1. m₁+m₂/m₁m₂l² 2. (m₁+m₂)l² 3. √(m₁m₂) l² 4. m₁m₂/m₁+m₂l²

Answer: 4. m₁m₂/m₁+m₂l² Take the two masses at distances x₁ and x₂ from the centre of mass, with x₁+x₂=l and m₁x₁=m₂x₂ . This gives x₁=m₂ l/m₁+m₂ and x₂=m₁ l/m₁+m₂ . I=m₁x₁²+m₂x₂²=m₁m₂²+m₂m₁²/(m₁+m₂)²l²=m₁m₂/m₁+m₂l² This is μ l² , with μ the reduced mass. Option (4).

Topic: Moment of inertia, radius of gyration, MI of simple objects· Rotational Motion

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

1. $\frac{m_1+m_2}{m_1m_2}l^2$

2. $(m_1+m_2)l^2$

3. $\sqrt{m_1m_2}\,l^2$

4. $\frac{m_1m_2}{m_1+m_2}l^2$

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Answer: 4. $\frac{m_1m_2}{m_1+m_2}l^2$

Take the two masses at distances $x_1$ and $x_2$ from the centre of mass, with $x_1+x_2=l$ and $m_1x_1=m_2x_2$.

This gives $x_1=\dfrac{m_2 l}{m_1+m_2}$ and $x_2=\dfrac{m_1 l}{m_1+m_2}$.

$$I=m_1x_1^2+m_2x_2^2=\frac{m_1m_2^2+m_2m_1^2}{(m_1+m_2)^2}l^2=\frac{m_1m_2}{m_1+m_2}l^2$$

This is $\mu l^2$, with $\mu$ the reduced mass. Option (4).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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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

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