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Angular momentum

  1. Nov 1, 2006 #1
    Hi i have a question with this problem
    A thin, uniform, metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 m below the ceiling by a small 3.00-kg ball, initially traveling horizontally at 10.0 m/s. The ball rebounds in the opposite direction with a speed of 6.00 m/s.

    Find the angular speed of the bar just after the collision.

    I understand that angular momentum is conserved.
    what i have done is found the moment of inertia of the bar and the ball
    I_bar=(1/3)ml^2
    I_ball=m1.5^2
    i then added the 2 together to get the total moment of inertia
    I_bar+I_ball=I
    i then know that the angular momentum is equal to I(w)_init=I(w)_f
    therefore
    w=(mvl)/I
    which means angular velocity is equal to the massof the ball times the velocity (im am not sure which velocity i should use though, should i use both ?)times the distance from the pivot point. all this over I

    is this methodology correct?
     
    Last edited: Nov 2, 2006
  2. jcsd
  3. Nov 2, 2006 #2

    OlderDan

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    The ball and the bar are not moving with the same angular velocity at any time. There is no ball contribution to I. You can still compute the total angular momentum of the system before and after collision. Just calculate the angular momentum of the ball about the pivot point before and after the collision, and require that the total angular momentum (bar plus ball) be conserved.
     
  4. Nov 2, 2006 #3
    hmm, im not quite sure i understand exactly what you are saying. because before the collision the angular momentum of both the bar and ball would be 0 since it does not state that the ball is spinning. sorry im a little confused on what you mean
     
  5. Nov 2, 2006 #4

    OlderDan

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    Angular momentum does not always involve spinning. What is the definition of angular momentum, and I do not mean the equation for rotating objects that is derived from the definition?

    I need to leave, so here is a bit more information. If there is angular momentum after the colision, then there had to be angular momentum before the collision. The rod has none, so it must have been the angular mometum of the ball. The definition is

    L = r x p

    This is a vector equation with the cross product. Find the angular momentum of the ball relative to the pivot point of the rod before and after the collision.
     
    Last edited: Nov 2, 2006
  6. Nov 2, 2006 #5
    ok so then L(before collision)=m10(1.5) and L(after collision)=m(6)(1.5)?
     
  7. Nov 2, 2006 #6

    OlderDan

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    For the ball, yes, except that you have left out one very important detail, and you have not written the units.
     
  8. Nov 2, 2006 #7
    oh right the units agree and would be
    L(before collision)=m(kg)*10(m/s)*1.5(m)
    L(after collision)=m(kg)*6(m/s)*1.5(m)
    so then how can i relate this to the angular momentum of the bar
    would i add these together because they act in the same direction and then set them equal to the angular momentum of the door.
    hence
    L(beforecollision)+L(aftercollision)=L(of bar)
     
  9. Nov 2, 2006 #8

    OlderDan

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    Actually it is
    L_ball(beforecollision) + 0 = L_ball(aftercollision) + L(of bar) because the bar is initially at rest. But you must be careful of the signs when you do the calculation. What is the ball doing after the collision?
     
  10. Nov 2, 2006 #9
    the ball bounced and is now travelling in the opposite direction therefore the angular momentum would be negative
    L_ball(beforecollision) + 0 = -L_ball(aftercollision)+L(of bar)
     
  11. Nov 2, 2006 #10

    OlderDan

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    I know what you mean, but the negative of a negative is a positive. The numerical value of the first term on the right has to be negative. If you put the - sign in front, and the term represented by the words is negative, you wind up with a positive. If you take the initial angular mometum of the ball as positive the left side is positive and the last term is positive.
     
  12. Nov 2, 2006 #11
    oh i get what your saying, sorry thats what it was meant to signify. one last question it gives me the weight of the door in Newtons however i dont think that this is needed so do i just solve for the mass.
    i guess what i am trying to say is the weight is unnecessary for the problem right?
    and thanks for all your help
     
  13. Nov 2, 2006 #12

    OlderDan

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    Are you by any chance confusing this with a similar problem? What door? You do need the mass of the bar in this problem, and you can get that from the weight.
     
  14. Nov 2, 2006 #13
    yes i am sorry i meant to say bar, for some reason ive been using door all morning. haha
    well thanks so much for your help
     
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