How is the motion of a rolling sphere down a slope? Write the general formula for the kinetic energy of a rolling body down a slope.

(N/A) The motion of a sphere rolling down a slope is a combination of translational motion (of the center of mass) and rotational motion (about the center of mass). As the sphere rolls without slipping,it accelerates due to the component of gravity acting along the incline $(mg \sin \theta)$.
The total kinetic energy $(K)$ of a rolling body is the sum of its translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$:
$K = K_t + K_r$
$K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since the body rolls without slipping,$v = r\omega$,where $v$ is the linear velocity,$r$ is the radius,and $\omega$ is the angular velocity. Substituting $I = mk^2$ (where $k$ is the radius of gyration) and $\omega = v/r$:
$K = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(v/r)^2$
$K = \frac{1}{2}mv^2 (1 + \frac{k^2}{r^2})$