JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 \({r} \sqrt{\frac{3}{2 g h}}\)
Answer & Solution
Correct Answer
(C) \( \frac{1}{r} \sqrt{\frac{4 g h}{3}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\frac{3}{4} \mathrm{mv}^{2}\) \(\mathrm{u}=\sqrt{\frac{4}{3} \mathrm{gh}}\) \(\omega=\frac{V}{r}\) \(\omega= \frac{1}{r} \sqrt{\frac{4 g h}{3}}\)
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