A uniformly thick wheel with moment of inertia l and radius R is free to rotate about its centre of mass (see fig). A massless string is wrapped over its rim and two blocks of masses $m_{1}$ and $m_{2} (m_{1} > m_{2})$ are attached to the ends of the string . The system is released from rest . The angular speed of the wheel when $m_{1}$ descents by a distance h is

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${[\frac{(m_{1} + m_{2})}{(m_{1} - m_{2}) R^{2} + I}]}^{- 1 / 2} gh$

${[\frac{2 (m_{1} + m_{2}) gh}{(m_{1} - m_{2}) R^{2} - I}]}^{- 1 / 2}$

${[\frac{(m_{1} - m_{2})}{(m_{1} + m_{2}) R^{2} + I}]}^{- 1 / 2} gh$

${[\frac{2 (m_{1} - m_{2}) gh}{(m_{1} + m_{2}) R^{2} + I}]}^{- 1 / 2}$

answer is A.

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Detailed Solution

$\begin{array}{l} (m_{1} - m_{2}) g h = \frac{1}{2} (m_{1} + m_{2}) v^{2} + \frac{1}{2} I ω^{2} \end{array}$

$\begin{array}{l} (m_{1} - m_{2}) g h = \frac{1}{2} (m_{1} + m_{2}) (ω R)^{2} + \frac{1}{2} I ω^{2} \end{array}$

$(m_{1} - m_{2}) g h = \frac{ω^{2}}{2} ((m_{1} + m_{2}) R^{2} + I)$

$ω = \sqrt{\frac{2 (m_{1} - m_{2}) g h}{(m_{1} + m_{2}) R^{2} + I}}$

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