$A$ disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in the figure. At any instant,for the lowermost point of the disc -

$A$ body is rolling without slipping on a horizontal plane. If the rotational kinetic energy of the body is $50 \%$ of its total kinetic energy, then the body is

$A$ 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 $g = 10\,m/s^2$).

If a solid sphere is rolling without slipping on a horizontal plane,then the ratio of its rotational and total kinetic energies is

$A$ solid sphere is in rolling motion. In rolling motion,a body possesses translational kinetic energy $(K_t)$ as well as rotational kinetic energy $(K_r)$ simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere is

$A$ solid sphere of mass $2 \ kg$ and radius $0.5 \ m$ is rolling without slipping on a horizontal surface. The ratio of the rotational and translational kinetic energies of the sphere is (in $: 5$)