# Hollow cube - moment of inertia

1. Jan 12, 2009

### blr

1. The problem statement, all variables and given/known data
How to calculate moment of inertia of hollow cube.

2. Relevant equations

3. The attempt at a solution
I guess that subtracting the moment of inertia of the inner cube from the moment of inertia of the outer cube is wrong.
Even it is close to solution, what mass to put in formula?

2. Jan 12, 2009

### Delphi51

A good idea to subtract away the inner cube!
The mass of the hollow cube will appear in your answer; if you don't know it just leave it as m.

3. Jan 13, 2009

### blr

I know mass of hollow cube, and it is m...
Moment of inertia of cube is m a^2/6.
So moment of inertia of hollow cube is m1 a^2/6 - m2 b^2/6....
I need help with m1 and m2...

4. Jan 13, 2009

### Carid

The density of the initial solid cube is a constant, so the mass is proportional to the volume. The volume of the removed cube and the volume of the remaining cubic shell are calculable in terms of a and b.

5. Jan 13, 2009

### blr

I know that, but I don't know how to do it :(

6. Jan 13, 2009

### Staff: Mentor

What is volume of the cube with an edge length a?

7. Jan 13, 2009

### blr

??? a^3

8. Jan 13, 2009

### Staff: Mentor

Good. What is its mass (assuming density d)?

9. Jan 13, 2009

### blr

d a^3

10. Jan 13, 2009

### Staff: Mentor

So where is the problem?

11. Jan 14, 2009

### blr

So...
I(of hollow cube) = m * $$\frac{a^3}{a^3-b^3}$$ * $$\frac{a^2}{6}$$ - m * $$\frac{b^3}{a^3-b^3}$$ * $$\frac{b^2}{6}$$

that is

$$\frac{m(a^5-b^5)}{6(a^3-b^3}$$

Is that ok?

12. Jan 15, 2009

### blr

anyone?

13. Jan 15, 2009

### Staff: Mentor

I don't get it. What is ma^3/(a^3-b^3)?

14. Jan 15, 2009

### blr

Mass of hollow cube is given. So that is mass of full cube, before removing inner cube with edge length b.

15. Jan 15, 2009

### Staff: Mentor

Ah, OK.

Looks OK to me, but I made so many stupid mistakes lately that I don't trust myself these days.

16. Jan 16, 2009

Thank you...