# Moment of inertia

1. Apr 14, 2005

### johnnyb

I have to show what the moment of inertia of a rectangular plate with mass M and sides A and B is about its centre of mass. I have come up with
$$\frac{1}{12}M(a^2 + b^2)$$

Now I have to show what the moment of inertia of the same plate is except this time about an axis perpedicular to the plate and passes through one corner. I know it is:
$$\frac{1}{3}M(a^2 + b^2)$$ But having some problems proving it

2. Apr 14, 2005

### whozum

3. Apr 14, 2005

### johnnyb

Yes it is.

I = I (at centre of mass) + $$Md^2$$

But I can't see how that can get me from the first to second equation

4. Apr 14, 2005

### ramollari

Easy,

$$I_{cm} = \frac{1}{12}M(a^2 + b^2)$$

The displacement is half the diagonal, that is,

$$\frac{a^2 + b^2}{4}$$

So add them up and you get:

$$I = \frac{1}{3}M(a^2 + b^2)$$

5. Apr 14, 2005

### dextercioby

Steiner's theorem.That's the name i learnt once with the theorem itself...

Daniel.