A) \[\frac{\omega M}{M+m}\]
B) \[\frac{\omega \left( M-2M \right)}{\left( M+2m \right)}\]
C) \[\frac{\omega \left( M+2m \right)}{M}\]
D) \[\frac{\omega M}{M+2m}\]
Correct Answer: D
Solution :
Initial angular momentum of ring\[L=I\omega =M{{R}^{2}}\omega \]Final angular momentum of ring and two particle system \[L=(M{{R}^{2}}+2m{{R}^{2}})\omega '\] As there is no torque on the system, therefore angular momentum remains constant, \[\ \therefore \] \[M{{R}^{2}}\omega =(M{{R}^{2}}+2m{{R}^{2}})\omega '\]\[\Rightarrow \] \[\omega '=\frac{M\omega }{M+2m}\]
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