A uniformly thick wheel with moment of inertia $I$ and radius $R$ is free to rotate about its centre of mass, which is fixed. A massless string is wrapped over its rim and two blocks of masses $m_1$ and $m_2$ $(m_1 > m_2)$ hang from the two ends of the string. The system is released from rest. The angular speed of the wheel when $m_1$ descends by a distance $h$ is:
Answer: (A) $\left[\dfrac{2(m_1 - m_2)gh}{(m_1 + m_2)R^2 + I}\right]^{1/2}$
The string does not slip, so both blocks move with $v = \omega R$. When $m_1$ goes down by $h$, $m_2$ goes up by $h$.
Energy conservation:
$$(m_1 - m_2)gh = \frac{1}{2}(m_1 + m_2)\omega^2R^2 + \frac{1}{2}I\omega^2$$
$$\omega = \left[\frac{2(m_1 - m_2)gh}{(m_1 + m_2)R^2 + I}\right]^{1/2}$$
Solution by Sreeraj P, M.Sc Physics