see in order to apply angular momentum conservation that axis needs to be chosen on which no external torque is acting
if you look at that point where the particle sticks to the rod(which i am assuming to be the bottom) then there is no angular impulse acting on the combined system at that point so you can conserve te angular momentum at that point.
but there is one special point in the combined system about which also you can apply angular momentum conservation and that point is the centre of mass os the entire system.
step 1 --- find the location of centre of mass
assume origin to be at the bottom of the rod
location of centre of mass = ((M * L/2) + (m * 0))/(M+m) = ML / 2(M+m) = d (say)
step 2 --------- initial angular momentum about the centre of mass
it include contribution towards angular momentum from the particle as well as the rod
contribution from the particle ------ since the particle is d below the centre of mass
ang1 = ( - d )j * (mv)i (j - vector in the +y dir
i - vector in the +x dir
* cross product)
contribtion from the rod ---------- since the rod is stationary
ang2 = 0
=> total angular momentum initial = ang1 + ang2
finally the rod and the particle will start rotating about the centre of mass and centre of mass will be translating with a velocity which it had earlier
ie V = mv / (M + m)
don't worry about the translation just take the final angular momentum as
I * w
where I is the moment of inertia of the combind system about the centre of mass which needs to be calculated as follows
moment of inertio of the particle I1 = m * d2
moment of inertia of the rod about centre = M * L2 / 12
moment of inertia about the centre of mass = I2 = M * L2 / 12 + M * (0.5L - d)2
=> I = I1 + I2
you know I ,you know initial angular momentum
hence you can calculate w, the final angular velocity