A light rod of length l has two masses `m_(1)andm_(2)` attached to its two ends. The moment of inertia of the system about and axis perpendicular to the rod and passing through the centre of mass is ………

Options

A. `(m_(1)+m_(2))l^(2)`
B. `sqrt(m_(1)m_(2))l^(2)`
C. `(m_(1)m_(2))/(m_(1)+m_(2))l^(2)`
D. `(m_(1)+m_(2))/(m_(1)m_(2))l^(2)`

Answer

C

Explanation

Position vector of centre of mass
`r_(cm)=(m_(1)-r_(1)+m_(2)r_(2))/(m_(1)+m_(2))`
`0=(-m_(1)r_(1)+m_(2)r_(2))/(m_(1)+m_(2))`
`therefore m_(1)r_(1)=m_(2)r_(2)`
`therefore (m_(1))/(m_(2))=(r_(2))/(r_(1))...(1)`
Now,
`(m_(1)+m_(2))/(m_(2))=(r_(1)+r_(2))/(r_(1))`
`therefore (m_(1)+m_(2))/(m_(2))=(l)/(r_(1)) [because r_(1)+r_(2)=l]`
`therefore r_(1)=(m_(2)l)/(m_(1)+m_(2)),r_(2)=(m_(1)l)/(m_(1)+m_(2))`
`therefore` Moment of inertia about an axis passing through centre of mass and perpendicular to it for this arrangement
`I_(cm)=m_(1)r_(1)^(2)+m_(2)r_(2)^(2)`
`=m_(1)xx(m_(2)^(2)l^(2))/((m_(1)+m_(2))^(2))+m_(2)xx(m_(2)^(2)l^(2))/((m_(1)+m_(2))^(2))`
`=m_(1)m_(2)l^(2)[(m_(2)+m_(1))/((m_(1)+m_(2))^(2))]`
`=(m_(1)m_(2)l^(2))/(m_(1)+m_(2))`
Note : `(m_(1)m_(2))/(m_(1)+m_(2))mu` is known as reduce mass `therefore I=mur^(2)`