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

Moment of Inertia Calculation 

We are given a system with two masses, \( m_1 \) and \( m_2 \), and their respective distances from a pivot point. We need to calculate the total moment of inertia of the system.

Step 1: Relating the Distances

The problem starts with the equation that relates the two distances \( r_1 \) and \( r_2 \) for the two masses:

\(m_1 r_1 = m_2 (l - r_2)\)

Where: - \( r_1 \) is the distance of mass \( m_1 \), - \( r_2 \) is the distance of mass \( m_2 \), - \( l \) is the total distance between the two masses.

By rearranging the equation, we get:

\((m_1 + m_2) r_1 = m_2 r\)

Solving for \( r_1 \), we get:

\(r_1 = \frac{m_2 l}{m_1 + m_2}\)

Step 2: Calculating \( r_2 \)

We can find \( r_2 \) by subtracting \( r_1 \) from the total distance \( l \):

\(r_2 = r - r_1 = r - \frac{m_2 l}{m_1 + m_2} = \frac{m_1 l}{m_1 + m_2}\)

Step 3: Total Moment of Inertia

The total moment of inertia \( I \) for the system is the sum of the individual moments of inertia of the two masses:

\(I = I_1 + I_2\)

Where: - \( I_1 = m_1 r_1^2 \), - \( I_2 = m_2 r_2^2 \).

Now, substituting the expressions for \( r_1 \) and \( r_2 \) into the formula for the total moment of inertia:

\(I = m_1 r_1^2 + m_2 r_2^2\)

Substitute the values for \( r_1 \) and \( r_2 \):

\(I = m_1 \left(\frac{m_2 l}{m_1 + m_2}\right)^2 + m_2 \left(\frac{m_1 l}{m_1 + m_2}\right)^2\)

Now, simplify the expression:

\(I = \frac{m_1 m_2 l^2}{m_1 + m_2}\)

Conclusion:

The total moment of inertia of the system is:

\(I = \frac{m_1 m_2 l^2}{m_1 + m_2}\)