Consider a cylinder of mass $M$ resting on a rough horizontal rug that is pulled out from under it with acceleration $a$ perpendicular to the axis of the cylinder. What is the friction force $F_{friction}$ at point $P$? It is assumed that the cylinder does not slip.

$A$ spherical shell of $1 \, kg$ mass and radius $R$ is rolling with angular speed $\omega$ on a horizontal plane (as shown in the figure). The magnitude of the angular momentum of the shell about the origin $O$ is $\frac{a}{3} R^{2} \omega$. The value of $a$ will be ..............

$A$ hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is $\frac{x}{5}$. The value of $x$ is . . . . . . .

$A$ disc is rolling (without slipping) on a horizontal surface. $C$ is its centre and $Q$ and $P$ are two points on the same horizontal line passing through $C$,such that $Q$ is at a distance $r$ from $C$ and $P$ is at a distance $r$ from $C$ on the opposite side. Let $V_P, V_Q$ and $V_C$ be the magnitudes of velocities of points $P, Q$ and $C$ respectively,then:

$A$ solid sphere of radius $R$ is rolled by a force $F$ acting at the top of the sphere as shown in the figure. The sphere rolls without slipping on a rough stationary surface. Initially,the sphere is at rest. Then:

$A$ solid sphere is in rolling motion. In rolling motion,a body possesses translational kinetic energy $(K_t)$ as well as rotational kinetic energy $(K_r)$ simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere is