JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2020
- Hard
- MCQ
A uniformly thick wheel with moment of inertia \(I\) and radius \(R\) is free to rotate about its centre of mass (see fig). A massless string is wrapped over its rim and two blocks of masses \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\left(\mathrm{m}_{1}>\mathrm{m}_{2}\right)\) are attached to the ends of the string. The system is released from rest. The angular speed of the wheel when \(\mathrm{m}_{1}\) descents by a distance \(h\) is
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Answer & Solution
Correct Answer
(B) \(\left[\frac{2\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{gh}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{R}^{2}+\mathrm{I}}\right]^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
by using work energy theorem
\(\mathrm{Wg}=\Delta \mathrm{KE}\)
\(\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{gh}=\frac{1}{2}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{V}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\)
\(\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{gh}=\frac{1}{2}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)(\omega \mathrm{R})^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\)
\(\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{gh}=\frac{\omega^{2}}{2}\left[\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{R}^{2}+\mathrm{I}\right]\)
\(\omega=\sqrt{\frac{2\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{gh}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{R}^{2}+\mathrm{I}}}\)
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