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Conservation of angular momentum of disk and bullet system

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  1. Nov 30, 2015 #1
    I know that the angular momentum is conserved in the below example, but intuitively I am struggling. Anyway, here goes!
    1. The problem statement, all variables and given/known data
    A bullet that is traveling without spinning hits and sticks into a filled disk tilted on its side an arbitrary distance "d" away from the center of mass. The disk and the bullet spin at an arbitrary angular velocity "w" right after the collision. Linear momentum is conserved, so the disk travels to the same height just like it was hit in the center. The system includes the bullet and the disk.

    Q: How is the angular momentum conserved?

    2. Relevant equations
    This is more conceptual, but momentum equations:

    Linear momentum: p=mv

    Angular momentum: L = r x mv = I*w

    3. The attempt at a solution
    A. What is really confusing is that the only way that the angular momentum can be conserved is if the system before the collision has some sort of "relative" angular velocity, which is confusing as hell. It's easier when I think of it as the cross product of the lever arm and linear momentum, but nothing is really rotating before!! A buddy told me it has angular momentum relative to the center of the mass of the disk, and that I just multiply the mass and velocity of the bullet by the radius at which the bullet hits the disk relative to the disk's center of mass. I believe it, but still, this seems counter-intuitive to me. Any conceptual help would be appreciated if anyone has the time! :)
     
  2. jcsd
  3. Nov 30, 2015 #2

    haruspex

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    The description of the set-up is far from clear to me (tilted which way in relation to the bullet's trajectory? How do these relate to the axis of the disk? Is the disk free standing or mounted on an axle?...)
    But to answer your question, given an object of mass m moving with velocity ##\vec v## at position ##\vec r## relative to a point P, the angular momentum of the mass about P is given by ##m\vec r \times\vec v##.
     
  4. Nov 30, 2015 #3
    Thanks for the response! Yeah, the setup looks like this image I found using google. I know when you say is true, I just am having a hard time thinking about angular momentum as a quantity about something. I understand it better now though!

    untitled_435.jpg
     
  5. Nov 30, 2015 #4

    haruspex

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    Imagine intercepting a tennis ball struck straight at your chest by holding a racket head right in front of you. You feel the impulse, but there's no tendency to spin you around. Intercepting a parallel ball at the furthest you can reach out to the side with extended arm and racket would be quite another story. That's angular momentum for you.
     
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