Question

A hoop rolls down an inclined plane. The fraction of its total kinetic energy that is associated with rotational motion is

• A. $1:2$
• B. $1:3$
• C. $1:4$
• D. $2:3$

Answer 1

The correct option is A $1:2$

Moment of inertia of hoop (or ring), $I=M{R}^2$

Also, $v=Rw$ (due to pure rolling)

Now, Rotational K.E $K.E_R=\frac{1}{2}I{w}^2=\frac{1}{2}M{R}^2{w}^2=\frac{1}{2}M{v}^2$

Total K.E, $K.E=\frac{1}{2}I{w}^2+\frac{1}{2}M{v}^2=\frac{1}{2}M{R}^2{w}^2+\frac{1}{2}M{v}^2$

$K.E=M{v}^2$

Thus $K.E_R:K.E=1:2$