Question

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 $m_1$ descents by a distance $h$ is :

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

Answer 1

Answer: (D)

by using work energy theorem
$Wg = \Delta KE$
$\left(m_{1}-m_{2}\right)gh= \frac{1}{2}\left(m_{1}+m_{2}\right)V^{2} + \frac{1}{2}I\omega^{2}$
$\left(m_{1}-m_{2}\right)gh= \frac{1}{2}\left(m_{1}+m_{2}\right)\left(\omega R\right)^{2} + \frac{1}{2}I\omega ^{2}$
$\left(m_{1}-m_{2}\right)gh= \frac{\omega^{2}}{2}\left[\left(m_{1}+m_{2}\right) R^{2}+I\right]$
$\omega = \sqrt{\frac{2\left(m_{1}-m_{2}\right)gh}{\left(m_{1}+m_{2}\right)R^{2} +1}}$