$A$ solid sphere is in purely rotational motion about its diameter. The ratio of its angular momentum $(L)$ and kinetic energy $(K)$ is $\frac{\pi}{22}$. Find the angular velocity $(\omega)$ of the sphere. (Take $\pi = \frac{22}{7}$) (in $rad/s$)

$A$ uniform circular disc of radius $R$,lying on a frictionless horizontal plane,is rotating with an angular velocity $\omega$ about its own axis. Another identical circular disc is gently placed on top of the first disc coaxially. The loss in rotational kinetic energy due to friction between the two discs,as they acquire a common angular velocity,is ($I$ is the moment of inertia of the disc).

Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre) and rotating with angular speeds $\omega_1$ and $\omega_2$ are brought into contact face to face with their axes of rotation coincident.
$(a)$ Does the law of conservation of angular momentum apply to the situation? Why?
$(b)$ Find the angular speed of the two-disc system.
$(c)$ Calculate the loss in kinetic energy of the system in the process.
$(d)$ Account for this loss.

$A$ uniform rod $AB$ of mass $2 \ kg$ and length $30 \ cm$ is at rest on a smooth horizontal surface. An impulse of force $0.2 \ Ns$ is applied to end $B$. The time taken by the rod to turn through a right angle will be $\frac{\pi}{X} \ s$,where $X = \text{ . . . . . . }$.

$A$ uniform circular disc placed on a rough horizontal surface has initially a velocity $v_0$ and an angular velocity $\omega_0$ as shown in the figure. The disc comes to rest after moving some distance in the direction of motion. Then $\frac{v_0}{r\omega_0}$ is

$A$ solid sphere of mass $m$,radius $R$,having moment of inertia about an axis passing through its center of mass as $I$ is recast into a disc of thickness $t$ whose moment of inertia about an axis passing through the rim (edge) and perpendicular to its plane remains $I$. Then the radius of the disc is: