A) 2L
B) 4L
C) \[\frac{L}{2}\]
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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