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A solid sphere of mass ' $M$ ' and radius ' $R$ ' is rotating about its diameter. A solid cylinder of same mass and radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation ( $\mathrm{K}_{\text {sphere }}$ to $\mathrm{K}_{\text {cylinder }}$ ) will be
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Verified Answer
The correct answer is:
1: 5
Kinetic energy of sphere is given by:
$$
\begin{aligned}
& E_{\text {sphere }}=\frac{1}{2} I_{\text {sphere }} \omega_{\text {sphere }}^2 \\
& E_{\text {sphere }}=\frac{1}{2} \times \frac{2}{5} m R^2 \omega_{\text {sphere }}^2
\end{aligned}
$$
Kinetic energy of cylinder is given by:
$$
\begin{aligned}
& E_{\text {cylinder }}=\frac{1}{2} I_{\text {cylinder }} \omega_{\text {cylinder }}^2 \\
& E_{\text {cylinder }}=\frac{1}{2} \times \frac{1}{2} m R^2 \omega_{\text {cylinder }}^2 \\
& \text { Given, } \omega_{\text {cylinder }}=2 \omega_{\text {sphere }} \\
& \frac{E_{\text {sphere }}}{E_{\text {cylinder }}}=\frac{1}{5}
\end{aligned}
$$
$$
\begin{aligned}
& E_{\text {sphere }}=\frac{1}{2} I_{\text {sphere }} \omega_{\text {sphere }}^2 \\
& E_{\text {sphere }}=\frac{1}{2} \times \frac{2}{5} m R^2 \omega_{\text {sphere }}^2
\end{aligned}
$$
Kinetic energy of cylinder is given by:
$$
\begin{aligned}
& E_{\text {cylinder }}=\frac{1}{2} I_{\text {cylinder }} \omega_{\text {cylinder }}^2 \\
& E_{\text {cylinder }}=\frac{1}{2} \times \frac{1}{2} m R^2 \omega_{\text {cylinder }}^2 \\
& \text { Given, } \omega_{\text {cylinder }}=2 \omega_{\text {sphere }} \\
& \frac{E_{\text {sphere }}}{E_{\text {cylinder }}}=\frac{1}{5}
\end{aligned}
$$
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