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Discs/Hoops Rotational Physics Question

  1. May 8, 2013 #1
    1. A hoop with mass 2.0 kg and radius 1.5 m is rotating with initial angular velocity of 5.4 s-1. A disc with the same mass and radius is dropped on top of the hoop so that the centers coincide. What conservation law applies here? What assumptions must you make to apply it? What is the final angular velocity?



    2. Kinetic energy 1 = kinetic energy 2

    Ke hoop = m x v^2
    ke disc = 1/2 m x v^2




    3. Conservation of momentum
    There must be no additional external forces like air resistance or friction

    ke hoop = 2.0 kg x 5.4 s-
    Ke hoop initial = 10.8 kg m s-

    Now i dont know if this works like a simple collision where I would factor in the new mass of the disc placed on it... but I am sure that I would have to factor in the moment of inertia of said disc so 1/2 m x v^2 to determine the change in angular velocity.
     
  2. jcsd
  3. May 8, 2013 #2

    Simon Bridge

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    What conservation law applies here?
    -- conservation of angular momentum.

    What assumptions must you make to apply it?
    -- no friction? What is holding the two objects together then?

    What is the final angular velocity?
    -- your stated conservation law was "momentum" so you should be using the momentum equations here and not the KE ones.

    The way to tackle conservation problems is to think in terms of "before" and "after" stuff happens - ignore the "during". The math works much the same as the linear case you've done before but with the rotational equations - so mass becomes "moment of inertia" and velocity becomes "angular velocity" so ##p=mv## (linear momentum) becomes ##L=I\omega## (angular momentum)
     
  4. May 9, 2013 #3

    haruspex

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    Boognish did specify external forces.
     
  5. May 9, 2013 #4
    so understanding what you have told me i think that it resolves to:

    Li= m r^2 w
    Li = 2.0 x 1.5^2 x 5.4-s
    Li = 24.3

    Lf = Li

    Lf = (Ii + If) x wf

    Lf = ((m R^2 + (1/2 m R^2)) x Wf

    24.3 = ((2.0 x 1.5^2) + (.5 x 2.0 x 1.5^2)) x wf
    wf = 3.65s-

    Any thoughts???
     
  6. May 9, 2013 #5

    Simon Bridge

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    Oh yes, my bad. "Friction was an example of the kind of thing. Question still stands though.
    Objects held together by internal forces?

    @boognish: that's what I described all right.
    I'd watch the labels myself though ... one may expect that Lf = If.wf where If = I(hoop)+I(disk)
    You wouldn't want to do all that work and get marked down because the marker misread you.
     
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