Question

A light rod of length $l$ has two masses $m_1$ and $m_2$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is -

• A. $\frac{m_1 m_2}{m_1 + m_2} l^2$
• B. $\frac{m_1 + m_2}{m_1 m_2} l^2$
• C. $(m_1 + m_2) l^2$
• D. $\sqrt{m_1 m_2} l^2$

Answer 1

The Correct Option is A

To solve the problem of finding the moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass, we need to follow these steps:

1. Understand the System Configuration: We have a rod of length $l$ with two masses, $m_1$ and $m_2$, attached at its ends. The rod is light, implying its mass is negligible compared to the masses at the ends.
2. Locate the Centre of Mass (COM): The centre of mass $x_{cm}$ is given by the formula: $$ x_{cm} = \frac{m_1 \cdot 0 + m_2 \cdot l}{m_1 + m_2} = \frac{m_2 \cdot l}{m_1 + m_2} $$
3. Moment of Inertia Calculation: The moment of inertia $I$ about the centre of mass is obtained by summing the moments of inertia of both masses about the COM using the parallel axis theorem.
   • For $m_1$, located at a distance $x_{cm}$ from the COM: $$ I_1 = m_1 \cdot (x_{cm})^2 = m_1 \cdot \left( \frac{m_2 \cdot l}{m_1 + m_2} \right)^2 $$
   • For $m_2$, located at a distance $(l - x_{cm})$ from the COM: $$ I_2 = m_2 \cdot (l - x_{cm})^2 = m_2 \cdot \left( \frac{m_1 \cdot l}{m_1 + m_2} \right)^2 $$
4. Sum the Moments of Inertia: Combine $I_1$ and $I_2$ to find the total moment of inertia: $$ I = I_1 + I_2 = m_1 \cdot \left( \frac{m_2 \cdot l}{m_1 + m_2} \right)^2 + m_2 \cdot \left( \frac{m_1 \cdot l}{m_1 + m_2} \right)^2 $$ $$ = \frac{m_1 m_2^2 l^2}{(m_1 + m_2)^2} + \frac{m_2 m_1^2 l^2}{(m_1 + m_2)^2} $$ $$ = \frac{m_1 m_2 (m_1 + m_2) l^2}{(m_1 + m_2)^2} $$ $$ = \frac{m_1 m_2}{m_1 + m_2} l^2 $$
5. Conclusion: The moment of inertia of the system about the axis through the centre of mass is $\frac{m_1 m_2}{m_1 + m_2} l^2$.