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 notthe same. Solve:Applying Newton’s second law to 1,m2,mand the pulley yields the three equations: ()( )1G11G2222 1120.50 N mTF maF Tma TRTRIα−=−+=−−=Noting that 21,aaa−==21p2,ImR=and /,aR=the above equations simplify to 211122221pp110.50 N m10.50 N m220.060 maTmgma mgT ma TTmRmaRRR⎛⎞⎛⎞−=−=+=+⎜⎟⎜
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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.