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Conservation of Angular Momentum help

  1. Nov 29, 2013 #1
    1. The problem statement, all variables and given/known data
    For an assignment, I was shown a video where two identical pucks were launched at each other. They were not spinning when launched. They had Velcro on their edges so they stuck to each other when they collided. They hit off-center from each other. Due to conservation of angular momentum, when they collided, they stayed in place at the point of collision and just rotated.

    Then I was asked, if the mass of each puck was doubled, what would happen to the angular speed, ω.


    2. Relevant equations
    ----eqn1--------- m(r^2)ω
    ----eqn2--------- m(r^2)ω + m(r^2)ω = (m(r^2) + m(r^2))ω



    3. The attempt at a solution
    First, I tried using the simple equation for angular momentum:
    m(r^2)ω
    If the mass is doubled, then angular speed is halved.

    But then I saw another equation for conservation of momentum:
    m(r^2)ω + m(r^2)ω = (m(r^2) + m(r^2))ω

    With this equation, when mass is doubled, angular speed stays the same.
    Which one do I use? Or am I even going about this correctly?
     
  2. jcsd
  3. Nov 29, 2013 #2
    Haha ignore your second equation. If you factor in omega on the right side you will see that it doesn't really tell you anything.

    Angular momentum = [itex]m r^2 ω[/itex]
    However, we must remember that there are TWO pucks spinning. Since each puck has a mass of m, the total mass is 2m.
    Total angular momentum: [itex]L = 2mr^2ω[/itex]
    It is common to call angular momentum L.
    This is the total angular momentum of two pucks spinning.
    If you double the mass then angular momentum would be:
    [itex]2(2m)r^2ω_2[/itex] where ω2 is the new angular speed. We replaced m with 2m and since we did that, the angular speed is unkown, so we represent it with ω2 and solve.
    [itex]L=2mr^2\omega=2(2m)r^2\omega_2[/itex]
    [itex]\omega=2\omega_2[/itex]
    [itex]\omega_2=\frac{1}{2}\omega[/itex]

    So, when you double the mass, the angular speed ω gets halved.
     
  4. Nov 29, 2013 #3

    haruspex

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    But you (and Hertz) are assuming angular momentum is the same initially. Why should that be? Are you reading it as the masses magically being doubled while they're spinning? I doubt that's what's intended.
    I don't understand how you would be applying that here.
    There is a very quick way of answering a question like this: you notice that mass, as a dimension, can be 'factored out'. There are no dimensional magic constants involved (like G, say), and mass does not feature in the dimension of the quantity of interest. It follows that you can scale all the masses involved without affecting the answer.
    But let's also analyse it from first principles. First, you have to pick a reference point for the angular momentum. The common mass centre would be a good choice. Since it's an oblique collision, the lines of travel of the two pucks do not pass through the reference point. This means that each has an angular momentum about the reference point before collision. If they approach at speed v and (if they could pass through each other) would each miss the common mass centre by distance d, what is the total angular momentum about the common mass centre before collision?
     
  5. Nov 29, 2013 #4
    Thanks for the reply Hertz.

    Maybe I described the scenario/question wrong, because your answer was incorrect. =(
    They spin at the same rate.

    Here is their explanation:

    By doubling the mass but keeping the velocities unchanged, we doubled the angular momentum of the two-puck system. However, we also doubled the moment of inertia. Since L=I ω, the rotation rate of the two-puck system must remain unchanged.
     
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