A ring and a disc of same mass and same diameter are rolling without slipping. Their linear velocities are same, then the ratio of their kinetic energy is

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Aicharya · Accepted Answer

For ring, $\frac{\mathrm{K}^2}{\mathrm{R}^2}=1$ For disc, $\frac{\mathrm{K}^2}{\mathrm{R}^2}=\frac{1}{2}$ $\begin{aligned} & \frac{(\mathrm{K} . \mathrm{E})_{	ext {ring }}}{(\mathrm{K} . \mathrm{E})_{	ext {disc }}}=\frac{\left[\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}ight)ight]_{	ext {ring }}}{\left[\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}ight)ight]_{	ext {disc }}} \ & =\frac{(1+1)}{\left(1+\frac{1}{2}ight)}=\frac{2 	imes 2}{3}=1.33 \end{aligned}$

A ring and a disc of same mass and same diameter are rolling without slipping. Their linear velocities are…

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