$A$ light rod carries three equal masses $A$,$B$,and $C$ as shown in the figure. If the rod is released from the horizontal position,what will be the velocity of mass $B$ when the rod reaches the vertical position?

Explain the work done by torque.

$A$ uniform bar of length $6l$ and mass $8m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ moving in the same horizontal plane with speeds $2v$ and $v$ respectively,strike the bar (as shown in the figure) and stick to the bar after the collision. The total energy after the collision (about the center of mass,$c$) will be:

$A$ thin uniform rod of mass $1 \ kg$ and length $1 \ m$ is hinged at one end to the ground. It originally stands vertically and is allowed to fall to the ground. If the rod hits the ground with angular speed $\omega$, then the correct statement is (Assume $g = 10 \ m \ s^{-2}$):

When $12000 \ J$ of work is done on a flywheel,its frequency of rotation increases from $10 \ Hz$ to $20 \ Hz$. The moment of inertia of the flywheel about its axis of rotation is (take $\pi^2=10$):

Write the formula of work done by torque in a rotational rigid body about a fixed axis.