Three rods each of mass m and length l are joined together to form an equilateral triangle as shown in figure. Find the moment of inertial of the system about an axis passig through its centre of mass and perpendicular to the plane of the particle.

Explanation

Moment of inertia of rod BC about an axis perpendicular to plane of triangle ABC and passing through the mid-point of rod BC (i.e., D) is

`I_(1)=(ml^(2))/(12)rArrAD =sqrt(l^(2)-(l^(2))/(4))=(sqrt(3))/(2)l`
`r=OD=(AD)/(2)=(sqrt(3)l)/(3xx2)=(l)/(sqrt(3))`
From theorem of parallel axes, moment of inertia of this rod about the axis passing through CM and perpendicular to plain ABC is
`I_(2)=I_(1)+mr^(2)=(ml^(2))/(12)+m((l)/(2 sqrt(3)))^(2)=(ml^(2))/(6)`
`:.` Moment of inertia of all the three rod is
`I=3I_(2)=3((ml^(2))/(6))=(ml^(2))/(2)`.