$A$ wheel of radius $R$ rolls on the ground with a uniform velocity $v$. The relative acceleration of the topmost point of the wheel with respect to the bottommost point is

$A$ circular hoop of radius $50 \ cm$ and mass $1 \ kg$ 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. Let $v$ be the velocity of the centre of the hoop when it ceases to slip. The ratio $v / \omega_0$ will be (in $cm$)

$A$ solid sphere spinning about a horizontal axis with an angular velocity $\omega$ is placed on a horizontal surface. Subsequently,it rolls without slipping with an angular velocity of

$A$ thin circular ring and a circular disc of equal mass are rolling without sliding. If their linear velocities are equal and the total kinetic energy of the disc is $6 \ J$,then the total kinetic energy of the ring is (in $J$)

$A$ body is rolling on a horizontal surface without slipping. Its rotational kinetic energy is equal to its translational kinetic energy. The body is a .......

At time $t=0$,a disk of radius $1 \text{ m}$ starts to roll without slipping on a horizontal plane with an angular acceleration of $\alpha = \frac{2}{3} \text{ rad s}^{-2}$. $A$ small stone is stuck to the disk. At $t=0$,it is at the contact point of the disk and the plane. Later,at time $t=\sqrt{\pi} \text{ s}$,the stone detaches itself and flies off tangentially from the disk. The maximum height (in $\text{m}$) reached by the stone measured from the plane is $\frac{1}{2} + \frac{x}{10}$. The value of $x$ is. . . . . . . [Take $g=10 \text{ m s}^{-2}$.]