JEE MAIN - Physics (2025 - 24th January Evening Shift - No. 16)
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 :
$\frac{3}{4}$
$\frac{4}{3}$
$\frac{5}{2}$
$\frac{2}{5}$
Explanation
$$\begin{aligned}
& \frac{\text { Linear KE }}{\text { Rotational K.E }}=\frac{\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^2}{\frac{1}{2} \mathrm{I} \omega^2} \\
& \frac{\mathrm{mv}_{\mathrm{cm}}^2}{\frac{2}{5} \mathrm{mR}^2 \omega^2}=\frac{5}{2} \quad(\mathrm{~V}=\omega \mathrm{R})
\end{aligned}$$
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