Consider two object with `m_1 gt 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 strech. The pulley turns without friction. 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.
.

Options

A. `sqrt((2(m_1 + m_2)gh)/(m_1 + m_2 + (I)/(R^2)))`
B. `sqrt((2(m_1 - m_2)g h)/(m_1 + m_2 + (I)/(R^2)))`
C. `sqrt(((m_1 - m_2)gh)/(m_1 + m_2 + (I)/(R^2)))`
D. `sqrt(((m_1 + m_2)gh)/(m_1 + m_2 + (I)/(R^2)))`

Answer

B

Explanation

(b) Take the two objects pulley, and Earth as the system. If we neglect friction in the system, then mechanical energy is conserved and we can state 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
`Delta K = K_f = K_i`
=`(1)/(2) m_1 v^2 + (1)/(2) m_2 v^2 + (1)/(2) I omega^2`
where `m_1` and `m_2` have a common speed. But
`v = R omega ` so that `Delta K = (1)/(2) (m_1 + m_2 + (I)/(R^2))V^2`
From FIG. we see that the system loses potential energy because of the motion of `m_1` and gains potential energy because of the motion of `m_2`. Applying the law of conservation of energy,
`Delta K +_ Delta U = 0`, gives
`(1)/(2) (m_1 + m_2 + (I)/(R^2))v^2 + m_2 gh - m_1 gh = 0`
`v = sqrt((2(m_1 - m_2)gh)/(m_1 + m_2 + (I)/(R^2)))`.