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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:

[tex]I_{i}\omega_{i}=I_{f}\omega_{f}[/tex]

[tex]rmv=\frac{1}{12}\left(m+M\right)d^{2}\omega_{f}[/tex]

[tex]\omega_{f}=\frac{12rmv}{\left(m+M\right)d^{2}}[/tex]

[tex]\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}}[/tex]

[tex]\omega_{f}=18.319\frac{rad}{s}[/tex]

Wrong answer. Any thoughts? Thanks in advance!

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# Homework Help: Angular momentum problem

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