Angular speed around a bar/clay system's center of mass after impact

[pixelized]

On a frictionless table, a glob of clay of mass 0.740 kg strikes a bar of mass 1.740 kg perpendicularly at a point 0.140 m from the center of the bar and sticks to it. If the bar is 0.660 m long and the clay is moving at 9.600 m/s before the impact, what is the final speed of the center of mass?

I have solved this part. The next part is the one I'm stuck on.

At what angular speed does the bar/clay system rotate about its center of mass after the impact in radians/second?

I have tried conservation of energy 1/2m1v1 = 1/2m2v2 + 1/2 Iw^2 but my answer is wrong.

Final equation. w = (m1v1-m2v2)/(1/2ML^2)

Could someone tell me where I went wrong?

 

[michael376071]

Might just be a typo, for energy, I believe its $$\frac{1}{2}mv^2$$ But you might already have that, think you are on the right track with both angular energy and translational though.

 

[thepatientmental]

It looks like you have set up the equation correctly, but you may have made a mistake in your calculations. It's always a good idea to double check your work and make sure all units are consistent throughout the equation. Additionally, make sure you are using the correct values for the moments of inertia for the bar and clay. If you are still getting an incorrect answer, you can also try using the conservation of angular momentum equation, which states that the initial angular momentum must equal the final angular momentum. This equation can also be used to solve for the final angular speed.