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,mbut a downward force on the outer edge of the pulley. Similarly the string exerts a force on block 1mto the right, but a leftward force on the outer edge of the pulley. Solve:(a)Newton’s second law for 1mand 2mis 11Tma=and 222.gma−=Using the constraint 21,aaa−=+=we have 1a=and .a−+=Adding these equations, we get 2(),mgm m a=+or 2112mmgaTmamm=⇒==++(b)When the pulley has mass m, the tensions ( and)TTin the upper and lower portions of the string are different. Newton’s second law for 1mand the pulley are: 12andaTRTR Iα=−We are using the minus sign with because the pulley accelerates clockwise. Also, .aR=Thus, a=and 2Ia aIRRR=Adding these two equations gives
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