MHT CET · Physics · Rotational Motion
A rotating body has angular momentum 'L'. If its frequency of rotation is halved and rotational kinetic energy is doubled, its angular momentum becomes
- A \(2 \mathrm{~L}\)
- B \(\frac{\text { L }}{4}\)
- C \(4 \mathrm{~L}\)
- D \(\frac{\mathrm{L}}{2}\)
Answer & Solution
Correct Answer
(C) \(4 \mathrm{~L}\)
Step-by-step Solution
Detailed explanation
Kinetic energy \(k=\frac{1}{2} I \omega^{2}\)
\(\therefore \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{I}_{2} \mathrm{w}_{2}^{2}}{\mathrm{I}_{1} \omega_{1}^{2}}\) \(\therefore 2=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}\left(\frac{\mathrm{l}}{2}\right)^{2} \quad \therefore \frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=8\)
Angular momentum \(\quad \mathrm{L}=\mathrm{I} \omega\) \(\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}=\frac{\mathrm{I}_{2} \omega_{2}}{\mathrm{I}_{1} \omega_{1}}=8 \times \frac{1}{2}=4\) \(\therefore \mathrm{L}_{2}=4 \mathrm{~L}_{1}=4 \mathrm{~L} \quad\)
\(\therefore \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{I}_{2} \mathrm{w}_{2}^{2}}{\mathrm{I}_{1} \omega_{1}^{2}}\) \(\therefore 2=\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}\left(\frac{\mathrm{l}}{2}\right)^{2} \quad \therefore \frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=8\)
Angular momentum \(\quad \mathrm{L}=\mathrm{I} \omega\) \(\frac{\mathrm{L}_{2}}{\mathrm{~L}_{1}}=\frac{\mathrm{I}_{2} \omega_{2}}{\mathrm{I}_{1} \omega_{1}}=8 \times \frac{1}{2}=4\) \(\therefore \mathrm{L}_{2}=4 \mathrm{~L}_{1}=4 \mathrm{~L} \quad\)
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