In the figure shown the cart of mass 6m is initially at rest. A particle of mass m is attached to the end of the light rod which can rotate freely about A. If the rod is released from rest in a horizontal position shown, determine the velocity $v_{r e l}$ of the particle with respect to the cart when the rod is vertical.

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Solution

The correct option is C

When the rod becomes vertical

$m v_{1} = 6 m v_{2} \Rightarrow v_{1} = 6 v_{2}$ .........(i)

$\frac{1}{2} m v_{1}^{2} + \frac{1}{2} 6 m v_{2}^{2} = m g l$ .........(ii)

Solve (i) and (ii) to get $V_{r e l} = v_{1} + v_{2} = \sqrt{\frac{7}{3} g l}$

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