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 :
Solution
<p>$$\begin{aligned}
& \frac{\text { Linear KE }}{\text { Rotational K.E }}=\frac{\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^2}{\frac{1}{2} \mathrm{I} \omega^2} \\
& \frac{\mathrm{mv}_{\mathrm{cm}}^2}{\frac{2}{5} \mathrm{mR}^2 \omega^2}=\frac{5}{2} \quad(\mathrm{~V}=\omega \mathrm{R})
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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