A uniformly thick wheel with moment of inertia I 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 $m_{1} a n d m_{2} (m_{1} > m_{2})$ 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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${[\frac{2 (m_{1} + m_{2}) g h}{(m_{1} + m_{2}) R^{2} + I}]}^{\frac{1}{2}}$

${[\frac{m_{1} + m_{2}}{(m_{1} + m_{2}) R^{2} + I}]}^{\frac{1}{2}} g h$

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

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

answer is D.

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Detailed Solution

According to work energy theorem

$W_{T o t a l} = Δ K E$

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

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