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 $\theta$.

$A$ rod of mass $M$ and length $l$ is at rest on a smooth horizontal surface. $A$ particle of the same mass $M$ strikes one end of the rod with velocity $u$ perpendicular to the rod,elastically. Just after the collision,what is the kinetic energy of the upper half part of the rod?

$A$ homogeneous disc of mass $2 \ kg$ and radius $15 \ cm$ is rotating about its axis (which is fixed) with an angular velocity $4 \ rad/s$. The linear momentum of the disc is

$A$ uniform sphere of mass $m$ and radius $R$ is placed on a rough horizontal surface. The sphere is struck horizontally at a height $h$ from the floor. Match the following:

$(a)$ $h = \frac{R}{2}$	$(i)$ Sphere rolls without slipping with a constant velocity and no loss of energy.
$(b)$ $h = R$	$(ii)$ Sphere spins clockwise,loses energy by friction.
$(c)$ $h = \frac{3R}{2}$	$(iii)$ Sphere spins anti-clockwise,loses energy by friction.
$(d)$ $h = \frac{7R}{5}$	$(iv)$ Sphere has only a translational motion,loses energy by friction.

Two bodies have moments of inertia $I_1 = I$ and $I_2 = 2I$ about their axes of rotation. If their rotational kinetic energies are equal,what is the ratio of their angular momenta?

$A$ $2 \, kg$ steel rod of length $0.6 \, m$ is clamped on a table vertically at its lower end and is free to rotate in a vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end,the speed of the free end of the rod when it passes through its lowest position is $\ldots \ldots \ldots \ldots \, ms^{-1}$. (Take $g = 10 \, ms^{-2}$)