A) \[\frac{\omega \,(M-2\,m)}{(M+2m)}\]
B) \[\frac{\omega M}{(M+2m)}\]
C) \[\frac{\omega \,M}{(M+m)}\]
D) \[\frac{\omega \,(M+2\,m)}{M}\]
Correct Answer: B
Solution :
Key Idea: Angular momentum remains conserved in the universe. According to conservation of angular momentum L = constant or \[I\omega =\text{constant}\] \[\therefore {{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}\] Initial moment of inertia \[{{I}_{1}}=M{{R}^{2}}\] and angular velocity \[{{\omega }_{1}}=\omega \] Hence, \[{{I}_{1}}{{\omega }_{1}}=M{{R}^{2}}\omega \] When two objects of mass m are attached to opposite ends of a diameter, the final readings are \[{{I}_{2}}=M\,{{R}^{2}}+m\,{{R}^{2}}+m\,{{R}^{2}}\] \[=(M+2m)\,{{R}^{2}}\] So, \[{{I}_{2}}{{\omega }_{2}}=(M+2m){{R}^{2}}{{\omega }_{2}}\] ....(iii) \[\therefore \] From Eqs. (i), (ii) and (iii) \[M{{R}^{2}}\,\omega \,=\,(M+2m)\,{{R}^{2}}\,{{\omega }_{2}}\] \[\Rightarrow {{\omega }_{2}}=\frac{\omega M}{M+2m}\]
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