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

Answer: (A)

$\frac{m_1 m_2}{m_1 + m_2} l^2$

Answer 2

$m_{1}r_{1} = m_{2 } \left(l -r_{2}\right)$
$\left(m_{1} + m_{2}\right)r_{1} = m_{2}r$
$r_{1} = \frac{m_{2}l}{m_{1} + m_{2}}$
and $r_{2} =r - r_{1} = r - \frac{m_{2}l}{m_{1}+m_{2}} = \frac{m_{1}l}{m_{1} +m_{2}}$
So $I = I_{1} + I_{2}$
$=m_{1}r_{1}^{2} + m_{2}r_{2}^{2}$
$=m_{1} \left(\frac{m_{2}l}{m_{1}+m_{2}}\right)^{2} + m_{2}\left(\frac{m_{1}l}{m_{1}+m_{2}}\right)^{2} = \frac{m_{1}m_{2}l^{2}}{m_{1}+m_{2}}$