Homework Help: Proving generality of moment of inertia

1. Dec 19, 2016

TubbaBlubba

1. The problem statement, all variables and given/known data
Consider a circular disc (that might be a tautology), free to rotate and translate in the xy-plane only. It has radius R, but its mass is unhomogeneously distributed in some unknown fashion. All we know is that the center of mass is displaced from the centre of the disc by some distance a, and that it has mass m.

We can model a disc with such a center of mass by superimposing a disc and a point mass on its edge. After finding expressions for their masses in terms of m and a/R, some algebraic manipulation, and summing the respective known-from-theorem moments of inertia for the two ideal objects, we arrive at the moment of inertia for rotation around the center...
2. Relevant equations
I = m * (1/2) * R^2 * (1 + a/R).
3. The attempt at a solution
So, we get a moment of inertia that is a weighted mean between that of a disc and a point mass orbiting at distance R. My intuition is that under these constraints, this is the MoI for ALL circular discs characterized by radius R, displaced CoM ratio a/R, and mass m. I feel like there should be a relatively simple argument that shows this, that as long as the shape and radius of the object and the CoM remain the same, the rest of the mass distribution doesn't matter under the provided constraints. But I can't think of one, and that is driving me crazy...

Anyone here that could help?

2. Dec 19, 2016

haruspex

To the extent that it has the same mass, same outline, same mass centre (in relation to the disc outline) and same moment of inertia about the disc's centre? Or in what sense?
Since the mass distribution can be anything with the right mass centre, I do not believe the MoI is predictable. Just consider the special case a=0. At one extreme, all the mass is at the disc's centre, the disc being reduced to a notional region. At another, it is all at the periphery.

Maybe I have misunderstood your post.

3. Dec 19, 2016

TubbaBlubba

But the equation I derived for the moment of inertia works for both those cases. As a->0, we get the MoI of a disc (solid cylinder, if you will) that is homogeneous. As a->r, we get that of a point particle orbiting the centre. I'm NOT talking about hollow cylinders!

I am considering the CoM being displaced radially by a, yes, in some direction, while the mass remains the same (which is crucial!). Does it matter if the mass is mostly in a lump around the CoM or if it's a solid cylinder of some light material with a small heavy weight on the outside?

If not, what are some alternative moments of inertia for the problem posed, expressed in the parameters R, a and M?

4. Dec 19, 2016

haruspex

Sure, but your specification allows for all the mass being concentrated at the centre, leading, in the limit, to zero MoI.

Edit: I can offer this: given a mass M, a moment of inertia Mk2 and a radius R, you can construct a lamina from a homogeneous disc radius R>4k/3, mass m, and a point mass M-m at its periphery, such that its moment of inertia about its mass centre is Mk2. If my calculations are right, $\frac mM=\frac 34\pm\sqrt{\frac 9{16}-\frac{k^2}{R^2}}$.

Edit 2: ... But I think I made a mistake somewhere...

Last edited: Dec 19, 2016
5. Dec 19, 2016

TubbaBlubba

Hm, that's right. So we need another condition of constraint I suppose...

I'm doing work on a Lagrangian related to this. Shape, R, a and m are the only quantities given for the disc. The MoI just gives a scalar factor in the denominator (from acceleration due to kinetic energy) for the Eqs of Motion and aren't that important here but I still want to make sure I don't miss any interesting cases, and make sure to nail down how general the MoIs I choose are. I guess I should limit myself to a "perturbed disc" that is a superposition of a homogeneous disc and some other mass distribution. I wonder if my equation is general enough then?

Alternatively I could write an expression that covers the entire range between the extremes of point mass in middle and point mass at rim. But then I'd need an additional weight factor... Hm.

Last edited by a moderator: Dec 20, 2016