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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 α 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 ª 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 ˆ ) × ( ( 30 ) ( 9. 8 N ) y ˆ ) + ( 1. 5 m x ˆ ) × ( ( 24 ) ( 9. 8 N ) y ˆ ) T net = 58. 8 N · m z ˆ The angular acceleration is α = T net I = 58. 8 N · m z ˆ 1 21. 5 kg · m 2 = ( . 484 / s 2 ) z ˆ 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 α 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 α , F E = ( m E g E + m E r E α ) y ˆ 1 A similar analysis with ( a T = + r T α ) leads to the following expression for the force that Thomas exerts on the board, F T = ( m T g E m T r T α ) y ˆ In interpreting these force take care to note that for the given case α turns out to be negative. Applying T = Iα to the board proceeds as (with α = α z ˆ ), ( + r T x ˆ ) × (( m T g E m T r T α ) y ˆ ) + ( r E x ˆ ) × (( m E g E + m E r E α ) y ˆ ) = 1 1 2 ML 2 α z ˆ ( r T m T g E m T r T 2 α + r E m E g E m E r E 2 α ) z ˆ = 1 12 ML 2 α z ˆ r T m T g E + r E m E g E = µ 1 1 2 ML 2 + m T r T 2 + m E r E 2 ¶ α α = 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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This note was uploaded on 11/06/2010 for the course PHYSICS 1220 taught by Professor Yung during the Fall '09 term at Missouri (Mizzou).
 Fall '09
 yung
 Physics

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