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Angular Momentum of Rod Question

  1. Nov 4, 2006 #1
    [​IMG]

    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?
     
    Last edited: Nov 4, 2006
  2. jcsd
  3. Nov 4, 2006 #2

    ehild

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    Incorrect. Use the angular momentum of the bullet instead of its translational momentum.

    ehild
     
  4. Nov 4, 2006 #3
    OK, so is this correct:

    [tex]L_b=m(\frac{L}{2})^2\frac{V_b\sin(\theta)}{\frac{L}{2}}=L_d[/tex]

    So, you cannot convert straight from Translational to Rotational Momentum? Why not?

    Also, is the angular momentum equation for the rod+bullet correct from my first post?
     
    Last edited: Nov 4, 2006
  5. Nov 5, 2006 #4

    ehild

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    You can not compare angular momentum with translational momentum. They are different quantities. The conservation of angular momentum means that the angular momentum of a system does not change when no external torgue is applied. The initial angular momentum is that of the bullet, as the rod is in rest. How do you calculate the angular momentum of the bullet before collision?

    ehild
     
  6. Nov 5, 2006 #5
    The angular momentum of the bullet relative to the centre of mass of the rod just before the collision is

    [tex]L_b=m(\frac{L}{2})^2\frac{V_b\sin(\theta)}{\frac{L}{2}}[/tex]

    I've got the answer, thanks for the help.
     
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