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A uniform thin rod of length 0.40 m and mass 3.0 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is a rest when a 3.0-g bullet traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet velocity makes an angle of 60° with the rod (see the figure). If the bullet lodges in the rod and the angular velocity of the rod is 12.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

What I've done:

As I read this I thought of three ideas:

1. Conservation of Angular Momentum, [tex]P_b=L_r[/tex]

2. Inelastic Collision

3. Parrallel Axis Theorem (with the Mass of the Bullet at the end of the rod after it hits it.

First thing I tried was quite simple. I simply set [tex]P_b=L_r[/tex] but the velocity that I found was incorrect. Suspecting that I was neglecting using the angle, I then tried [tex]P_b\sin(\theta)=L_r[/tex]. This, however, also left me with an incorrect result. As I don't see anything wrong in my methods for finding velocity I think I may have made a mistake in calculating the Moment of Inertia of the rod in it's final state which I have as:

[tex]I_f=\frac{(m_r)(L^2)}{12}+\frac{(m_b)(L^2)}{4}[/tex]

and thus:

[tex]V_b=\frac{[\frac{(m_r)(L^2)}{12}+\frac{(m_b)(L^2)}{4}]\omega}{(m_b)\sin(\theta)}[/tex]

Where is the mistake here? I assume it is legal to convert from translational momentum to rotational momentum like this?

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# Homework Help: Angular Momentum of Rod Question

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