
#1
Jan1910, 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] 



#2
Jan2010, 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 



#3
Jan2110, 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.




#4
Jan2110, 06:07 PM

P: 4,664

moment of inertia of a rectangular prismWhere ρ is the density. So ρ = mass/xy You can take it from here. Bob S 



#5
Jan2110, 06:12 PM

P: 124

could you explain how this come about?




#6
Jan2210, 12:14 PM

P: 1

(1/2)*M((1/2)*a)^2+(1/4)*M*(b^4*arccos((2*a^2b^2)/b^2)*sqrt(a^2*(a^2b^2)/b^4)+2*a^42*a^2*b^2)/(sqrt(a^2*(a^2b^2)/b^4)*b^2*ab)+4*M*(int(((1/2)*Piarccos((1/2)*b/r)arccos((1/2)*a/r))*r, r = (1/2)*b .. sqrt(a^2+b^2)))/ab




#7
Jan2210, 12:36 PM

P: 4,664

Where ρ is the density. So ρ = mass/xy I =(2m/xy)[ x·y^{3}/(3·8) + y·x^{3}/(3·8)] =(m/12)(y^{2} + x^{2}) If x = y, then I = m·x^{2}/6 Compare this to a solid disk of diameter d: I = md^{2}/8 Bob S 



#8
Jan2210, 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.




#9
Jan2210, 01:32 PM

P: 4,664

Bob S 



#10
Jan2210, 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? 



#11
Jan2210, 02:57 PM

P: 4,664

Looks good. Here is a slightly different way:
Use r = d/2 and m= pi·d^{2}/4 I = ∫_{o}^{2pi}∫_{o}^{r}ρ·r^{2} r·dr dθ I = 2·pi ∫_{o}^{r}ρ·r^{2} r·dr I = 2·pi·ρ·r^{4}/4 = pi·ρ·d^{4}/32 = m·d^{2}/8 Bob S 



#12
Jan2210, 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.




#13
Jan2210, 07:58 PM

P: 124





#14
Jan2210, 09:45 PM

P: 510





#15
Jan2210, 10:05 PM

P: 4,664

I = ∫_{y/2}^{+y/2}∫_{x/2}^{+x/2}ρ r^{2} dx dy = ∫_{y/2}^{+y/2}∫_{x/2}^{+x/2}ρ[x^{2}+y^{2}] dx dy
I = ∫_{y/2}^{+y/2}∫_{x/2}^{+x/2}ρ·x^{2} dx dy Integrating by y first, then by x: I = ∫_{x/2}^{+x/2} ρ·y·x^{2} dx = 2∫_{o}^{+x/2} ρ·y·x^{2} dx = [2ρ·y/3][x/2]^{3}=2ρ·y·x^{3}/24 = ρ·y·x^{3}/12 Now using ρ = m/xy we get I = m·x^{2}/12 (just the first term; second term is m·y^{2}/12 ) Bob S 



#16
Jan2210, 10:38 PM

P: 124

Cool. Thanks Bob!!!



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