Question

A loop and a disc roll without slipping with same linear velocity $v$. The mass of the loop and the disc is same. If the total kinetic energy of the loop is $8J$, find the kinetic energy of the disc (in $J$)

Answer 1

MI of loop $I=M{R}^2$

So $KE=\frac{I{\omega }^2}{2}+\frac{M{v}^2}{2}=M{v}^2=8J$ ...(1)

Now MI of disc $I=\frac{M{R}^2}{2}$

So KE of disc $KE=\frac{I{\omega }^2}{2}+\frac{M{v}^2}{2}=\frac{3M{v}^2}{4}=\frac{3\times 8}{4}=6J$