A uniform sphere with mass 28.0 kg and radius 0.380 m is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is 236 J, what is the tangential velocity of a point on the rim of the sphere?

Solution 39E Step 1: Mass of sphere M = 28 kg Radius of sphere R = 0.380 m Kinetic energy K = 236 J Let us find the moment of the sphere It is given by 2 I = (2/5) × MR Substituting values we get I = (2/5) × 28 kg × (0.380 m) 2 2 I = 2.16 kg m Step 2 : Using kinetic energy relation, we shall find the angular velocity of the sphere K = 1/2 I 2 Rearranging to find angular velocity 236 J = 1/2 × 2.16 kg m × 2 2 236 J 2 1/2×1.61 kg m= = 217.8 m /s 2 2 = 14.76 m/s Velocity of the sphere is 14.76 m/s