Question

A hollow, thin-walled sphere of radius R = 12.0 cm rolls down an inclined plane from a height of 23.0 cm. How fast will its center of gravity be moving when it reaches the bottom?

Ans: 1.64 m/s

Answer 1

Since it is rolling, there will be two kinetic energy one is rotational and the other one is due to translation motion.

So KE = 0.5*m*v^2 + 0.5*I*w^2, but in rolling as we know there is no slipping and the relation between the angular speed and the velocity is given by , v = w*R. The moment of Inertia of a hallow sphere is (2/3).m.r^2.

Then KE = 0.5.m.v^2+0.5.(2/3).m.r^2.(v/r)^2 = (5/6)*m*v^2.

Here friction and normal does no work, only gravity force is working here. So the work done by gravity is,

W = mgh = m*9.8*0.23 J.

From work energy theorem we know that, m*2.254 = (5/6)*m*v^2.

Hence v = 1.644 m/s.