Moment of Inertia of a rectangle?

1. Oct 9, 2011

GreenPrint

1. The problem statement, all variables and given/known data

I was trying to find the moment of inertia of a rectangle with width a and height b were axis of rotation is through it's center of mass

2. Relevant equations

3. The attempt at a solution

I = integral r^2 dm
rho = dm/dA
dm = rho dA

If I take an infinitely small area of an infinitely thin rectangle a distance dr from the point of rotation, it would have a width of dr and a height of b, hence

dA = b dr

dm = rho b dr

I = rho b integral[-a/2,a/2] r^2 dr

I = (rho b)/3 * r^3 |[-a/2,a/2]

I = (rho b)/3 * [ a^3 /8 - (- a^3/8) ]

I = (rho b)/3 * [ a^3/8 + a^3 / 8]

I = (rho b a^3)/12

rho = M/A
A = ab

I = (M b a^3)/(12 a b)
I = (M a^2)/12

I don't see what I'm doing wrong, my book tells me it's I = [ M(a^2 + b^2) ] /12

thanks for any help

2. Oct 9, 2011

WJSwanson

I'm assuming that this is asking about an axis of rotation normal to the surface, through the center of mass. In that case, with the differential elements of width da (not dr, since you're dealing with the thickness of thin linear elements instead of the thickness of a ring whose center is along the axis of rotation), the distance r between the axis of rotation and a given point on that thin strip is (a2 + b2)1/2. When you use dr as your differential width, you're essentially setting up a series of infinitesimal rings whose points are all equidistant from the axis of rotation.

Try reevaluating your integral using this information and see if this yields the answer you expected.

3. Oct 9, 2011

GreenPrint

when we say

rho = dm/dA

is this necessarily correct? Like when we sub back in we subsitute the mass of the whole thing and the area of the whole rectangle not dm/dA (after evaluating the integral)

it just seems backwards to me, shouldn't it be the density of the infinitely small strip and not the whole object itself

4. Oct 9, 2011

GreenPrint

also so

sqrt(a^2 + b^2)
is the radius from any infinitely small point on any infinitely small thin strip to the point of rotation? I'm having a sort of difficult time visualizing this. What is the need to take infinitely small points on the infinitely thin strip

Does this derivation require multivariable calculus? I haven't taken it yet and my professor said to just go over the derivation of this chart in my book with that lists the moments of inertia and this was one of them and I have seen some derivations on line that get pretty complicated that have like multiple integrals and what have you and is beyond my current level of math education.

5. Oct 9, 2011

WJSwanson

Well, as a personal convention I usually use $\sigma$ to denote area densities of mass and charge, but yes, by definition the area density of mass is the derivative of mass with respect to area, that is, differential mass divided by differential area. That is to say, when you take an infinitesimal element (of any shape, but usually analogous to the geometry involved) of a lamina, the mass of that element -- which is the differential mass -- will be equal to the product of your area-density of the object and the area of that element (i.e. the differential area).

6. Oct 9, 2011

GreenPrint

ah that is why we substitute back in the density for mass of the object divided by the area of the object (of the whole object) after integrating, of that makes much more sense.

7. Oct 9, 2011

8. Oct 9, 2011

WJSwanson

Some of them do require multivariable calculus. The good news is that multiple integrals are fairly straightforward. When you have a function such that

F = ∫∫A f(x,y) dx dy

you perform what's called an iterated integral, where you integrate first with respect to y (i.e., you use y as the variable of integration and treat any function of x that is independent of y as a constant) and evaluate it, then integrate the result with respect to x -- or vice versa. You needn't perform those operations yourself at this point, just be aware that that's basically what's being done in a lot of these processes. (I oversimplified the process a bit.)

Anyway, here's basically the MS Paint rundown of why your r2 = x2 + y2:

and so basically what this means is that the differential moment of inertia dI over that differential area element is the square of the distance between the axis of rotation & that little differential area element, times your area mass density. This means that

$I = \int\int_{A} \sigma d(r^{2}) dA = \int^{y = a/2}_{y = -a/2} \int^{x = b/2}_{x = -b/2} \sigma(x^{2} + y^{2}) dx dy$

That's the basic idea.

Similarly, when you have a three-dimensional mass configuration, your moment of inertia is the volume integral of the product of the mass density, the square of the distance from a differential volume element to the axis of rotation, and the differential volume. It will make more sense after you've had some experience with multivariable calculus, but hopefully this gives you at least a vague sense of the way these things are derived.

NOTE: The limits of integration above assume that you've chosen the coordinate axes such that the rectangle has sides parallel to the x and y axes and the axis of rotation is the z axis.

Last edited: Oct 10, 2011
9. Oct 10, 2011

Staff: Mentor

Impressive artwork. But doesn't look right. a and b are supposed to be constants, the dimensions of the block. Your diagram should have used x and y, as you used in the algebra in the following paragraph.

10. Oct 10, 2011

WJSwanson

Thanks for catching that! lol. Good call.