$A$ circular disc of mass $M$ and radius $R$ is rotating about its axis with angular speed $\omega_{1}$. If another stationary disc having radius $\frac{R}{2}$ and same mass $M$ is dropped co-axially onto the rotating disc,gradually both discs attain a constant angular speed $\omega_{2}$. The energy lost in the process is $p \%$ of the initial energy. The value of $p$ is:

$A$ thin uniform circular disc of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity $\omega$. Another disc of same dimension but of mass $M/4$ is placed gently on the first disc coaxially. The angular velocity of the system now is

The angular velocity of a body changes from $\omega_1$ to $\omega_2$ without the application of any external torque,but this change occurs due to a change in its moment of inertia. The ratio of the radii of gyration in the two cases will be:

The position vector of a particle is given by $\vec{r} = 2\hat{i} - 6\hat{j} - 12\hat{k}$ units. $A$ force $\vec{F} = p\hat{i} + 3\hat{j} + 6\hat{k}$ units acts on it. For what value of $p$ is the angular momentum conserved?

$A$ smooth uniform rod of length $L$ and mass $M$ has two identical beads of negligible size,each of mass $m$,which can slide freely along the rod. Initially,the two beads are at the center of the rod and the system is rotating with angular velocity ${\omega _0}$ about an axis perpendicular to the rod and passing through the midpoint of the rod (see figure). There are no external torques. When the beads reach the ends of the rod,the angular velocity of the system is

Angular momentum of a particle rotating under a central force is constant due to