A thin, light string is wrapped around the outer rim of a uniform hollow cylinder of mass 4.75 kg having inner and outer radii as shown in ?Fig. E10.25?. The cylinder is then released from rest. (a) How far must the cylinder fall before its center is moving at 6.66 m/s? (b) If you just dropped this cylinder without any string, how fast would its center be moving when it had fallen the distance in part (a)? (c) Why do you get two different answers when the cylinder falls the same distance in both cases?
Solution 27E Step 1: Initially, the kinetic energy of the system was zero since, the cylinder was at rest. Then, the energy will be completely potential and it will be, PE initialmgH The kinetic energy of the system can be calculated when the speed is 6.66 m/s. The kinetic energy at a speed v can be given by, KE = KE translational Rotational 2 KE translational½ mv Where, m is th mass of the system. 2 KE rotational½ I Where, I - moment of inertia of the system and - Angular velocity
