Two coaxial discs, having moments of inertia $I_1$ and $\frac{I_1}{2}$ are a rotating with respectively angular velocities $\omega_1$ and $\frac{\omega_1}{2}$, about their common axes. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If $E_f$ and $E_i$ are the final and initial total energies, then $(E_f -E_i)$ is
$\frac{{{I_1}\omega _1^2}}{6}$
$\frac{3}{8}{I_1}\omega _1^2$
$ - \frac{{{I_1}\omega _1^2}}{{12}}$
$ - \frac{{{I_1}\omega _1^2}}{{24}}$
A solid sphere rolls down without slipping on an inclined plane, then percentage of rotational kinetic energy of total energy will be ........ $\%.$
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
$A$ ring of mass $m$ is rolling without slipping with linear velocity $v$ as shown is figure. $A$ rod of identical mass is fixed along one of its diameter. The total kinetic energy of the system is :-
A particle performs uniform circular motion with an angular momentum $L.$ If the angular frequency of the particle is doubled and kinetic energy is halved, its angular momentum becomes