ABC is a triangular framwork of three uniform rods each of mass `m` and length `2l`. It is free to rotate in its own plane about a smooth horizontal axis through A which is perpendicular to ABC. If it is released from rest when AB is horizontal and C is above AB. Find the maximum velocity of C in the subsequent motion.

Explanation

`I_(A)=I_(AB)+I_(AC)+I_(BC)`
`=(4)/(3)ml^(2)+(4)/(3)ml^(2)+[(1)/(3)ml^(2)+m(lsqrt(3))^(2)]`
`=6ml^(2)`
if `omega` is the angular velocity in the second position. Then using conservation of mechanical energy have

FOR COM `h_(i)=+(sqrt(3)l)/(3)`
and `h_(f)=-(2sqrt(3))/(3)l`
`3mg((lsqrt(3))/(3))=(1)/(2)(ml^(2))omega^(2)+3mg((-2//sqrt(3))/(3))`
or `omega=sqrt((gsqrt(3))/(l))`
Now, velocity of C at this instant is `2lomega` or `2sqrt(glsqrt(3))` and maximum