Question

Consider two masses with $m_1>m_2$ connected by a light inextensible string that passes over a pulley of radius $R$ and moment of inertia $I$ about its axis of rotation. The string does not slip on the pulley and the pulley turns without friction. The two masses are released from rest separated by a vertical distance $2h$ . When the two m asses pass each other, the speed of the masses is proportional to

• A. $\sqrt{\frac{m_1-m_2}{m_1+m_2+\frac{I}{R^2}}}$
• B. $\sqrt{\frac{\left(m_1+m_2\right)\left(m_1-m_2\right)}{m_1+m_2+\frac{I}{R^2}}}$
• C. $\sqrt{\frac{m_1+m_2+\frac{I}{R^2}}{m_1-m_2}}$
• D. $\sqrt{\frac{\frac{I}{R^2}}{m_1+m_2}}$

Answer 1

Answer: (A)

Loss of potential energy of$m_1$ appears as kinetic energies of $m_1$,$m_2$ and pulley and also as potential energies of $m_1$ and $m_2$. So, by energy conservation, we get
$m_{1gh}=\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2+\frac{1}{2}l{\omega }^2+m_{2}gh$
Here, $m_1$ falls by distance $h,m_2$ rises bydistance $h$,
$v$ = speed of $m_1$=speed of $m_2$
as they passes each other and $\omega$ = angular speed of pulley $=\frac{v}{R}.$ So, $\left(m_1-m_2\right)gh=\frac{v^2}{2}\left(m_1+m_2+\frac{I}{R^2}\right)$
$\therefore v=\frac{2gh\left(m_1-m_2\right)}{\left(m_1+m_2+\frac{I}{R^2}\right)}$
or $v\alpha \sqrt{\left(\frac{m_1-m_2}{m_1+m_2+\frac{I}{R^2}}\right)}$