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
$m_{1 gh}=\frac{1}{2}m_{1}v^{2}+\frac{1}{2}m_{2} v^{2}+\frac{1}{2}l\omega^{2}+m_{2} gh$
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 $v \alpha\sqrt{\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}+\frac{I}{R^{2}}}\right)}$