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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
$7:10$
$5:7$
$2:5$
$10:7$
Solution
Translational kinetic energy, ${K_t} = \frac{1}{2}m{v^2}$
Rotational kinetic energy, ${K_r} = \frac{1}{2}I{\omega ^2}$
$\begin{array}{ccccc}
\therefore \,{K_t} + {K_r} = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}m{v^2} + \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\left( {\frac{v}{r}} \right)^2}\\
\therefore \,{K_t} + {k_r} = \frac{7}{{10}}m{v^2}\,\,\,\,\,\,\left[ {I = \frac{2}{5}m{r^2}\left( {for\,sphere} \right)} \right]\\
So,\,\frac{{{K_t}}}{{{K_t} + {K_r}}} = \frac{5}{7}
\end{array}$
