$A$ hoop of radius $r$ and mass $m$ rotating with an angular velocity $\omega_0$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

Consider a disc rotating in the horizontal plane with a constant angular speed $\omega$ about its centre $O$. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown,two pebbles $P$ and $Q$ are simultaneously projected at an angle towards $R$. The velocity of projection is in the $y-z$ plane and is same for both pebbles with respect to the disc. Assume that $(i)$ they land back on the disc before the disc completes $\frac{1}{8}$ rotation,$(ii)$ their range is less than half disc radius,and $(iii)$ $\omega$ remains constant throughout. Then

$A$ particle executes uniform circular motion with angular momentum $L$. Its rotational kinetic energy becomes half,when the angular frequency is doubled. Its new angular momentum is

$A$ ring of mass $m$ and radius $R$ has three particles attached to the ring as shown in the figure. The centre of the ring has a speed $v_0$. The kinetic energy of the system is: (Slipping is absent) (in $, mv_0^2$)

Find the minimum height $h$ of the obstacle so that the sphere of radius $R$ can stay in equilibrium on an inclined plane of angle $\theta$.

Two point-like objects of masses $20 \text{ g}$ and $30 \text{ g}$ are fixed at the two ends of a rigid massless rod of length $10 \text{ cm}$. This system is suspended vertically from a rigid ceiling using a thin wire attached to its center of mass,as shown in the figure. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is $1.2 \times 10^{-8} \text{ N m rad}^{-1}$. The angular frequency of the oscillations is $n \times 10^{-3} \text{ rad s}^{-1}$. The value of $n$ is