$A$ sphere is rolling without slipping on a fixed horizontal plane surface. In the figure,$A$ is the point of contact,$B$ is the centre of the sphere,and $C$ is its topmost point. Then:
$(i) \vec{V}_C - \vec{V}_A = 2(\vec{V}_B - \vec{V}_C)$
$(ii) \vec{V}_C - \vec{V}_B = \vec{V}_B - \vec{V}_A$
$(iii) |\vec{V}_C - \vec{V}_A| = 2|\vec{V}_B - \vec{V}_C|$
$(iv) |\vec{V}_C - \vec{V}_A| = 4|\vec{V}_B|$

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 $g=10 \text{ m s}^{-2}$.]

Obtain the necessary condition $v_{cm} = R\omega$ for a body rolling without slipping.

$A$ wheel is rolling straight on ground without slipping. If the axis of the wheel has speed $v$,the instantaneous velocity of a point $P$ on the rim,defined by angle $\theta$ relative to the vertical,relative to the ground will be:

An object is rolling on a horizontal surface without slipping. If its rotational kinetic energy and translational kinetic energy are equal,the object is a:

$A$ sphere is rolling without slipping on a fixed horizontal plane surface. $A$ is the point of contact,$B$ is the centre of the sphere,and $C$ is the topmost point. Then: