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

Options

A. `(m_1 m_2 )/(m_(1) + m_(2)) "" l^(2)`
B. `(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

A

Explanation

Here , `l_(1) + l_(2) = l`
Centre of mass of the system ,
`I_(1) = (m_(1) xx 0 + m_(2) xx l)/(m_(1) + m_(2)) = (m_(2) l)/(m_(1) + m_(2))`
`l_(2) = l - l_(1) = (m_(1) l)/(m_(1) + m_(2))`

Required moment of inertia of the system ,
`I = m_(1) l_(1)^(2) + m_(2) l_(2)^(2) = (m_(1) m_(2)^(2) + m_(2) m_(1)^(2)) (l^(2))/((m_(1) + m_(2))^(2))`
`= (m_(1) m_(2) (m_(1) + m_(2))l^(2))/((m_(1) + m_(2))^(2)) = (m_(1) m_(2))/(m_(1) + m_(2))""l^(2)`