# Moment of inertia of a rectangular prism

by benhou
Tags: inertia, moment, prism, rectangular
 P: 123 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: $$\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr$$ $$\int (\frac{\pi}{2}-arccos\frac{b/2}{r}-arccos\frac{a/2}{r})rdr$$
 P: 4,663 Are you trying to calculate one of these?: http://en.wikipedia.org/wiki/List_of_moments_of_inertia Bob S
 P: 123 Yeah, the cuboid. Do you know how to do it? My formula doesn't look like it is going to work.
P: 4,663
Moment of inertia of a rectangular prism

 Quote by benhou Yeah, the cuboid. Do you know how to do it? My formula doesn't look like it is going to work.
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
 P: 123 could you explain how this come about?
 P: 1 (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
P: 4,663
 Quote by benhou could you explain how this come about?
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
 P: 123 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.
P: 4,663
 Quote by 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.
Do the integral for a solid disk of diameter d.
Bob S
 P: 123 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?
 P: 4,663 Looks good. Here is a slightly different way: Use r = d/2 and m= pi·d2/4 I = ∫o2pi∫orρ·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
 P: 123 Thanks, it works as it came out to be. I guess if I practise it more, the idea will come clearer.
P: 123
 Quote by Bob S ∫-y/2+y/2∫-x/2+x/2ρ[x2+y2] dx dy =2∫ox/2ρyx2dx +2∫oy/2ρxy2dy
wait, how did the dy become y and dx become x? This was the derivation of the rectangular prism formula.
P: 513
 Quote by benhou wait, how did the dy become y and dx become x?
It's a property of definite integrals.Integrating from -y/2 to 0 gives the same result as integrating from 0 to y/2
P: 4,663
I = ∫-y/2+y/2-x/2+x/2ρ r2 dx dy = ∫-y/2+y/2-x/2+x/2ρ[x2+y2] dx dy
 Quote by benhou wait, how did the dy become y and dx become x? This was the derivation of the rectangular prism formula.
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
 P: 123 Cool. Thanks Bob!!!

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