Question

Two particles of masses $${m_1}$$ and ${m_2}$ are connected by a rigid massless rod of length $r$ to constitute a dumbbell. The moment of inertia of the dumbbell about an axis perpendicular to the rod and passing through the center of mass is:
A. $\dfrac{{{m_1}{m_2}{r^2}}}{{{m_1} + {m_2}}}$
B. $\left( {{m_1} + {m_2}} \right){r^2}$
C. $\dfrac{{{m_1}{m_2}{r^2}}}{{{m_1} - {m_2}}}$
D. $\left( {{m_1} - {m_2}} \right){r^2}$

Answer 1

Hint:We will first calculate the center of mass of the dumbbell from masses $${m_1}$$ and ${m_2}$. Then we will calculate the moment of inertia of both the masses $${m_1}$$ and ${m_2}$ about the center of mass of both the masses. Now, the moment of inertia of the dumbbell about an axis perpendicular to the rod and passing through the center of mass can be calculated by adding the center of mass of masses ${m_1}$ and ${m_2}$.

Complete step by step answer:
In this question, there are two masses $${m_1}$$ and ${m_2}$ that are connected by a rigid massless rod of length $r$.
The position of center of mass of the dumbbell from mass ${m_1}$ is given below
${M_1} = \dfrac{{{m_2}r}}{{{m_1} + {m_2}}}$
Also, the position of center of mass of the dumbbell from mass ${m_2}$ is given below
${M_2} = \dfrac{{{m_1}r}}{{{m_1} + {m_2}}}$
Now, the moment of inertia of the mass ${m_1}$ about the center of mass of the dumbbell from mass ${m_1}$ is given below
${I_1} = {m_1} \times {\left( {\dfrac{{{m_2}r}}{{{m_1} + {m_2}}}} \right)^2}$
$ \Rightarrow \,{I_1} = \dfrac{{{m_1}m_2^2{r^2}}}{{{{\left( {{m_1} + {m_2}} \right)}^2}}}$
Also, the moment of inertia of mass ${m_2}$ about the center of mass of the dumbbell from mass ${m_2}$ is given below
${I_2} = {m_2} \times {\left( {\dfrac{{{m_1}r}}{{{m_1} + {m_2}}}} \right)^2}$
$ \Rightarrow \,{I_2} = \dfrac{{{m_2}m_1^2{r^2}}}{{{{\left( {{m_1} + {m_2}} \right)}^2}}}$
Now, the moment of inertia of the dumbbell about an axis perpendicular to the rod and passing through the center of mass can be calculated by adding the center of mass of masses ${m_1}$ and ${m_2}$ as given below
$I = {I_1} + {I_2}$
$ \Rightarrow \,I = \dfrac{{{m_1}m_2^2{r^2}}}{{{{\left( {{m_1} + {m_2}} \right)}^2}}} + \dfrac{{{m_2}m_1^2{r^2}}}{{{{\left( {{m_1} + {m_2}} \right)}^2}}}$
$ \Rightarrow \,I = \dfrac{{{m_1}{m_2}\left( {{m_1} + {m_2}} \right){r^2}}}{{{{\left( {{m_1} + {m_2}} \right)}^2}}}$
$ \therefore\,I = \dfrac{{{m_1}{m_2}{r^2}}}{{{m_1} + {m_2}}}$
Therefore, the moment of inertia of the dumbbell about an axis perpendicular to the rod and passing through the center of mass is $\dfrac{{{m_1}{m_2}{r^2}}}{{{m_1} + {m_2}}}$.

Hence, option A is the correct option.

Note:Here remember that the center of mass of mass ${m_1}$ will be in terms of ${m_2}$. On the other hand, the center of mass of ${m_2}$ will be in terms of ${m_1}$. Also, the moment of inertia of mass ${m_1}$ will be the product of mass and the center of mass of ${m_1}$. On the other hand, the moment of inertia of mass ${m_2}$ will be the product of mass and the center of mass of ${m_2}$.