A) \[\frac{(M+4m)\,\omega }{M}\]
B) \[\frac{(M-4m)\,\omega }{M+4m}\]
C) \[\frac{M\,\omega }{4m}\]
D) \[\frac{M\,\omega }{M+4m}\]
Correct Answer: D
Solution :
| Key Idea: If external torque acting on the system is zero, hence angular momentum remains conserved. |
| \[{{\tau }_{\text{ext}}}=0\] |
| or \[\frac{dL}{dt}=0\] |
| or \[L=\]constant |
| or \[I\omega =\] constant |
| \[\therefore \] \[{{I}_{1}}\,{{\omega }_{1}}={{I}_{2}}\,{{\omega }_{2}}\] (i) |
| Here, \[{{I}_{1}}=M{{r}^{2}},\,\,{{\omega }_{1}}=\omega ,\,\,{{I}_{2}}=M{{r}^{2}}+4m{{r}^{2}}\] |
| Hence, Eq. (i) can be written as |
| \[M{{r}^{2}}\omega =(M{{r}^{2}}+4m{{r}^{2}})\,{{\omega }_{2}}\] |
| \[\therefore \] \[{{\omega }_{2}}=\frac{M\omega }{M+4m}\] |
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