Moments of inertia

[PWrong]

I've been thinking about finding the moments of inertia of 4D rotatopes. This might have been done before, but I'll show them anyway.

A moment of inertia about an axis, or plane is
I = Integral (r^2 dm),
which is a constant for any solid.

The r^2 is derived from K =1/2 mv^2, and v = f cross r
where f is the angular frequency. So assuming we keep the same definition for kinetic energy, we shouldn't change this to r^3.

Here's the derivations for a cylinder and a sphere
http://hyperphysics.phy-astr.gsu.edu/hbase/icyl.html#icyl2
http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html#sph2

For a cubinder with radius R, length L1 and tridth L2, rotating about the two linear axes, we can follow the same method as for the cylinder, and end up with the same inertia.

We can find the inertia of a spherinder by integrating the inertia of a sphere over the length. The length is included in the mass, so it gets cancelled out, and the inertia is identical to a sphere.

These are encouraging results, but the glome and the duocylinder might be more difficult. It's getting late, so I'll try them tomorrow.

[PWrong]

We can find the inertia of a spherinder by integrating the inertia of a sphere over the length. The length is included in the mass, so it gets cancelled out, and the inertia is identical to a sphere.

I should mention that I'm talking about a rolling spherinder, not a "twirling" spherinder. That is, it's rotating about the plane of it's linear axis and one "round" axis.

Anyway, I've just figured out the inertia of a glome, using the same method as the sphere derivation (except I ignore the density by setting it to 1, and just subsitute the mass at the end).

That is, divide the glome (radius R) into tarrow spherinders of radius r, along the w axis. Then integrate the moments of the spherinders about the zw plane.

For each spherinder, dI = 2/5 r^2 dm
and dm = 4/3 pi r^3 dw, so
dI = 8/15 pi r^5 dw
using r^2 = R^2 - w^2, we have

I = Integral ( 8/15 pi (R^2 - w^2) ^(5/2) dw) from -R to R

Now the integral is a lot messier than for the sphere, but it does cancel out nicely. integrator.com gives

pi/90 *(w sqrt[R^2-w^2]*(33R^4 - 26R^2 w^2 + 8w^2)
+ 15 R^6 arctan [w/sqrt[R^2 -w^2]] )

We have to take the limit as w --> R and w --> -R, so we don't divide by zero. Finally, we get
I = pi^2/6 R^6, and substituting the mass, we get another nice formula:

I = 1/3 m R^2

I'll start the duocylinder next, unless someone else wants to try. Remember a duocylinder can have two different radii, r1 and r2.

[PWrong]

I just found the inertia of a duocylinder rotating about the zw plane, but my computer crashed before I could post. So rather than write it out again I'll leave it as an exercise to the reader to show that:

I = 1/4 m r₁^2
where r₁^2 = x^2 + y^2

Here's a table of the moments I've found so far. In each case, I = kmr^2.
The lower the value of k, the faster the solid will roll down an inclined plane/realm.

k(cylinder) = 1/2
k(cubinder) = 1/2
k(sphere) = 2/5
k(spherinder) = 2/5
k(glome) = 1/3
k(duocylinder) = 1/4

It's interesting that the duocylinder is faster than the glome. Any comments on this? Please tell me if I've made a mistake somewhere.