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If a spherical ball rolls on a table without slipping the friction of its total energy associated with rotational energy is :
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$2 / 7$
Rotational energy of sphere
$E_R=\frac{1}{2} I \omega^2$
For sphere, moment of inertia
$\begin{aligned} I & =\frac{2}{5} m R^2 \\ \therefore \quad E_R & =\frac{1}{2}\left(\frac{2}{5} m R^2\right)\left(\frac{v}{R}\right)^2 \\ & =\frac{1}{5} m v^2\end{aligned}$
Translational kinetic energy $E_r=\frac{1}{2} m v^2$
$\begin{aligned} \therefore \text { Total energy } & =\frac{1}{5} m v^2+\frac{1}{2} m v^2 \\ & =\frac{7}{10} m v^2\end{aligned}$
$\therefore$ Required fraction $=\frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2}$
$=\frac{2}{7}$
$E_R=\frac{1}{2} I \omega^2$
For sphere, moment of inertia
$\begin{aligned} I & =\frac{2}{5} m R^2 \\ \therefore \quad E_R & =\frac{1}{2}\left(\frac{2}{5} m R^2\right)\left(\frac{v}{R}\right)^2 \\ & =\frac{1}{5} m v^2\end{aligned}$
Translational kinetic energy $E_r=\frac{1}{2} m v^2$
$\begin{aligned} \therefore \text { Total energy } & =\frac{1}{5} m v^2+\frac{1}{2} m v^2 \\ & =\frac{7}{10} m v^2\end{aligned}$
$\therefore$ Required fraction $=\frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2}$
$=\frac{2}{7}$
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