12_P71InstructorSolution - 12.71 Model Assume the string...

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12.71. Model: Assume the string does not slip on the pulley. Visualize: The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward force on the block 2 , m but a downward force on the outer edge of the pulley. Similarly the string exerts a force on block 1 m to the right, but a leftward force on the outer edge of the pulley. Solve: (a) Newton’s second law for 1 m and 2 m is 11 Tm a = and 22 2 . gm a = Using the constraint 21 , aa a −= += we have 1 a = and . a −+ = Adding these equations, we get 2 () , mg m m a =+ or 2 1 12 mmg aT m a mm =⇒ = = ++ (b) When the pulley has mass m , the tensions ( a n d ) TT in the upper and lower portions of the string are different. Newton’s second law for 1 m and the pulley are: 1 2 and a T RT R I α = We are using the minus sign with because the pulley accelerates clockwise. Also, . aR = Thus, a = and 2 Ia a I R RR = Adding these two equations gives
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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 UGA.

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