In the system shown in Fig. 9.17, a 12.0-kg mass is released from rest and falls, causing the uniform 10.0-kg cylinder of diameter 30.0 cm to turn about a frictionless axle through its center. How far will the mass have to descend to give the cylinder 480 J of kinetic energy?

Solution to 89P Step 1 Mass of the object =12 kg Mass of the cylinder =10kg Diameter of cylinder =0.30m Radius of the cylinder =0.15m Kinetic energy of cylinder =480J Moment of inertia of cylinder =MR /2 2 Step 2 Kinetic energy of a rotating body =K r K =r(½)I 2 Kr(¼)10xr 2 2 2 (480x4)/(10x(0.15) = 2 =92.37 rad/s Velocity of the block =tangential velocity of the cylinder v=r v=0.15x92.37 v=13.9m/s