$A$ thin circular ring of mass $m$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two objects each of mass $M$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $\omega '$ equal to:

The total angular momentum of all particles of a system is conserved if:

$A$ force $\overrightarrow{F} = \alpha \hat{i} + 3\hat{j} + 6\hat{k}$ is acting at a point $\overrightarrow{R} = 2\hat{i} - 6\hat{j} - 12\hat{k}$. The value of $\alpha$ for which angular momentum about the origin is conserved is

$Assertion$ : For a system of particles under a central force field,the total angular momentum is conserved.
$Reason$ : The torque acting on such a system is zero.

$A$ swimmer curls their body before entering the water so that...

$A$ uniform circular disc of mass $50 \ kg$ and radius $0.4 \ m$ is rotating with an angular velocity of $10 \ rad \ s^{-1}$ about its own axis,which is vertical. Two uniform circular rings,each of mass $6.25 \ kg$ and radius $0.2 \ m$,are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in $rad \ s^{-1}$) of the system is: