(B) By using the work-energy theorem,the work done by gravity equals the change in kinetic energy of the system.$W_{g} = \Delta KE$$(m_{1} - m_{2}) gh

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

(B) By using the work-energy theorem,the work done by gravity equals the change in kinetic energy of the system.$W_{g} = \Delta KE$$(m_{1} - m_{2}) gh = \frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} v^{2} + \frac{1}{2} I \omega^{2}$Since the string does not slip,$v = \omega R$. Substituting this into the equation:$(m_{1} - m_{2}) gh = \frac{1}{2} (m_{1} + m_{2}) (\omega R)^{2} + \frac{1}{2} I \omega^{2}$$(m_{1} - m_{2}) gh = \frac{\omega^{2}}{2} [(m_{1} + m_{2}) R^{2} + I]$Solving for $\omega$:$\omega = \sqrt{\frac{2(m_{1} - m_{2}) gh}{(m_{1} + m_{2}) R^{2} + I}}$

Class 11 Physics System of Particles and Rotational Motion Rotation Motion Basic, Motion of Connected Mass

Difficult

$A$ 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 $m_{1}$ 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 $m_{1}$ descends by a distance $h$ is

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A
$\left[\frac{m_{1}+m_{2}}{\left(m_{1}+m_{2}\right) R^{2}+I}\right]^{\frac{1}{2}} gh$
B
$\left[\frac{2\left(m_{1}-m_{2}\right) gh}{\left(m_{1}+m_{2}\right) R^{2}+I}\right]^{\frac{1}{2}}$
C
$\left[\frac{2\left(m_{1}+m_{2}\right) gh}{\left(m_{1}+m_{2}\right) R^{2}+I}\right]^{\frac{1}{2}}$
D
$\left[\frac{\left(m_{1}-m_{2}\right)}{\left(m_{1}+m_{2}\right) R^{2}+I}\right]^{\frac{1}{2}} gh$