Question 493: Centre of Mass and Rotational Motion

Question

A solid sphere of mass ' $m$ ' and radius ' $R$ ' is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with angular speed twice that of sphere. The ratio of kinetic energy of sphere to kinetic energy of cylinder will be

Accepted Answer

$$ I_{	ext {sphere }}=I_{\mathrm{s}}=\frac{2}{5} m R^2 $$
Let $\omega_s$ be angular speed of sphere,
$$ \begin{aligned} 	herefore \quad E_{	ext {sphere }} & =\frac{1}{2} I_s \omega_s^2 \ & =\frac{1}{2}\left(\frac{2}{5} m R^2ight) \omega_s^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $$
Similarly,
$$ \mathrm{I}_{	ext {cylinder }}=\mathrm{I}_{\mathrm{c}}=\frac{1}{2} \mathrm{mR}^2 $$
Let $\omega_c$ be the angular speed of cylinder, Then it is given
$$ \begin{aligned} \omega_c=2 \omega_s & \ 	herefore \quad \mathrm{E}_{	ext {cylinder }} & =\frac{1}{2} \mathrm{I}_c \omega_c^2 \ & =\frac{1}{2}\left(\frac{1}{2} \mathrm{mR}^2ight)\left(2 \omega_sight)^2 \quad \ldots 	ext { (ii) } \ 	herefore \quad \frac{\mathrm{E}_{	ext {sphere }}}{\mathrm{E}_{	ext {cylinder }}} & =\frac{\frac{1}{2}\left(\frac{2}{5} \mathrm{mR}^2ight) \omega_s^2}{\frac{1}{2}\left(\frac{1}{2} \mathrm{mR}^2ight)\left(4 \omega_s^2ight)} \quad \ldots[	ext { From (i) and (ii) }] \ & =\frac{1}{5} \end{aligned} $$

Answer

$2: 3$

Answer

$1: 5$

Answer

$3: 1$

Answer

$1: 4$

Centre of Mass and Rotational Motion

A solid sphere of mass ' m ' and radius ' R ' is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with angular speed twice that of sphere. The ratio of kinetic energy of sphere to kinetic energy of cylinder will be

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