Question

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 $v_{rel}$ of the particle with respect to the cart when the rod is vertical, then

• A. $\sqrt{\frac{14}{3}gl}$
• B. $\sqrt{\frac{7}{6}gl}$
• C. $\sqrt{\frac{7}{3}gl}$
• D. $\sqrt{\frac{8}{3}gl}$

Answer 1

The correct option is C $\sqrt{\frac{7}{3}gl}$

Let $v_1,v_2$ be velocities of $m,6m$ respectively when rod is vertical


Conservation of momentum,
When the rod becomes vertical
$m{v}_1=6m{v}_2\Rightarrow v_1=6v_2$

Conservation of energy at this instant,
$\frac{1}{2}m{v}_1^2+\frac{1}{2}6m{v}_2^2=mgl$
$\frac{1}{2}m(6v_2{)}^2+\frac{1}{2}6m{v}_2^2=mgl$
$21v_2^2=gl$
$v_2=\sqrt{\frac{gl}{21}}$
$v_1=6\sqrt{\frac{gl}{21}}$
Relative velocity
$v_{rel}=v_1+v_2=7\sqrt{\frac{gl}{21}}=\sqrt{\frac{7}{3}gl}$