# Angular velocity doubt

Here is a rod of mass $M$ and of length $l$ resting on a smooth frictionless horizontal table. Now a small bullet of mass $m$ comes and collides inelastically with this rod and here I wanted to find the velocity of the point H just after the bullet gets embedded in the rod. Now, here first of all as per the image 2, I used the conservation of momentum.I can get the velocity of centre of mass of rod just after collision to be

$$v=mu/(m+M)$$

Also since the body will rotate about the centre of mass with angular velocity $\omega$ (say) and I also know that omega can be easily calculated by using conservation of angular momentum using

$$L_{initial}=L_{final}$$

Taking the point of COM to be point $O$

$$mu(OH)=(I_{rod, O})\omega + m(OH)^2 \omega$$ $I_{rod, O}$ means moment of inertia about point $O$.

Now I have the angular velocity of the bottommost point after the collision and its distance from the COM, but since the COM is itself moving ahead,I suspect that I can't use $v_{o}=(OH) \omega$. Kindly help me in this regard.

You know the angular velocity of the rod and bullet and the linear velocity of the centre of mass of the rod and bullet.

Considering the rotation about the combined centre of mass you can find the linear velocity relative to that centre of mass of any position on the rod.

Now add the two velocities vectorially.

$I_t$ of rod and bullet immediately after the impact $= I+ m(Ml/2 +m.0)/(m+M))^2$

and center of rotation $is = (I (L/2)^2 + 0)/I_t$

Conservation of momentum: $mu= I_tw + (m+M)v$

v is the horizontal velocity immediately after impact.

Also $w. X_{\text{distance to center of rotation}}= v$

So we can solve for v.