Moment of inertia of a rectangular prism

[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:

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

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

 

[Bob S]

Are you trying to calculate one of these?:
http://en.wikipedia.org/wiki/List_of_moments_of_inertia
Bob S

 

[benhou]

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

 

[Bob S]

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∫_o^x/2ρyx²dx +2∫_o^y/2ρxy²dy

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

You can take it from here.

Bob S

 

[benhou]

could you explain how this come about?

 

[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

 

[Bob S]

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∫_o^x/2ρyx²dx +2∫_o^y/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

 

[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.

 

[Bob S]

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]

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?

 

[Bob S]

Looks good. Here is a slightly different way:

Use r = d/2 and m= pi·d²/4

I = ∫_o^2pi∫_o^rρ·r² r·dr dθ

I = 2·pi ∫_o^rρ·r² r·dr

I = 2·pi·ρ·r⁴/4 = pi·ρ·d⁴/32 = m·d²/8

Bob S

 

[benhou]

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

 

[benhou]

∫_-y/2^+y/2∫_-x/2^+x/2ρ[x²+y²] dx dy =2∫_o^x/2ρyx²dx +2∫_o^y/2ρxy²dy

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

 

[vin300]

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]

I = ∫_-y/2^+y/2∫_-x/2^+x/2ρ r² dx dy = ∫_-y/2^+y/2∫_-x/2^+x/2ρ[x²+y²] dx dy

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

 

[benhou]

Cool. Thanks Bob!