How do you calculate the work needed to stop a rotating circle?

Galadirith
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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: {\omega}_{v}[/tex] <br /> Final: {\omega}_{u}<br /> <br /> Now I don&#039;t know where to go from here. Obviously I can&#039;t 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&#039;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 &quot;circular motion&quot;, 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 can&#039;t find the specific area of circular motion dealing with this sort of mechanics. And help would be appreciated, thanks guys. <br /> <br /> This is all out of curiosity hence the slight ambiguity in not asking a question with actual values to deal with, thanks :D
 
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Galadirith said:
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.

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:
 
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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