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Derive KE of rotating ring and disk

  1. Apr 1, 2017 #1
    <Mentor's note: Moved from a technical forum and thus no template.>

    A ring's kinetic energy is integral of 0.5v2 dm. Distance X is rΘ, and Θ is defined as distance traveled/radius, so X is r*distance traveled / r. Velocity V is X divided by time, so V is r*distance traveled / rt, and I define omega w as distance traveled / rt.

    Plugging into integral of 0.5v2dm, I get 0.5(r*distance traveled / rt)2dm, and I get 0.5mr2(distance traveled2/r2t2), which is equal to 0.5mr2w2.

    I find that my KE of the ring is 0.5mr2w2.

    For the approach to a disk, I can add the kinetic energy of concentric rings.

    So KE of ring is 0.5mr2w2, and w is distance traveled / rt, so I get 0.5mr2(distance traveled2/r2t2). Canceling the r2 gives me 0.5m(distance traveled2/t2). Integrating this from radius of 0 to r gives me 0.5mr(distance traveled2/t2).

    If I multiply by r2/r2, I get 0.5mr*r2(distance traveled2/t2r2), and (distance traveled2/t2r2) is w2, so my KE of the disk is 0.5mr3w2.

    Is this correct?
    Last edited by a moderator: Apr 1, 2017
  2. jcsd
  3. Apr 1, 2017 #2
    You are close, but not quite right. I believe the answer should be (pi)(w^2)(area-density)(r^3)/(4). By the way, someone should check that, I often make algebra mistakes. So you're first step is right on. The KE = 1/2mv^2, or the integral of 1/2v^2*dm. Velocity, for a rotating bit of mass, equals omega (w) times the radius r, so we can replace v^2 with w^2*r^2. Now, what is dm? Well, it's an infinitesimal bit of mass. I think the easiest way to think of this is to say that it's the mass of a infinitely skinny ring a radius r away from the center of rotation. That ring has a thickness of dr, and a length )if you laid it out straight) of 2*pi*r*dr. So the mass of that ring is the density per unit area, which I'll call k, times the little tiny area, 2pi*r*dr.

    So our integral is now 2*pi*r*k*w^2*r^2*.5*dr, or simplifying things, pi*k*w^2*r^3 dr. Integrate that from 0 to R, and we get (pi*w^2*k*R^4)/(4)
  4. Apr 1, 2017 #3
    Oh, I see. I need to add up areas, not lines, to get the final area.
  5. Apr 1, 2017 #4
    exactly. The main skill in doing integrals like that is to accurately find your infinitesimal quantity. It might be a tiny force or a tiny mass or a tiny length, but it can't be zero. It just has to approach zero. That's why it's so helpful to think of integrals first as Riemann sums, where we approximate a value by adding up lots of small subdivisions.
  6. Apr 1, 2017 #5
    Yeah, what threw me off was that I'm too used to doing integrals of y(x) functions that I forget dx is the little width of each piece and eventually it just became embedded in my mind that I was adding up little lines to get area, when actually I was adding up little areas to get area under the curve.
  7. Apr 1, 2017 #6
    The way that calculus generalizes to other things is incredibly cool. When you get a chance, make sure you learn multi-variable calculus and stuff like that. Triple integrals and unit conversions are so fun.
  8. Apr 2, 2017 #7
    Yeah, I've gone through all of that, but it's been awhile since I've done it so I forgot some stuff.

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