A mass M is kept hanging by rod of length L. What tangential velocity must be given to it so that it can just reach the top of vertical circle?

Suppose the mass of rod is ‘m’ then MI of the system is

I1=MI of the rod about the axis passing through point of suspension of the rod +MI of the mass ‘M’ about the same axis

=[(mL^2)/3+ML^2]

If V is the velocity of mass ‘M’required for the given condition then angular velocity'w’ at the lowest point is

w=V/L

Thus KE 'K’ of the system is

K=(1/2)[(mL^2)/3+ML^2] (w^2)=(1/2)[m/3+M] V^2

[Since V=L w]

Position of CM of the system'y’ below the point of suspension of the rod is given as

y=[mL/2+ML]/(m+M)=L [m/2+M]/(m+M)

At the highest point potential energy ‘U’[K is zero] is

U=(m +M)g L[m+2M]/(m+M)

=(m+2M) gL

Applying conservation law of energy we get

( 1/2)[m/3+M] V^2=(m+2M) j g L

Or [m+3M] V^2=6 (m+2M) g L

Or. V^2=6 (m+2M) gL/(m+3M)

Or. V=[6 (m+2M)gL /(m+3M)]^(1/2)

If mass of the rod ‘m’ is negligible then

V=[4gL]^(1/2)