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6.System of Particles and Rotational Motion
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As shown in the figure, a bob of mass $\mathrm{m}$ is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius $\mathrm{r}$ and mass $m$. When released from rest the bob starts falling vertically. When it has covered a distance of $h$. the angular speed of the wheel will be
A
$\frac{1}{r} \sqrt{\frac{2 g h}{3}}$
B
$ r \sqrt{\frac{3}{4 g h}}$
C
$ \frac{1}{r} \sqrt{\frac{4 g h}{3}}$
D
${r} \sqrt{\frac{3}{2 g h}}$
(JEE MAIN-2020)
Solution
$\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\frac{3}{4} \mathrm{mv}^{2}$
$\mathrm{u}=\sqrt{\frac{4}{3} \mathrm{gh}}$
$\omega=\frac{V}{r}$
$\omega= \frac{1}{r} \sqrt{\frac{4 g h}{3}}$
Standard 11
Physics
