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Angular Momentum of a turntable

  1. Dec 2, 2005 #1


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    I have tried this problem many times, but I can't even get it started really. Any hints are appreciated!!

    A 60.0 kg woman stands at the rim of a horizontal turntable having a moment of inertia of 500kg/m^2 and a radius of 2 meters. The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (birdseye view) at a constant speed of 1.5m/s relative to earth.

    (a) In what direction and with what speed does the turntable rotate?
    (b) How much work does the woman do to set herself and the turntable into motion?

  2. jcsd
  3. Dec 2, 2005 #2
    Question is incomplete.

    The simple answer might be that she applies the same forces that she would on Earth and the disk would move in the opposite direction.

    What a mind bender, I have not done the math on this but, just by looking at it, she may not overcome the inertia of the disk. If she does then the disk should turn counter-clockwise. I know this is not a complete answer. I had a much better one but turns out I did not read your question correctly. Sorry
    Last edited: Dec 2, 2005
  4. Dec 2, 2005 #3


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    Of course the governing principle behind this is the conservation of angular momentum, so since both the woman and the turn table are both at rest initially, they have a combined 0 angular momentum to start out with, so at any point in time afterward, it must stay zero.

    Angular momentum, L, = the sum of all the pieces of the system,
    L = I * w, where I is the rotational inertia and w (or some purists like to call it omega) is the angular velocity.

    I would like to think this 60 kg woman could be treated as a point mass, and thus, you can find her rotational inertial to be, I = mass*radius^2
    For the turn table, you already know the rotational inertial, 500 kg/m^2

    When the woman begins to walk around the turn table, she has some angular velocity. You can calculate it by dividing the linear velocity the radius. w = v / r.
    Now the woman has some, non zero, angular momentum, so the turn table must too (in the opposite direction) to cancel it out.
    Calculate the angular momentum of the woman, set it equal to the angular momentum of the turn table, then you can solve for the angular velocity of the turn table.

    The work done by the woman to get the turn table and herself moving is equal to the difference in the final and initial kinetic energies.
    KE = 1/2*I*w^2
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