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

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

The translational kinetic energy is $K_{t r an s} = \frac{1}{2} m v^{2}$ .

The rotational kinetic energy is $K_{ro t} = \frac{1}{2} I ω^{2}$ .

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

$K_{ro t} = \frac{1}{2} (\frac{2}{5} m R^{2}) (\frac{v}{R})^{2} = \frac{1}{5} m v^{2}$

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

$K_{t o t a l} = K_{t r an s} + K_{ro t} = \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2} = \frac{7}{10} m v^{2}$

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

$\frac{K _{t o t a l}}{K _{ro t}} = \frac{\frac{7}{10} m v ^{2}}{\frac{1}{5} m v ^{2}} = \frac{7}{2}$

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

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