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

`m_(1)r_(1)=m_(2)(l-r_(2))`
`(m_(1)+m_(2))r_(1)=m_(2)r`
`r_(1)=(m_(2)l)/(m_(1)+m_(2))`
and `r_(2)=r-r_(1)=r-(m_(2)l)/(m_(1)+m_(2))=(m_(1)l)/(m_(1)+m_(2))`
So `I=I_(1)+I_(2)` ltBrgt `=m_(1)r_(1)^(2)+m_(2)r_(2)^(2)`
`=m_(1)((m_(2)l)/(m_(1)+m_(2)))^(2)+m_(2)((m_(1)l)/(m_(1)+m_(2)))^(2)=(m_(1)m_(2)l^(2))/(m_(1)+m_(2))`