A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :

Question

Aicharya · Accepted Answer

$\begin{aligned}
& \mathrm{KE}_{(T)}=\frac{1}{2} m v^2 \\ & \mathrm{KE}_{(R)}=\frac{1}{2} \cdot \frac{2}{5} m R^2 \cdot \frac{v^2}{R^2}=\frac{1}{2} m v^2\left(\frac{2}{5}\right)
\end{aligned}$
So, $\frac{\mathrm{KE}_{(T)}}{\mathrm{KE}_{(R)}}=\frac{5}{2}$

A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of…

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