### Problem

A solid sphere of mass $m$ and radius $r$ is spinning about a horizontal axis with an angular velocity $\omega_1$ is placed on a rough horizontal surface. Subsequently, it

rolls without slippingwith an angular velocity of $\omega_2$. Find $\frac{\omega_2}{\omega_1}$

### Attempt at solution

I applied the conservation of energy. Initially the energy due to rotational motion only, can be calculated by $$\frac{1}{2}I\omega_1^2$$ where $I$ is the moment of inertia of the sphere.

After placing the sphere on the table, the energy of the sphere is due to its rotational and transnational motion. This can be calculated by

$$\frac{1}{2}I\omega_2^2 + \frac{1}{2}mv_{cm}^2$$ where $v_{cm}$ is the velocity of the centre of mass of the sphere. Since the problem states the sphere rolls without slipping, $v_{cm} = \omega_2\cdot r$ On equating, I calculated $\frac{\omega_2}{\omega_1}$ as $\sqrt\frac{2}{7}$

### Issue

The answer given is $\frac{2}{7}$. I'm unable to find where I went wrong.