = = ( ) 2 2 2 2 2 1 1 1 1 1 2 rod b rod rod 2 2 2 3 16 ( ) (6 /19 ) K I I I m L m L v L ω ω = = + = + ( ) ( ) 2 2 2 19 6 3 1 2 rod rod 2 48 19 152 K m v m v = = Then 2 3 rod 2 152 2 1 1 rod 8 3/19. m v K K m v = = E VALUATE : The collision is inelastic and 2 1 . K K < 10.88. I DENTIFY : Apply Eq.(10.29). S ET U P : The door has 2 1 3 I ml = . The torque applied by the force is av rF , where /2 r l = . E XECUTE : av av, av and . rF L rF t rJ τ Σ = Δ = Δ = The angular velocity ω is then ( ) av av av 2 1 3 2 3 , 2 l F t L rF t F t I I ml ml ω Δ Δ Δ Δ = = = = where l is the width of the door. Substitution of the given numeral values gives 0.514rad s. ω = E VALUATE : The final angular velocity of the door is proportional to both the magnitude of the average force and also to the time it acts. 10.89. (a) I DENTIFY : Apply conservation of angular momentum to the collision between the bullet and the board: S ET U P : The system before and after the collision is sketched in Figure 10.89a. Figure 10.89a E XECUTE : 1 2 L L = 3 2 1 sin (1.90 10 kg)(360 m/s)(0.125 m) 0.0855 kg m /s L mvr mvl φ = = = × = 2 2 2 L I ω = 2 2 1 2 board bullet 3 I I I ML mr = + = + 2 3 2 2 1 2 3 (0.750 kg)(0.250 m) (1.90 10 kg)(0.125 m) 0.01565 kg m I = + × = Then 1 2 L L = gives that 2 1 2 2 2 0.0855 kg m /s 5.46 rad/s 0.1565 kg m L I ω = = = (b) I DENTIFY : Apply conservation of energy to the motion of the board after the collision. S ET U P : The position of the board at points 1 and 2 in its motion is shown in Figure 10.89b. Take the origin of coordinates at the center of the board and y + to be upward, so cm,1 0 y = and cm,2 , y h = the height being asked for. 1 1 other 2 2 K U W K U + + = + E XECUTE : Only gravity does work, so other 0. W = 2 1 1 2 K I ω = 1 cm,1 0 U mgy = = 2 0 K = 2 cm,2 U mgy mgh = = Figure 10.89b

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10-32 Chapter 10 Thus 2 1 2 . I mgh ω = 2 2 2 3 2 (0.01565 kg m )(5.46 rad/s) 0.0317 m 3.17 cm 2 2(0.750 kg 1.90 10 kg)(9.80 m/s ) I h mg ω = = = = + × (c) I DENTIFY and S ET U P : The position of the board at points 1 and 2 in its motion is shown in Figure 10.89c. Apply conservation of energy as in part (b), except now we want cm,2 0.250 m. y h = = Solve for the ω after the collision that is required for this to happen. Figure 10.89c E XECUTE : 2 1 2 I mgh ω = 3 2 2 2 2(0.750 kg 1.90 10 kg)(9.80 m/s )(0.250 m) 0.01565 kg m mgh I ω + × = = 15.34 rad/s ω = Now go back to the equation that results from applying conservation of angular momentum to the collision and solve for the initial speed of the bullet. 1 2 L L = implies bullet 2 2 m vl I ω = 2 2 2 3 bullet (0.01565 kg m )(15.34 rad/s) 1010 m/s (1.90 10 kg)(0.125 m) I v m l ω = = = × E VALUATE : We have divided the motion into two separate events: the collision and the motion after the collision. Angular momentum is conserved in the collision because the collision happens quickly. The board doesn°t move much until after the collision is over, so there is no gravity torque about the axis. The collision is inelastic and mechanical energy is lost in the collision. Angular momentum of the system is not conserved during this motion, due to the external gravity torque. Our answer to parts (b) and (c) say that a bullet speed of 360 m/s causes the board to swing up only a little and a speed of 1010 m/s causes it to swing all the way over.
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336_PartUniversity Physics Solution - Dynamics of...

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