Moment of inertia of a rectangular prism

In summary, Bob 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: ∫-y/2+y/2∫-x/2+x/2ρ r2 dx dy = 2∫ox/2ρyx2dx +2∫oy/2ρxy2dy.
  • #1
benhou
123
1
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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  • #3
Yeah, the cuboid. Do you know how to do it? My formula doesn't look like it is going to work.
 
  • #4
benhou said:
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
 
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  • #5
could you explain how this come about?
 
  • #6
(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
benhou said:
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
 
  • #8
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
benhou said:
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
 
  • #10
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
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
Thanks, it works as it came out to be. I guess if I practise it more, the idea will come clearer.
 
  • #13
Bob S said:
-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.
 
  • #14
benhou said:
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
 
Last edited:
  • #15
I = ∫-y/2+y/2-x/2+x/2ρ r2 dx dy = ∫-y/2+y/2-x/2+x/2ρ[x2+y2] dx dy
benhou said:
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
 
  • #16
Cool. Thanks Bob!
 

1. What is moment of inertia of a rectangular prism?

The moment of inertia of a rectangular prism, also known as rotational inertia, is a measure of an object's resistance to changes in its rotation. It depends on the mass distribution and the axis of rotation.

2. How is moment of inertia calculated for a rectangular prism?

The moment of inertia for a rectangular prism can be calculated by multiplying the mass of the object by the square of the distance between the axis of rotation and the center of mass, which is different for different axes of rotation.

3. What is the formula for moment of inertia of a rectangular prism?

The formula for moment of inertia of a rectangular prism is I = (1/12) * m * (a^2 + b^2), where I is the moment of inertia, m is the mass of the prism, and a and b are the dimensions of the prism along the two perpendicular axes passing through its center of mass.

4. How does the shape of a rectangular prism affect its moment of inertia?

The shape of a rectangular prism affects its moment of inertia as it determines the distribution of mass around the axis of rotation. A longer or wider prism will have a higher moment of inertia than a shorter or narrower one of the same mass, as the mass is distributed farther from the axis of rotation.

5. What are the practical applications of knowing the moment of inertia of a rectangular prism?

The moment of inertia of a rectangular prism is an important parameter in understanding the behavior of objects in rotational motion, such as pendulums, spinning tops, and gymnastics equipment. It also plays a role in the design and stability of structures and vehicles, such as bridges and airplanes.

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