Question

a body is rolling without slipping on horizontal plane .the rotational energy of the body is 40 % of the total kinetic energy .identify the body ????????????????????? option .ring ,disc ,hollow cylinder or hollow sphere

Answer 1

A body is rolling without slipping on horizontal plane. The rotational energy of the body is 40 % of the total kinetic energy. Identify the body?

Option: ring, disc, hollow cylinder or hollow sphere.

Solution

A body is rolling without slipping, so $v = r\omega$, where $v$ – velocity, $\omega$ – angular velocity, $r$ – radius.

Total kinetic energy:

$$K_{\mathrm{total}} = K_{\mathrm{rotational}} + \frac{m v^2}{2}.$$

We know that $\frac{K_{\mathrm{rotational}}}{K_{\mathrm{total}}} = 0.4$, then $K_{\mathrm{rotational}} = \frac{0.4}{0.6} * \frac{m v^2}{2} = \frac{2}{3} \frac{m r^2 \omega^2}{2}$.

But on the other hand

$$K_{\mathrm{rotational}} = \frac{l \omega^2}{2},$$

where $l$ – moment of inertia of body.

So

$$l = \frac{2}{3} m r^2 = l_{\mathrm{hollow\ sphere}}.$$

Answer: hollow sphere.