question_answer

A diatomic molecule is formed by two atoms which may be treated as mass points \[{{m}_{1}}\]and\[{{m}_{2}},\]joined by a massless rod of length r. Then the, moment of inertia of the molecule about an axis passing through the centre of mass and perpendicular to rod is :

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. \[I=\sum\limits_{i=1}^{n}{{{I}_{i}}}\] where  \[{{I}_{1}}={{m}_{1}}r_{1}^{2},\,{{I}_{2}}={{m}_{2}}r_{2}^{2}\] Given,           \[r={{r}_{1}}+{{r}_{2}}\]                 \[{{m}_{1}}{{r}_{1}}={{m}_{2}}{{r}_{2}}\] \[\therefore \]  \[{{m}_{1}}{{r}_{1}}={{m}_{2}}(r-{{r}_{1}})\] \[\Rightarrow \]               \[{{m}_{1}}{{r}_{1}}+{{m}_{2}}{{r}_{1}}={{m}_{2}}r\] \[\Rightarrow \]               \[{{r}_{1}}({{m}_{1}}+{{m}_{2}})={{m}_{2}}r\] \[\Rightarrow \]               \[{{r}_{1}}=\frac{{{m}_{2}}r}{({{m}_{1}}+{{m}_{2}})}\] Also,    \[{{r}_{2}}=r-{{r}_{1}}\]                 \[{{r}_{2}}=r-\frac{{{m}_{2}}r}{({{m}_{1}}+{{m}_{2}})}=\frac{{{m}_{1}}r}{{{m}_{1}}+{{m}_{2}}}\] \[\therefore \]  \[I={{I}_{1}}+{{I}_{2}}={{m}_{1}}r_{1}^{2}+{{m}_{2}}r_{2}^{2}\] \[I={{m}_{1}}\frac{m_{2}^{2}{{r}^{2}}}{{{({{m}_{1}}+{{m}_{2}})}^{2}}}+{{m}_{2}}\frac{m_{1}^{2}{{r}^{2}}}{{{({{m}_{1}}+{{m}_{2}})}^{2}}}\] \[I=\frac{{{m}_{1}}{{m}_{2}}{{r}^{2}}}{({{m}_{1}}+{{m}_{2}})}=\frac{{{m}_{1}}{{m}_{2}}}{({{m}_{1}}+{{m}_{2}})}.{{r}^{2}}\]