question_answer

Two particles of masses \[{{m}_{1}}\] and \[{{m}_{2}}\] are connected to a rigid massless rod of length r to constitute a dumb bell which is free to move in the plane. The moment of inertia of the dumb bell about an axis perpendicular to the plane passing through the centre of mass is:

A) \[\frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{r}^{2}}\]                

B) \[({{m}^{1}}+{{m}^{2}}){{r}^{2}}\]

C) \[\frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}-{{m}_{2}}}{{r}^{2}}\]                 

D) \[({{m}^{1}}-{{m}^{2}}){{r}^{2}}\]

Correct Answer: A

Solution :

[a] C is centre of mass of the dumb belt, \[{{r}_{1}}\]and \[{{r}_{2}}\]are distance of \[{{m}_{1}},{{m}_{2}}\]from C. Then M.O.I of dumb bell about the given axis \[I={{m}_{1}}r_{1}^{2}+{{m}_{2}}r_{2}^{2}\] \[r={{r}_{1}}+{{r}_{2,}}\] \[{{m}_{1}}{{r}_{1}}={{m}_{2}}{{r}_{2}}\] \[{{m}_{1}}{{r}_{1}}={{m}_{2}}r-{{m}_{2}}{{r}_{1}}\] \[{{r}_{1}}=\frac{{{m}_{2}}r}{{{m}_{1}}+{{m}_{2}}},{{r}_{2}}=\frac{{{m}_{1}}r}{{{m}_{1}}+{{m}_{2}}}\] \[I={{m}_{1}}\left[ \frac{{{m}_{2}}r}{{{m}_{1}}+{{m}_{2}}} \right]+{{m}_{2}}\left[ \frac{{{m}_{1}}r}{{{m}_{1}}+{{m}_{2}}} \right]\] \[I=\frac{{{m}_{2}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}{{r}^{2}}\]