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JEE MAIN 2022
26-07-2022 S1
Question
A disc of mass 1 kg and radius R is free of rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $4 \sqrt{\frac{x}{3 R}} \operatorname{rads}^{-1}$ where $x=$ $\_\_\_\_$ .$\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
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Solution
using conservation of mechanical energy $$ \begin{aligned} & m g 2 R=\frac{1}{2} I_{u=\omega^2} \omega^2+\frac{1}{2} I_{\text {partac }} \omega^2 \\ & m g 2 R=\frac{\omega^2}{2}\left[\frac{m R^2}{2}+m R^2\right] \\ & m g 2 R=\frac{\omega^2}{2} \frac{3}{2} m R^2 \\ & \frac{3}{4} \omega^2=\frac{2 g}{R} \\ & \omega^2=\frac{8 g}{3 R} \\ & \omega=\sqrt{\frac{80}{3 R}} \end{aligned} $$ Given $$ \begin{aligned} & \omega=4 \sqrt{\frac{x}{3 R}} \\ & 16 \frac{x}{3 R}=\frac{80}{3 R} \\ & x=5 \end{aligned} $$
Question Tags
JEE Main
Physics
Medium
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