A solid sphere is in rolling motion. In rolling motion, a body possesses translational kinetic energy ($K_t$)

Question

A solid sphere is in rolling motion. In rolling motion, a body possesses translational kinetic energy ($K_t$) as well as rotational kinetic energy ($K_r$) simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere will be:

Accepted Answer

For a solid sphere in rolling motion:
Translational kinetic energy, $K_t = \frac{1}{2}mv^2$
Rotational kinetic energy, $K_r = \frac{1}{2}I\omega^2$
The moment of inertia of a solid sphere about its diameter is $I = \frac{2}{5}mr^2$. In pure rolling, $v = r\omega$.
Substituting these values:
$K_r = \frac{1}{2}\left(\frac{2}{5}mr^2ight)\left(\frac{v}{r}ight)^2 = \frac{1}{5}mv^2$
Total kinetic energy, $K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
The required ratio is $\frac{K_t}{K_t + K_r} = \frac{\frac{1}{2}mv^2}{\frac{7}{10}mv^2} = \frac{1}{2} 	imes \frac{10}{7} = \frac{5}{7}$.
Therefore, the ratio is $5:7$.

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