A solid sphere rolls without slipping on an inclied plane at an angle $\theta$. The ratio of total kinetic energy to its rotational kinetic energy is

For a solid sphere rolling without slipping, the velocity of the center of mass $v$ and angular velocity $\omega$ are related by $v = \omega R$.

The translational kinetic energy is $K_{trans} = \dfrac{1}{2} m v^2$.

The rotational kinetic energy is $K_{rot} = \dfrac{1}{2} I \omega^2$.

Substituting the moment of inertia of a solid sphere $I = \dfrac{2}{5} m R^2$ and $\omega = \dfrac{v}{R}$:

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

The total kinetic energy is the sum of translational and rotational kinetic energies:

$K_{total} = K_{trans} + K_{rot} = \dfrac{1}{2} m v^2 + \dfrac{1}{5} m v^2 = \dfrac{7}{10} m v^2$

The ratio of total kinetic energy to rotational kinetic energy is:

$\dfrac{K_{total}}{K_{rot}} = \dfrac{\dfrac{7}{10} m v^2}{\dfrac{1}{5} m v^2} = \dfrac{7}{2}$

Answer: $\dfrac{7}{2}$