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Work for a rotating circle

  1. Nov 23, 2008 #1
    Hi guys, I am trying to figure out how you can apply the concept of work to a rotating circle. I am trying to figure out how to calculate the energy required to bring a rotating circle (assumed to be a uniform lamina) to rest.

    The circle has the following properties:
    - Radius r (m)
    - Mass m (kg)
    - Now I will have the linear edge velocity ...
    Initial: v (ms-1)
    Final: u (ms-1) (which will me 0 for me :D)
    - ... and I assume I will need the angular velocity, so
    Initial: [itex]{\omega}_{v}[/tex]
    Final: [tex]{\omega}_{u}[/tex]

    Now I dont know where to go from here. Obviously I cant directly apply Wd = Fxd, well I say that, I suppose I could express d as a function of; the linear velocity at a point on the circle, the deceleration at that point and the time of deceleration, then integrate across all the circle, But I don't think my Maths would be good enough, maybe after like a few weeks I could crack it. Is there a simpler way to look at this problem, or and area of mechanics that deals with this, I know "circular motion", but I have googled and looked though all resources I have available (in upper sixth or final year of high school if your in America) and cant find the specific area of circular motion dealing with this sort of mechanics. And help would be appreciated, thanks guys.

    This is all out of curiosity hence the slight ambiguity in not asking a question with actual values to deal with, thanks :D
     
  2. jcsd
  3. Nov 23, 2008 #2

    tiny-tim

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    Hi Galadirith! :smile:

    Work done = force times (perpendicular) distance …

    if there's an ordinary linear force, like a brake on the rim of a wheel, then just multiply the force by the distance the rim moves …

    if there's a torque (about the axle, say), the torque is already a force times a distance, so just multiply it by the angle moved (angle is dimensionless :wink:) … if that bothers you, remember that a torque can always be replaced by two parallel equal and opposite forces, and that, for the same torque, the forces will be inversely proportional to the radius, but the arc-distance will be directly proportional to the radius … so the radius cancels out, and all you have left is the angle. :smile:
     
  4. Nov 23, 2008 #3
    Thanks tiny-tim, that helps alot, I actually have something to work from now, I actually just did had a quick look at some stuff online related to torque, and that looks great, we haven't actually studied torque yet so It wouldn't have been something I would have looked at :D thanks tiny-tim.
     
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