$A$ thin hollow cylinder open at both ends is moving with the same speed $v$ in two different cases:
$(i)$ It slides without rotating.
$(ii)$ It rolls without slipping.
Find the ratio of the kinetic energies in the two cases.

$A$ 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?

Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is $I$. If the same rod is bent into a ring and its moment of inertia about its diameter is $I^{\prime}$,then the ratio $\frac{I}{I^{\prime}}$ 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 the plane with angular velocity $\omega$. Another disc of same mass but half the radius is gently placed over it coaxially. The angular speed of the composite disc will be :

$A$ rod of mass $M$ and length $L$ is lying on a horizontal frictionless surface. $A$ particle of mass $m$ travelling along the surface hits at one end of the rod with velocity $u$ in a direction perpendicular to the rod. The collision is completely elastic. After the collision, the particle comes to rest. The ratio of masses $\left(\frac{m}{M}\right)$ is $\frac{1}{x}$. The value of $x$ will be ..... .

$A$ uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$,the rod makes an angle $\theta$ with it (see figure). To find $\theta$,equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that: