A solid sphere is in rolling motion. In rolli | vedclass

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(B) Translational kinetic energy is given by $K_t = \frac{1}{2}mv^2$.Rotational kinetic energy is given by $K_r = \frac{1}{2}I\omega^2$.For a solid sp

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$5:7$

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(B) Translational kinetic energy is given by $K_t = \frac{1}{2}mv^2$.Rotational kinetic energy is given by $K_r = \frac{1}{2}I\omega^2$.For a solid sphere,the moment of inertia is $I = \frac{2}{5}mr^2$ and the relation between linear and angular velocity is $v = r\omega$,so $\omega = \frac{v}{r}$.Substituting these into the rotational kinetic energy formula:$K_r = \frac{1}{2} \left( \frac{2}{5}mr^2 ight) \left( \frac{v}{r} ight)^2 = \frac{1}{5}mv^2$.The total kinetic energy is $K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \left( \frac{5+2}{10} ight)mv^2 = \frac{7}{10}mv^2$.The ratio $K_t : (K_t + K_r)$ is $\frac{\frac{1}{2}mv^2}{\frac{7}{10}mv^2} = \frac{1}{2} 	imes \frac{10}{7} = \frac{5}{7}$.

$A$ 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 is

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