• # question_answer A particle performs uniform circular motion with an angular momentum L. If the frequency of rotation is doubled and its KE is halved, the angular momentum becomes: A)  2L                  B)  4LC)  $\frac{L}{2}$                                  D) $\frac{L}{4}$

Correct Answer: D

Solution :

Angular momentum $=L=mvr=m{{r}^{2}}\omega$ Also kinetic energy is $K=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{r}^{2}}{{\omega }^{2}}$ or         $K=\frac{1}{2}\times \frac{L}{\omega }.\,{{\omega }^{2}}=\frac{L\omega }{2}$ $\therefore$    $L=\frac{2k}{\omega }$ Given, $\omega '=2\omega ,\,k'=\frac{1}{2}k$ $\therefore$    $L'=\frac{2k'}{\omega '}=\frac{2({\scriptstyle{}^{k}/{}_{2}})}{2\omega }=\frac{L}{4}$ Hence, the correction option is [d].

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