# Moment of inertia of a rectangular prism

1. Jan 19, 2010

### benhou

I was trying to derive the formula for the moment of inertia of a rectangular prism at its centre of mass. And ended up with this:

$$\frac{1}{2}M(a/2)^{2}+2\frac{M}{ab}\int^{b/2}_{a/2}arccos(\frac{a^{2}}{2r^{2}}-1)rdr+4\frac{M}{ab}\int^{\sqrt{a^{2}+b^{2}}}_{b/2}(\frac{\pi}{2}-arccos\frac{b/2}{r}-arccos\frac{a/2}{r})rdr$$

Due to the limitation of my knowledge in Calculus, I was stuck here. Can anyone help me solve these two integral:

$$\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr$$

$$\int (\frac{\pi}{2}-arccos\frac{b/2}{r}-arccos\frac{a/2}{r})rdr$$

2. Jan 20, 2010

### Bob S

3. Jan 21, 2010

### benhou

Yeah, the cuboid. Do you know how to do it? My formula doesn't look like it is going to work.

4. Jan 21, 2010

### Bob S

I = ∫-y/2+y/2-x/2+x/2ρ r2 dx dy = ∫-y/2+y/2-x/2+x/2ρ[x2+y2] dx dy =2∫ox/2ρyx2dx +2∫oy/2ρxy2dy

Where ρ is the density. So ρ = mass/xy

You can take it from here.

Bob S

Last edited: Jan 22, 2010
5. Jan 21, 2010

### benhou

could you explain how this come about?

6. Jan 22, 2010

### ngchou

(1/2)*M((1/2)*a)^2+(1/4)*M*(b^4*arccos((2*a^2-b^2)/b^2)*sqrt(-a^2*(a^2-b^2)/b^4)+2*a^4-2*a^2*b^2)/(sqrt(-a^2*(a^2-b^2)/b^4)*b^2*ab)+4*M*(int(((1/2)*Pi-arccos((1/2)*b/r)-arccos((1/2)*a/r))*r, r = (1/2)*b .. sqrt(a^2+b^2)))/ab

7. Jan 22, 2010

### Bob S

I = ∫-y/2+y/2-x/2+x/2ρ r2 dx dy = ∫-y/2+y/2-x/2+x/2ρ[x2+y2] dx dy =2∫ox/2ρyx2dx +2∫oy/2ρxy2dy

Where ρ is the density. So ρ = mass/xy

I =(2m/xy)[ x·y3/(3·8) + y·x3/(3·8)] =(m/12)(y2 + x2)

If x = y, then I = m·x2/6

Compare this to a solid disk of diameter d: I = md2/8

Bob S

8. Jan 22, 2010

### benhou

Oh, ya. It's just the idea that we have to use both dx and dy, and integrate them one after the other, also that even though x and y are in different intervals we only let the intervals on the intergral sign worry about it. I just can't grasp it.

9. Jan 22, 2010

### Bob S

Do the integral for a solid disk of diameter d.
Bob S

10. Jan 22, 2010

### benhou

ok, i would start like this: $$\sum r^{2}\Delta m$$
= $$\sum r^{2}\rho A$$
= $$\sum r^{2}\frac{M}{\pi(d/2)^{2}}2\pi r\Delta r$$
= $$8\frac{M}{d^{2}}\sum r^{3}\Delta r$$
= $$8\frac{M}{d^{2}}\int^{d/2}_{0}r^{3}dr$$
= $$8\frac{M}{d^{2}}\frac{(d/2)^{4}}{4}$$
= $$\frac{Md^{2}}{8}$$

Is this how you want me to do it? Or some other way?

11. Jan 22, 2010

### Bob S

Looks good. Here is a slightly different way:

Use r = d/2 and m= pi·d2/4

I = ∫o2piorρ·r2 r·dr dθ

I = 2·pi ∫orρ·r2 r·dr

I = 2·pi·ρ·r4/4 = pi·ρ·d4/32 = m·d2/8

Bob S

12. Jan 22, 2010

### benhou

Thanks, it works as it came out to be. I guess if I practise it more, the idea will come clearer.

13. Jan 22, 2010

### benhou

wait, how did the dy become y and dx become x? This was the derivation of the rectangular prism formula.

14. Jan 22, 2010

### vin300

It's a property of definite integrals.Integrating from -y/2 to 0 gives the same result as integrating from 0 to y/2

Last edited: Jan 22, 2010
15. Jan 22, 2010

### Bob S

I = ∫-y/2+y/2-x/2+x/2ρ r2 dx dy = ∫-y/2+y/2-x/2+x/2ρ[x2+y2] dx dy
Let's take just the first term:

I = ∫-y/2+y/2-x/2+x/2ρ·x2 dx dy

Integrating by y first, then by x:

I = ∫-x/2+x/2 ρ·y·x2 dx = 2∫o+x/2 ρ·y·x2 dx = [2ρ·y/3][x/2]3=2ρ·y·x3/24 = ρ·y·x3/12

Now using ρ = m/xy we get

I = m·x2/12 (just the first term; second term is m·y2/12 )

Bob S

16. Jan 22, 2010

### benhou

Cool. Thanks Bob!!!