A sphere of mass m rolls along a plane surface. Determine its...

The total kinetic energy of a rolling body is the sum of its translational and rotational kinetic energies.

$K = K_{trans} + K_{rot}$

$K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

For a solid sphere, the moment of inertia about its centre of mass is $I = \dfrac{2}{5}mR^2$.

For pure rolling, the relation between linear speed and angular speed is $v = \omega R$, which implies $\omega = \dfrac{v}{R}$.

Substituting the values of $I$ and $\omega$ into the rotational kinetic energy term:

$K_{rot} = \dfrac{1}{2} \left(\dfrac{2}{5}mR^2\right) \left(\dfrac{v}{R}\right)^2 = \dfrac{1}{5}mv^2$

Adding both kinetic energies:

$K = \dfrac{1}{2}mv^2 + \dfrac{1}{5}mv^2 = \dfrac{7}{10}mv^2$