Question

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 center of mass and perpendicular to rod is

• A. zero
• B. $\left(m_1+m_2\right)r^2$
• C. $\left(\frac{m_1+m_2}{m_1{m}_2}\right)r^2$
• D. $\left(\frac{m_1{m}_2}{m_1+m_2}\right)r^2$

Answer 1

Answer: (D)

Total moment of inertia will be equal to the sum of moment of inertia due to individual masses.
$I={\sum }_{l=1}^n{I}_i$
where $I_1=m_1{r}_1{}^2,I_2{m}_2{r}_2{}^2.$
Given, $r=r_1+r_2{m}_1{r}_1=m_2{r}_2$
$\therefore m_1{r}_1=m_2\left(r-r_1\right)$
$\Rightarrow m_1{r}_1+m_2{r}_1=m_{2}r$
$\Rightarrow r_1\left(m_1+m_2\right)=m_{2}r$
$\Rightarrow r_1=\frac{m_{2}r}{\left(m_1+m_2\right)}$
Also, $r_2=r-r_1{r}_2=r-\frac{m_{2}r}{\left(m_1+m_2\right)}=\frac{m_{1}r}{m_1+m_2}$
$\therefore I=I_1+I_2=m_1{r}_1{}^2+m_2{r}_2{}^2$
$I=m_1\frac{m_2^2{r}^2}{{\left(m_1+m_2\right)}^2}+m_2\frac{m_1{}^2{r}^2}{\left(m_1+m_2\right)}.$
$I=\frac{m_1{m}_2{r}^2}{\left(m_1+m_2\right)}=\frac{m_1{m}_2}{\left(m_1+m_2\right)}\cdot r^2$