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. $({m_1} + m{}_2){r^2}$
C. $(\dfrac{{{m_1} + {m_2}}}{{{m_1}{m_2}}}){r^2}$
D. $(\dfrac{{{m^{_{}}}_1{m_2}}}{{{m_1} + {m_2}}}){r^2}$

Answer 1

Hint: for this we need to have basic knowledge of center of mass and moment of inertia. Moment of Inertia can be defined as how easily a body can be rotated about a given axis. It is a rotational analogue of mass.

Step by step solution:

1. Let us first draw a figure which will help us in better understanding of the Question.
Below we have drawn a rod AB where $m_1$ is the mass at point A and $m_2$ is the mass at point B. At point O we have assumed the center of mass of the rod. We have drawn a perpendicular axis to the rod about which moment of inertia is to be calculated. Let $r_1$ be the distance between point A and O and $r_2$ be the distance between point B and O.

2. As we know that the position of center of mass of the particles is given by
$X = \dfrac{{{m_1}{x_1} + {m_2}{x_2} + .......{m_n}{x_n}}}{{{m_1} + {m_2} + .....{m_n}}} = \dfrac{{\sum {{m_i}{x_i}} }}{{\sum {{m_i}} }}$ …………….equation 1
So here according to our figure we have ${r_1} = \dfrac{{{m_1} \times 0 + {m_2} \times r}}{{{m_1} + {m_2}}}$ because $m_1$ is at the 0 distance from point A and $m_2$ is at the distance r from point A.
Again ${r_1} = \dfrac{{{m_2} \times r}}{{{m_1} + {m_2}}}$ ………………equation 2
And ${r_2} = r - \dfrac{{{m_2} \times r}}{{{m_1} + {m_2}}}$ ……………..equation 3

3. Now Moment of Inertia is given by $I = {m_1}{x_1}^2 + {m_2}{x_2}^2 + .........{m_n}{x_n}^2$
Here in our question $I = {m_1}{r_1}^2 + {m_2}{r_2}^2$
With the help of equations 2 and 3 we get
$I = \dfrac{{{m_1}{m_2}^2{r^2}}}{{{{({m_1} + {m_2})}^2}}} + \dfrac{{{m_1}^2{m_2}{r^2}}}{{{{({m_1} + {m_2})}^2}}}$
Now taking some common terms
$I = \dfrac{{{m_1}{m_2}{r^2}}}{{{{({m_1} + {m_2})}^2}}}({m_1} + {m_2})$
After doing the simplifications we get:
$I = \dfrac{{{m_1}{m_2}{r^2}}}{{{m_1} + {m_2}}}$
Or it can also be written as $I = (\dfrac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}){r^2}$
Hence option D is correct.

Note: In this question basic knowledge of moment of inertia and simple algebra is used. We can also solve this type of question using numerical values instead of variables.