Consider two objects with masses $m_{1}$ and $m_{2}$ ( $m_{1} > m_{2}$ ) connected by a light string that passes over a pulley having a moment of inertia of $I$ about its axis of rotation as shown in figure. The string does not slip on the pulley or stretch. The pulley turns without fiction. The two objects are released from rest separated by a vertical distance 2 $h$ . The translational speeds of the objects as they pass each other is

\sqrt{\frac{2 (m_{1} + m_{2}) g h}{m_{1} + m_{2} + \frac{1}{R^{2}}}}

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\sqrt{\frac{2 (m_{1} - m_{2}) g h}{m_{1} + m_{2} + \frac{1}{R^{2}}}}

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\sqrt{\frac{(m_{1} - m_{2}) g h}{m_{1} + m_{2} + \frac{1}{R^{2}}}}

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\sqrt{\frac{(m_{1} + m_{2}) g h}{m_{1} + m_{2} + \frac{I}{R^{2}}}}

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Solution

The correct option is B $\sqrt{\frac{2 (m_{1} - m_{2}) g h}{m_{1} + m_{2} + \frac{1}{R^{2}}}}$
If friction is neglected, then mechanical energy is conserved and we can say that the increase in kinetic energy of the system equals the decrease in potential energy. Since $K_{i} = 0$ (the system is initially at rest).
We have
$Δ K = K_{f} - K_{i}$
$= \frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} v^{2} + \frac{1}{2} I ω^{2}$
, where $m_{1}$ and $m_{2}$ have a common speed.
But $v = R ω$
$Δ K = \frac{1}{2} (m_{1} + m_{2} + \frac{I}{R^{2}}) v^{2}$
From the given figure, we see that the system loses potential energy when released because of the motion of masses and gains kinetic energy.
Change in P.E. $= m_{1} g .2 h - (m_{1} g h + m_{2} g h) = (m_{1} - m_{2}) g h$
Applying the law of conservation of energy,
Change in P.E = Change in K.E
$(m_{1} - m_{2}) g h = \frac{1}{2} (m_{1} + m_{2} + \frac{I}{R^{2}}) v^{2}$
$∴$ $v$ = $\sqrt{\frac{2 (m_{1} - m_{2}) g h}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$

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