$A$ thin uniform rod of length $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of:

$A$ 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:

$A$ roller is made by joining together two cones at their vertices $O$. It is kept on two rails $AB$ and $CD$,which are placed asymmetrically (see figure),with its axis perpendicular to $CD$ and its centre $O$ at the centre of the line joining $AB$ and $CD$ (see figure). It is given a light push so that it starts rolling with its centre $O$ moving parallel to $CD$ in the direction shown. As it moves,the roller will tend to

$A$ solid cube of wood of side $2a$ and mass $M$ is resting on a horizontal surface as shown below. The cube is free to rotate about the fixed axis $AB$. $A$ bullet of mass $m (< M)$ and speed $v$ is shot horizontally at the face opposite to $ABCD$ at a height $h$ above the surface to impart the cube an angular speed $\omega_{c}$,so that the cube just topples over. Then,$\omega_{c}$ is (Note: the moment of inertia of the cube about an axis passing through the centre of mass and parallel to the edge is $2Ma^{2}/3$)

According to the parallel axis theorem,$I = I_{cm} + Mx^2$. What will be the graph between $I$ and $x$?

$A$ hemisphere of mass $3m$ and radius $R$ is free to slide with its base on a smooth horizontal table. $A$ particle of mass $m$ is placed on the top of the hemisphere. If the particle is displaced with a negligible velocity,find the angular velocity of the particle relative to the centre of the hemisphere at an angular displacement $\theta$,when the velocity of the hemisphere is $v$.