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As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a flywheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance h, the angular speed of the wheel will be:-
 
Option: 1 r\sqrt{\frac{3}{2gh}}
 
Option: 2 r\sqrt{\frac{3}{4gh}}

Option: 3 \frac{1}{r}\sqrt{\frac{4gh}{3}}  

Option: 4 \frac{1}{r}\sqrt{\frac{2gh}{3}}
 

Answers (1)

best_answer
 

 

 

Conservation Of angular momentum -    

\\mg-T=ma\\T\times r=I\alpha\\T=\frac{mr^2}{2}\times\frac{a}{r}\times\frac{1}{r}\\T=\frac{ma}{2}\\mg=\frac{3ma}{2}\\a=\frac{2g}{3}\\Also,\ v=\sqrt{2as}=\sqrt{\frac{4gh}{3}}\\also, v=\omega r\\ \omega=\frac{v}{r}

\Rightarrow \sqrt{\frac{4gh}{3}}\times\frac{1}{r}=\frac{1}{r}\sqrt{\frac{4gh}{3}}

Hence option (3) is correct.

Posted by

Ritika Jonwal

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