Rotational Motion

JEE Main level, single correct, medium

261 of 280

Question 261 of 280



As shown in the figure, a bob of mass mm is tied to a massless string whose other end portion is wound on a fly wheel (disc) of radius rr and mass m.m. When released from rest the bob starts falling vertically. When it has covered a distance of h,h, the angular speed of the wheel will be:

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Answer: (A) 1r4gh3\dfrac{1}{r}\sqrt{\dfrac{4gh}{3}}

Using work-energy theorem
mgh=12mv2+12Iω2mgh=\dfrac{1}{2}{mv}^{2}+\dfrac{1}{2}{I\omega }^{2}
mgh=12mω2r2+12mr22ω2mgh=\dfrac{1}{2}{m\omega }^{2}{r}^{2}+\dfrac{1}{2}\dfrac{{mr}^{2}}{2}{\omega }^{2} ( since I=mr22I=\dfrac{{mr}^{2}}{2} and v=rωv=r\omega )
mgh=34mω2r2mgh=\dfrac{3}{4}{m\omega }^{2}{r}^{2}
Hence, ω=4gh3r2=1r4gh3\omega =\sqrt{\dfrac{4gh}{3{r}^{2}}}=\dfrac{1}{r}\sqrt{\dfrac{4gh}{3}}

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