# Rotational Inertia of a Cube

1. Jan 1, 2007

### montreal1775

Does anyone know what the rotational inertia of a cube of uniform density is when it is rotated about an edge? Any help is appreciated!

2. Jan 1, 2007

I suggest you start with the definition of rotational inertia of a rigid body of uniform density.

3. Jan 1, 2007

### montreal1775

I=integral(r^2,m)

but then what do I do?
I'd use the parallel axis theorem but I don't know how to find the rotational inertia for a cube about it's center of mass.

Last edited: Jan 1, 2007
4. Jan 1, 2007

### mbrmbrg

You need to be able to derive the equation; you can't look it up (in your textbook, etc.)?

5. Jan 1, 2007

### Saketh

$$I = \int r^2 \,dm = \rho \int r^2 \,dV$$
where $\rho$ is the density of the cube (assumed to be uniform).

You can then convert r into a Cartesian equivalent, then split dV into dx, dy, dz and do a triple integral.

6. Jan 1, 2007

### Ja4Coltrane

I think you should consider the cube from above a plane with greater mass--that is-- 1/12(m)(2L^2). Then parallel axis it.

7. Jan 2, 2007

### montreal1775

So would it be $$I = \iiint_{0}^{L} x^2+y^2+z^2 \,dxdydz$$ where L is the length of a side?

Last edited: Jan 2, 2007
8. Jan 2, 2007

### montreal1775

Also, Ja4Coltrane, I don't understand what you mean. Could you please elaborate?

9. Jan 2, 2007

### AlephZero

r is the distance from the axis of rotation, not from the centre. If you are rotating about the z axis, then r^2 = x^2 + y^2.

Re Ja4Coltane's post, possibly he means the inertia of the cube about one edge is essentially the same as the inertia of a square about one corner. The thickness contributes to the mass, but it doesn't affect the geometric part of the formula.

10. Jan 2, 2007

### montreal1775

So is $$I = \rho \iiint_{0}^{L} x^2+y^2+z^2 \,dxdydz$$ the rotational inertial of a cube rotated about a corner?

Last edited: Jan 2, 2007
11. Jan 2, 2007

### AlephZero

What do you mean by "the rotational inertial about the origin"? I know what rotation inertia about a line is, and I know what the rotation inertia tensor about a point is (and it's got 6 independent components, not just one).

The rotation inertia about the edge defined by (x=0, y=0) is $$I = \iiint_{0}^{L} x^2+y^2 \,dxdydz$$ That's very similar to the inertia of a square about one corner.

12. Jan 2, 2007