Derive expressions for the kinetic energy and velocity of a body rolling without sliding down an inclined plane of inclination $\theta$.

(N/A) Consider a body of mass $m$,moment of inertia about its geometric axis $I$,radius of gyration $k$,and geometric radius $R$ rolling down an inclined plane of inclination $\theta$ and height $h$ without slipping.
Since the body is rolling without slipping,its center of mass moves with linear velocity $v_{cm}$ and the body rotates about its axis with angular velocity $\omega$. The motion of the body is a combination of translation and rotation.
The total kinetic energy $K$ of the body is given by:
$K = K_{\text{translational}} + K_{\text{rotational}}$
$K = \frac{1}{2} m v_{cm}^{2} + \frac{1}{2} I \omega^{2}$
Given $I = m k^{2}$ and the condition for rolling without slipping $v_{cm} = R \omega$,we have $\omega = \frac{v_{cm}}{R}$.
Substituting these into the kinetic energy equation:
$K = \frac{1}{2} m v_{cm}^{2} + \frac{1}{2} (m k^{2}) \left( \frac{v_{cm}}{R} \right)^{2}$
$K = \frac{1}{2} m v_{cm}^{2} \left( 1 + \frac{k^{2}}{R^{2}} \right)$
This is the formula for the total kinetic energy of a rolling body.
To find the velocity at the bottom,we use the law of conservation of energy. The potential energy at the top is converted into the total kinetic energy at the bottom:
$m g h = K$
$m g h = \frac{1}{2} m v^{2} \left( 1 + \frac{k^{2}}{R^{2}} \right)$
Solving for $v$:
$v^{2} = \frac{2 g h}{1 + \frac{k^{2}}{R^{2}}}$
$v = \sqrt{\frac{2 g h}{1 + \frac{k^{2}}{R^{2}}}}$