A) \[\frac{M+2m}{M}\omega \]
B) \[\frac{M}{M+2m}\omega \]
C) \[\frac{M-2m}{M+2m}\omega \]
D) \[\frac{2M}{M+2m}\omega \]
Correct Answer: B
Solution :
Key Idea: external torque is acting on the system, then angular momentum of the system is conserved. As external torque on the system is zero i.e., \[\tau =0\] or \[\frac{dL}{dt}=0\] or \[L=\text{constant}\] or \[{{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}\] ?(i) Here,\[{{I}_{1}}=M{{R}^{2}}\](about its own axis),\[{{\omega }_{1}}=\omega \] \[{{I}_{2}}=(M+2m){{R}^{2}}\] Putting in Eq. (i), we find \[M{{R}^{2}}\omega =(M+2m){{R}^{2}}{{\omega }_{2}}\] or \[{{\omega }_{2}}=\left( \frac{M}{M+2m} \right)\omega \]You need to login to perform this action.
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