Question 484: Centre of Mass and Rotational Motion

Question

A bob of mass ' $m$ ' is tied by a massless string whose other end is wound on a flywheel (disc) of radius ' $R$ ' and mass ' $m$ '. When released from the rest, the bob starts falling vertically downwards. If the bob has covered a vertical distance ' $h$ ', then angular speed of wheel will be (There is no slipping between string and wheel, g - acceleration due to gravity)

Accepted Answer

$$ \begin{aligned} &	ext { According to law of conservation of energy, }\ &\begin{aligned} & 	ext { P.E }=	ext { K.E } \ 	herefore \quad & \mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2 \ 	herefore \quad \mathrm{mgh}= & \frac{1}{2} \mathrm{~m}(\omega r)^2+\frac{1}{2}\left(\frac{\mathrm{mr}^2}{2}ight) \omega^2 \ & \quad \ldots .\left(\because \mathrm{v}=\mathrm{r} \omega, \mathrm{I}=\frac{\mathrm{mr}^2}{2}ight) \end{aligned} \end{aligned} $$
$$ \begin{array}{ll} 	herefore & \mathrm{mgh}=\frac{3}{4} \mathrm{~m}^2 \mathrm{r}^2 \ 	herefore & \mathrm{gh}=\frac{3}{4} \omega^2 \mathrm{r}^2 \ 	herefore & \omega=\sqrt{\frac{4 \mathrm{gh}}{3 \mathrm{r}^2}}=\frac{2}{\mathrm{r}} \sqrt{\frac{\mathrm{gh}}{3}} \end{array} $$

Answer

$\frac{2}{\mathrm{R}} \sqrt{\frac{\mathrm{gh}}{3}}$

Answer

$\frac{1}{\mathrm{R}} \sqrt{\frac{2 \mathrm{gh}}{3}}$

Answer

$\mathrm{R} \sqrt{\frac{2 \mathrm{gh}}{3}}$

Answer

$2 R \sqrt{\frac{g h}{3}}$

Centre of Mass and Rotational Motion

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