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Unformatted text preview: Physics 206 Examples with Solutions Rotation Problem 1. Consider again Thomas and Elizabeth on the same teetertotter discussed in Example 1 2. 3 of Chapter 1 2 of the online text. Assume that the kids sit where they did in that example (Thomas 1. 5 m and Elizabeth 1 m from the pivot) but assume that Thomas has gained weight so that he now has a mass of 24 kg while Elizabeth still has a mass of 30 kg. Find the initial angular acceleration of the teetertotter board (assume they start sitting with the board level). Take the board to have a length of 3 m. Solution : There are two equivalent ways to work this problem. First way: The first is to apply T = I to the object consisting of the board and the two kids. In this case the net force on the system includes the weigh of the two kids directly. Consider the figure: The rotational inertia of the system (including the kids) about the pivot is (modeling the board as a long cylinder), I = I E + I T + I B oard = m E r E 2 + m T r T 2 + (1 / 1 2 ) M board L board 2 I = ( 30 ) (1 ) 2 + ( 24 ) ( 1 . 5 ) 2 + (1 / 12 ) ( 50 ) (3) 2 kg m 2 I = 1 21 . 5 kg m 2 The net torque of the board/Elizabeth/Thomas object (about the pivot) is the sum of the torques due to the weights of the board, Elizabeth, and Thomas. T net = T E + T T + T B T net = ( 1 m x ) ( ( 30 ) ( 9. 8 N ) y ) + ( 1. 5 m x ) ( ( 24 ) ( 9. 8 N ) y ) T net = 58. 8 N m z The angular acceleration is = T net I = 58. 8 N m z 1 21. 5 kg m 2 = ( . 484 / s 2 ) z y x Elizabeth Thomas ~ F E ~ W B ~ F T ~ N p ~ r T ~ r E ~ r E ~ r T + z out of page Second Way : Apply T = I to the board alone. In this case the force exerted by Thomas and Elizabeth on the board must first be calculated. Using the indicated coordinate system, m E g E + F B onE = m E a E F B onE = m E g E + m E a E Then the force F E that Elizabeth exerts on the board is, using the third law and noting a E = r E , F E = ( m E g E + m E r E ) y 1 A similar analysis with ( a T = + r T ) leads to the following expression for the force that Thomas exerts on the board, F T = ( m T g E m T r T ) y In interpreting these force take care to note that for the given case turns out to be negative. Applying T = I to the board proceeds as (with = z ), ( + r T x ) (( m T g E m T r T ) y ) + ( r E x ) (( m E g E + m E r E ) y ) = 1 1 2 ML 2 z ( r T m T g E m T r T 2 + r E m E g E m E r E 2 ) z = 1 12 ML 2 z  r T m T g E + r E m E g E = 1 1 2 ML 2 + m T r T 2 + m E r E 2 = r T m T g E + r E m E g E 1 1 2 ML 2 + m T r T 2 + m E r E 2 = 58. 8 N m 1 21 . 5 kg m 2 = . 484 / s 2 Problem 2. A thin rod of length L = 0. 5 m and mass M = 1 kg is is free to rotate about a fixed axis through one end of the rod. The net force on the rod has size 1 0 N and acts as shown in the figure below....
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 Fall '09
 yung
 Physics

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