The speed of rolling of a ring of mass $M$ changes from $V$ to $3\ V$. What is the change in its kinetic energy (in $,MV^2$)?

$A$ sphere cannot roll on 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 $g=10 \text{ m s}^{-2}$.]

$A$ uniform disc of mass $0.5\,kg$ and radius $r$ is projected with velocity $18\,m/s$ at $t = 0\,s$ on a rough horizontal surface. It starts off with a purely sliding motion at $t = 0\,s$. After $2\,s$ it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after $2\,s$ will be $..............J$ (given,coefficient of friction is $0.3$ and $g = 10\,m/s^2$).

$A$ sphere of mass $0.5 \ kg$ and diameter $1 \ m$ is rolling without slipping with a constant velocity of $5 \ m/s$. The ratio of its rotational kinetic energy to its total kinetic energy is:

$A$ wheel is rolling on the ground with a speed of $2 \ m/s$. What is the velocity of the endpoints of the horizontal diameter of the wheel?