Derive KE of rotating ring and disk

[yosimba2000]

<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?

 

[John Morrell]

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)

 

[yosimba2000]

Oh, I see. I need to add up areas, not lines, to get the final area.

 

[John Morrell]

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.

 

[yosimba2000]

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.

 

[John Morrell]

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.

 

[yosimba2000]

Yeah, I've gone through all of that, but it's been awhile since I've done it so I forgot some stuff.

Thanks!