12.67. Model: The hollow cylinder is a rigid rotating body. Visualize:We placed the origin of the coordinate system on the ground. Solve:(a)Newton’s second law for the block is GBB(),yFTma−+=where Tis the tension in the string, BFmg=is the gravitational force on the block, and yais the acceleration of block. The string tension exerts a negative (cw) torque on the cylinder, so the rotational form of Newton’s second law for the hollow cylinder is 2yyaIaTRIITRRτα=−==⇒ =−where we used the acceleration constraint .yaRα=With this expression for T, Newton’s second law for the block becomes BBB22B(/)yyIam gmgmaaRmIR−−−=⇒=+The moment of inertia of a hollow cylinder is 2C,ImR=so the equation for yasimplifies to 22BBC(3.0 kg)(9.8 m/s )5.88 m/s3.0 kg2.0 kgyamm===−++The speed of the block just before it hits the ground can now be found using kinematics:
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