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Conservation of Angular Momentum Question

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data
    A 2 meter long bar weighing 90N hangs vertically from a frictionless pivot. The ball is hit at a point 1.5 Meters below the ceiling by a ball that weighs 3Kg and is traveling with a velocity of 10 m/s. The ball bounces back with a velocity of -6 m/s. Find the angular speed of the bar just after the collision.


    2. Relevant equations
    L = Iw = mv x r


    3. The attempt at a solution
    I ended up finding the answer but I don't understand one step.
    At first I set up

    (3kg)*(10m/s)(1.5m) = (3kg)*(-6m/s)(1.5m) + (1/12)(9.18kg)((2^2)m)w
    Solving for w gives 23.5 rad/s, the WRONG answer.

    When I act like the ball's velocity ISN"T in the opposite direction as it was initially, I get the right answer, wtf!?

    (3kg)*(10m/s)(1.5m) = (3kg)*(6m/s)(1.5m) + (1/12)(9.18kg)((2^2)m)w
    w = 5.88m/s

    Is there some weird angular momentum absolute value stuff going on here? I don't get it... Part B asks "Why is the systems angular momentum conserved but the linear momentum is not," I hope it has something to do with the observation I just noticed.
     
  2. jcsd
  3. Oct 8, 2013 #2
    The rod in the problem is hanging by its end, not its center so factor of 1/12 is wrong here, 1/3 should be used. Because the moment of inertia in both cases are different and tend to give different answers. Moreover, mass of rod should be 90/9.81. solve again and you would arrive at the correct answer.

    and as far as linear momentum is concerned, since the rod is pivoted it experiences force at the pivot point linear momentum conservation goes out the window!!!:wink:

    Edit: and frictionless pivot just means that rotation is smooth, so angular momentum is conserved.
    And the second method isn't correct for the given condition, for the reason you already know.
     
  4. Oct 8, 2013 #3
    (With the help of NihalSh on the correction of the equations)

    When calculating angular momentum, it may be easier for you to put the ball on one side, and the bar on the other.

    (3kg)(10m/s)(1.5m) + (3kg)(-6m/s)(1.5m) = (1/3)(90/9.81kg)((2^2)m)w

    You can then take the negative out,

    (3kg)(10m/s)(1.5m) - (3kg)(6m/s)(1.5m) = (1/3)(90/9.81kg)((2^2)m)w

    now adding to the other side, we get your second equation,

    (3kg)*(10m/s)(1.5m) = (3kg)*(6m/s)(1.5m) + (1/3)(90/9.81kg)((2^2)m)w

    which appeared to give you the correct answer.

    For part B, consider the velocities of both the ball, and the pendulum. Velocity is a vector, and has both magnitude and direction. Which objects remain in the same direction, and which ones move?
     
    Last edited: Oct 8, 2013
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