moment of inertia of a rectangular prism


by benhou
Tags: inertia, moment, prism, rectangular
benhou
benhou is offline
#1
Jan19-10, 11:07 PM
P: 124
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:

[tex]\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[/tex]

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

[tex]\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr[/tex]

[tex]\int (\frac{\pi}{2}-arccos\frac{b/2}{r}-arccos\frac{a/2}{r})rdr[/tex]
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Bob S
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#2
Jan20-10, 07:30 PM
P: 4,664
Are you trying to calculate one of these?:
http://en.wikipedia.org/wiki/List_of_moments_of_inertia
Bob S
benhou
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#3
Jan21-10, 12:43 AM
P: 124
Yeah, the cuboid. Do you know how to do it? My formula doesn't look like it is going to work.

Bob S
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#4
Jan21-10, 06:07 PM
P: 4,664

moment of inertia of a rectangular prism


Quote Quote by benhou View Post
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
benhou
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#5
Jan21-10, 06:12 PM
P: 124
could you explain how this come about?
ngchou
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#6
Jan22-10, 12:14 PM
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
Bob S
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#7
Jan22-10, 12:36 PM
P: 4,664
Quote Quote by benhou View Post
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)[ xy3/(38) + yx3/(38)] =(m/12)(y2 + x2)

If x = y, then I = mx2/6

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

Bob S
benhou
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#8
Jan22-10, 01:15 PM
P: 124
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.
Bob S
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#9
Jan22-10, 01:32 PM
P: 4,664
Quote Quote by benhou View Post
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
benhou
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#10
Jan22-10, 01:59 PM
P: 124
ok, i would start like this: [tex]\sum r^{2}\Delta m[/tex]
= [tex]\sum r^{2}\rho A[/tex]
= [tex]\sum r^{2}\frac{M}{\pi(d/2)^{2}}2\pi r\Delta r[/tex]
= [tex]8\frac{M}{d^{2}}\sum r^{3}\Delta r[/tex]
= [tex]8\frac{M}{d^{2}}\int^{d/2}_{0}r^{3}dr[/tex]
= [tex]8\frac{M}{d^{2}}\frac{(d/2)^{4}}{4}[/tex]
= [tex]\frac{Md^{2}}{8}[/tex]

Is this how you want me to do it? Or some other way?
Bob S
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#11
Jan22-10, 02:57 PM
P: 4,664
Looks good. Here is a slightly different way:

Use r = d/2 and m= pid2/4

I = ∫o2piorρr2 rdr dθ

I = 2pi ∫orρr2 rdr

I = 2piρr4/4 = piρd4/32 = md2/8

Bob S
benhou
benhou is offline
#12
Jan22-10, 03:15 PM
P: 124
Thanks, it works as it came out to be. I guess if I practise it more, the idea will come clearer.
benhou
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#13
Jan22-10, 07:58 PM
P: 124
Quote Quote by Bob S View Post
-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.
vin300
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#14
Jan22-10, 09:45 PM
P: 510
Quote Quote by benhou View Post
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
Bob S
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#15
Jan22-10, 10:05 PM
P: 4,664
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 Quote by benhou View Post
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/2ρx2 dx dy

Integrating by y first, then by x:

I = ∫-x/2+x/2 ρyx2 dx = 2∫o+x/2 ρyx2 dx = [2ρy/3][x/2]3=2ρyx3/24 = ρyx3/12

Now using ρ = m/xy we get

I = mx2/12 (just the first term; second term is my2/12 )

Bob S
benhou
benhou is offline
#16
Jan22-10, 10:38 PM
P: 124
Cool. Thanks Bob!!!


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