A) zero
B) \[({{m}_{1}}+{{m}_{2}}){{r}^{2}}\]
C) \[\left( \frac{{{m}_{1}}+{{m}_{2}}}{{{m}_{1}}{{m}_{2}}} \right){{r}^{2}}\]
D) \[\left( \frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right){{r}^{2}}\]
Correct Answer: D
Solution :
Total moment of inertia will be equal to the sum of moment of inertia due to individual masses. \[I=\underset{l=1}{\mathop{\overset{n}{\mathop{\Sigma }}\,}}\,{{I}_{i}}\] where \[{{I}_{1}}={{m}_{1}}{{r}_{1}}^{2},\,\,\,\,\,\,\,\,{{I}_{2}}{{m}_{2}}{{r}_{2}}^{2}.\]You need to login to perform this action.
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