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 $m_2>m_2$ are attached to the ends of the string. The system is released from rest. The angular speed of the wheel, when $m_1$ descents through a distance $h$, is

• A. ${\left[\frac{2(m_1-m_2)gh}{(m_1+m_2)R^2+I}\right]}^{1/2}$
• B. ${\left[\frac{2(m_1+m_2)gh}{(m_1+m_2)R^2+I}\right]}^{1/2}$
• C. ${\left[\frac{(m_1-m_2)}{(m_1+m_2)R^2+I}\right]}^{1/2}gh$
• D. ${\left[\frac{(m_1+m_2)}{(m_1+m_2)R^2+I}\right]}^{1/2}gh$

Answer 1

The correct option is A ${\left[\frac{2(m_1-m_2)gh}{(m_1+m_2)R^2+I}\right]}^{1/2}$

Using the principal of conservation of energy,
$(m_1-m_2)gh=\frac{1}{2}(m_1+m_2)v^2+\frac{1}{2}I{\omega }^2$

As, $v=\omega R$

$\Rightarrow (m_1-m_2)gh=\frac{1}{2}(m_1+m_2)\left({\omega }^2{R}^2\right)+\frac{1}{2}I{\omega }^2$

$\Rightarrow (m_1-m_2)gh=\frac{{\omega }^2}{2}\left[\right(m_1+m_2)R^2+I]$

$\Rightarrow \omega =\sqrt{\frac{2(m_1-m_2)gh}{(m_1+m_2)R^2+I}}$

Hence, $\left(A\right)$ is the correct answer.