Find the angular speed of the system right after the collision

[Impathy]

The problem:
A projectile of mass m=1.8 kg moves to the right with speed v=24.8 m/s. THe projectile strikes and sticks to the end of a stationary rod of mass M=6.75 kg and length d=1.71 m that is pivoted about a frictionless axle through its center. Find the angular speed of the system right after the collision.
I know the I of a thin rod pivoting about its center is 1/12 Md^2 and that angular momentum is mrv. So with that, I tried this:
$$I_{i}\omega_{i}=I_{f}\omega_{f}$$
$$rmv=\frac{1}{12}\left(m+M\right)d^{2}\omega_{f}$$
$$\omega_{f}=\frac{12rmv}{\left(m+M\right)d^{2}}$$
$$\omega_{f}=\frac{12\left(\frac{1.71m}{2}\right)\left(1.8kg\right)\left(24.8\frac{m}{s}\right)}{\left(1.8kg+6.75kg\right)1.71m^{2}}$$
$$\omega_{f}=18.319\frac{rad}{s}$$
Wrong answer. Any thoughts? Thanks in advance!

 

[Fermat]

error is in the 2nd line, where you have taken the inertia as being for a mass of (m+M) rotating about the rod axle.

You should have the sum of two separate momenta here, one for M rotating about the rod axle and the other for a point mass, m, rotating about the rod axle, at a distance d/2, with the the same angular velocity.

 

[Impathy]

Awesome, got it. Thanks a million for your help!