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Moment of inertia question

  1. Aug 22, 2011 #1
    If you have a roundabout spinning with a man standing on it close to the centre, and then he walks out towards the edge of the roundabout, angular momentum is conserved, but kinetic energy is not (the roundabout rotates with a smaller angular speed). I'd like to know where the kinetic energy in this system goes to. Is the changed moment of inertia of the system some kind of equivalent to potential energy? As far as I know, ignoring all resistive forces, if the man returns to original position the kinetic energy will be as it was to start with. So surely energy is not being created or destroyed...surely there must be a constant total mechanical energy here, right?
    Thanks in advance
    Tom
     
  2. jcsd
  3. Aug 22, 2011 #2

    phyzguy

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    As the man moves outward, there is a centrifugal force pushing him outward. If he moves slowly, there must be some sort of damping or he will be accelerated and flung off the edge. This damping dissipates energy. Alternatively, if he 'free falls' as he moves outward, then he needs to be stopped when he reaches the edge of the roundabout. Whatever force stops him will dissipate his kinetic energy of motion in the radial direction. So the system has less kinetic energy when he is at the edge, which is what you said. When he is at the edge and moves back inward, he needs to do work against the centrifugal force to pull himself back in to the center. The work done adds to the kinetic energy of the system to get back to where you started.
     
  4. Aug 22, 2011 #3

    Drakkith

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    Isn't the kinetic energy conserved here? While the whole thing might spin slower, the greater momentum will mean equal energy right?
     
  5. Aug 22, 2011 #4

    Doc Al

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    No. Angular momentum is conserved, but not kinetic energy.
     
  6. Aug 22, 2011 #5
    I haven't carefully analyzed this, but I think it works like this:
    Near the center, the man has a very small tangential velocity. The tangential velocity increases as his approaches the edge. The kinetic energy and momentum of the roundabout must decrease by the amount that his increase, so it slows down.

    An aside: is a "roundabout" that playground toy that holds lots of kids and spins around? I've heard "roundabout" refer to what I call a traffic circle.
     
  7. Aug 22, 2011 #6
    Since potential energy is not changing and no external work is done, isn't kinetic energy conserved?
     
  8. Aug 22, 2011 #7

    Doc Al

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    No. External work isn't required for a kinetic energy change.

    But an external torque is required for a change in net angular momentum. Since there's no external torque here, angular momentum is conserved.
     
  9. Aug 22, 2011 #8

    Doc Al

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    Yes. It's also called a 'merry-go-round' or a carousel.
     
  10. Aug 22, 2011 #9

    Drakkith

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    Ah ok, I see.
     
  11. Aug 23, 2011 #10
    Thanks everyone. That's a lot clearer.

    Just to clarify, yes, where I'm from a roundabout is the object you described found in children's play parks, as well as a traffic device.
    Carousels and merry-go-rounds are the things found at funfairs, where you sit on a horse or a spaceship and you go round daintily to the tune of music.
     
  12. Aug 23, 2011 #11

    Doc Al

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    Right, the playground version of the roundabout is much simpler than the fairground versions with things to ride; they look something like this:
    ldroundabout_05.gif

    Or this:
    [PLAIN]http://www.mclaysplay.com/images/wicksteed/uploads/images/products/spiro-whirl-48.jpg [Broken]

    Lots of dangerous fun!

    See: http://en.wikipedia.org/wiki/Roundabout_(play [Broken])
     
    Last edited by a moderator: May 5, 2017
  13. Aug 23, 2011 #12
    I'm glad we sorted out the playground terminology, but I'd also like to clarify something in the physics. I see that this amounts to the spinning skater problem and that the rotational approach is elegant. Since tomwilliam asked about changes in kinetic energy, I'd like to understand it those terms.

    Please check my reasoning:
    Kinetic energy in general is not expected to be conserved, but total energy is. We are assuming there are no dissipative forces. Doesn't that mean that that the total energy is the sum of the kinetic and potential energies? And if the walking man never changes altitude, the potential is a constant, so the total kinetic energy is constant.

    The translational kinetic energy of the system at any moment depends on the mass and tangential velocity of each element of the system. The man's tangential velocity (and therefore his translational kinetic energy) increases as he walks from center to perimeter. The tangential velocity (and therefore the translational kinetic energy) of every point on the roundabout decreases. He gains kinetic energy, the roundabout loses kinetic energy, and energy is conserved.

    I know that's a clunky way to go about it, but are the errors in the reasoning?
     
  14. Aug 23, 2011 #13
    I think it's because you don't require force to go outside, but you need to do that when returning to center. So work is done while walking towards the center and thus negative work is done to the reversed process.
     
  15. Aug 23, 2011 #14

    Doc Al

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    If by 'total energy' you mean total mechanical energy (PE + KE), then total energy is not conserved. If you extend total energy to include thermal and chemical energy, then it is. You cannot assume there are no dissipative forces at work here--the person is not a rigid body, but has internal energy.

    Depending upon the mass of the roundabout, the man's KE may well increase as he moves toward the edge. But you cannot assume that KE is conserved. (You can easily show that KE and angular momentum cannot both be conserved.) If the rotational KE lost by the roundabout merely went into the man's translational KE, he would be moving too fast. He must dissipate that extra energy as he moves outward.
     
  16. Aug 23, 2011 #15
    Any time one moves against a force, work is done, as per work = integral force over distance times the cosine of the angle between the vectors. What this means in context is that when the man changes position (Displacement), work is done against centripetal acceleration (Force). If the man moves radially outwards, the angle between is zero, and thus the work done with centripetal acceleration adds to the man's kinetic energy.

    In practice, to return to his initial position, work must be done; that work doesn't disappear, it simply changes forms. Some of the work done as the man moves out may be lost as friction(he has to slow down to prevent from flying off). When he pulls himself back to the center, he must use energy, most likely in the form of stored chemical energy (his muscles).

    If anyone feels like doing the math, the merry-go-round would be approximated as a spinning disk with
    KE = 1/2 I [itex]\omega[/itex]2
    P = I [itex]\omega[/itex]
    (I is moment of inertia, [itex]\omega[/itex] is angular velocity)
    Where I for a disk is 1/2 m r2
    The man is a point mass on the disk with
    KE = 1/2 m V2
    P = I [itex]\omega[/itex]
    I = m r2
    V = r [itex]\omega[/itex]
    Realize then that angular momentum is conserved, IE: Initial Total Momentum is equal to Final Total Momentum. The change in kinetic energy of the system then turns into a different type of energy.
    Alternately, to simply find the work done by the man moving, you could integrate centripetal force through the (radial) distance traversed by the man. Of course, to do that accurately you'd still have to calculate a changing value for [itex]\omega[/itex]...

    Unfortunately either approach becomes rather complex when you realize that both [itex]\omega[/itex] and r change with distance. I think it's solvable, but I'm not feeling like it atm.
     
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