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. $\sqrt{m_1{m}_2}{l}^2$
• B. $\frac{m_1{m}_2}{m_1+m_2}{l}^2$
• C. $\frac{m_1+m_2}{m_1{m}_2}{l}^2$
• D. $(m_1+m_2)l^2$

Answer 1

The correct option is B $\frac{m_1{m}_2}{m_1+m_2}{l}^2$

The distance of center of mass from mass $m_1$ is $x_1=\frac{m_{2}l}{m_1+m_2}$
The distance of center of mass from mass $m_2$ is $x_2=\frac{m_{1}l}{m_1+m_2}$

Hence, moment of inertia of masses about the center of mass =$m_1{x}_1^2+m_2{x}_2^2$
$=m_1{\left(\frac{m_{2}l}{m_1+m_2}\right)}^2+m_2{\left(\frac{m_{1}l}{m_1+m_2}\right)}^2$
On simplifying above equation we get,
$=\frac{m_1{m}_2}{m_1+m_2}{l}^2$