A uniform rod of length `l` is spinning with an angular velocity `omega=2v/l` while its centre of mass moves with a velocity `v`. Find the velocity of the end of the rod.

Explanation

Velocity of end `A` `vecv_(A)=vecc_(A,C)=vecv_(C)`
Hence the velocity of `A` is
`v_(A)=sqrt(v_(AC)^(2)+v_(C)^(2)+2v_(AC)v_(C) costheta)`
we know `v_(C)=V,V_(AC)=l/2.omega` and `theta=90^@+30^@=120^@`
Hence `v_(A)=sqrt(v^(2)+(l^(2)omega^(2))/4+lvomegacos120^@)`

But `omega=(2v)/l`
`v_(AC)=l/2omega=l/2((2v)/l)=v`
which gives `v_(A)=v`
Let `vecv_(A)` make angle `phi` with the direction of `vecv_(C)`. Then
`phi=tan^(-1)(vsintheta)/(v_(AC)costheta+v_C)`
we get `phi=tan^(-1) (vsin120^@)/(vcos120^@+v)=(pi)/2`