Question

A hollow sphere of mass 1 kg and radius 10 cm is free to rotate about its diameter. If a force of 30 N is applied tangentially to it, its angular acceleration is

Answer 1

The torque acting on the sphere about its diameter = T = r×FF = 30 Nr = perpendicular distance = 10 cm = 0.1mtherefore,T = 30×0.1T = 3 N-mnow, Torque = Moment of inertia about the axis of rotation × angular accelerationT= I×αnow,I = 23×M×R2M = mass R = radiusI = 23×1×0.1I = 0.23Now,α = TI = 30.23 = 3×30.2 = 902 =45 rad/s2