12_P70InstructorSolution

12_P70InstructorSolution - 12.70. Model: The pulley is a...

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12.70. Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a disk and that the string does not slip. Visualize: Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same. Solve: Applying Newton’s second law to 1 , m 2 , m and the pulley yields the three equations: () ( ) 1G 1 1 G 22 2 2 1 12 0.50 N m TF m a F Tm a T R T R I α −= −+ = = Noting that 21 , aaa −== 2 1 p 2 , Im R = and /, aR = the above equations simplify to 2 11 1 2 2 2 2 1 p p 1 1 0.50 N m 1 0.50 N m 2 2 0.060 m a Tm gm a m gT m a TT m R m a RR R ⎛⎞ = = + = + ⎜⎟
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This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at University of Georgia Athens.

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