Ah, dang, fatalexception I think I finally figured out what exactly you wanted to know earlier. It might be too late, but anyways: Ultimately that page is just telling you that the moment of inertia of a rectangle through a given axis is bh^3/12. When determining what is the base and what is the height, the base is parallel to the axis in which you are calculating the moment of inertia from. Sometimes objects will not be such simple shapes like rectangles, squares or circles, so that's why they were showing you all the calculus steps. If you are calculating the moment of inertia of, say a quarter of a circle, after much integrating you're going to get: I_(x or y) = PI*r^4/16 . And say you want to find the moment of inertia about an axis which runs diagonally through a rectangle (for instance the axis runs from the top right to the bottom left corner of the rectangle). Your cartesian components of the moments of inertia again after integrating will become I=bh^3/3. So those first two examples were pointless. They show us the "proper" way to calculate the moments of inertia of a rectangle. However, we all know that moments of inertia of rectangles = bh^3/12 so those two examples just proved (through integrating) why it was what it was. Note that in these two examples, the shape was centered at the origin. example 3 just moved the reference axis to an edge of the rectangle, so ultimately u just multiply by 4. example 4 is like the first two examples, except that you subtract the smaller one from the bigger one. So if your beam is going to be a rectangle, then to calculate one component of the moment of inertia about an axis which runs straight through this beam should simply be I=bh^3/12 where b is the length of your beam and h is the width of the cross sectional area of your beam. But note that your beam is 3D so you have to account for one more dimension. But like mentioned, I highly doubt you'll need this

Ah, dang, fatalexception I think I finally figured out what exactly you wanted to know earlier. It might be too late, but anyways:

Ultimately that page is just telling you that the moment of inertia of a rectangle through a given axis is bh^3/12. When determining what is the base and what is the height, the base is parallel to the axis in which you are calculating the moment of inertia from. Sometimes objects will not be such simple shapes like rectangles, squares or circles, so that's why they were showing you all the calculus steps. If you are calculating the moment of inertia of, say a quarter of a circle, after much integrating you're going to get: I_(x or y) = PI*r^4/16 . And say you want to find the moment of inertia about an axis which runs diagonally through a rectangle (for instance the axis runs from the top right to the bottom left corner of the rectangle). Your cartesian components of the moments of inertia again after integrating will become I=bh^3/3.

So those first two examples were pointless. They show us the "proper" way to calculate the moments of inertia of a rectangle. However, we all know that moments of inertia of rectangles = bh^3/12 so those two examples just proved (through integrating) why it was what it was. Note that in these two examples, the shape was centered at the origin.
example 3 just moved the reference axis to an edge of the rectangle, so ultimately u just multiply by 4.
example 4 is like the first two examples, except that you subtract the smaller one from the bigger one.

So if your beam is going to be a rectangle, then to calculate one component of the moment of inertia about an axis which runs straight through this beam should simply be I=bh^3/12 where b is the length of your beam and h is the width of the cross sectional area of your beam. But note that your beam is 3D so you have to account for one more dimension.

But like mentioned, I highly doubt you'll need this

[color=#b5b5ff]I take it the example is assuming constant density?[/color]

Yes. Unless it explicitly says so, we always assume objects or shapes have constant density. At least that's how our professors want us to think :P

Yes. Unless it explicitly says so, we always assume objects or shapes have constant density. At least that's how our professors want us to think

That seems a bit complicated for a mid-year physics course. I was learning electricity and that kind of stuff while going over that.

That seems a bit complicated for a mid-year physics course. I was learning electricity and that kind of stuff while going over that.

[quote=Centreri]That seems a bit complicated for a mid-year physics course. I was learning electricity and that kind of stuff while going over that.[/quote] Yeah, well, my school is 1337. :P But anyway, thanks for figuring it out, MA, and I think I will need it, since I saw something involving bh^3/12 on the trial recording and analyzing paper we got...

Quote from Centreri

That seems a bit complicated for a mid-year physics course. I was learning electricity and that kind of stuff while going over that.

Yeah, well, my school is 1337.

But anyway, thanks for figuring it out, MA, and I think I will need it, since I saw something involving bh^3/12 on the trial recording and analyzing paper we got...

What if it's just a simple positive bar graph? There is no need for negative factors? Ooops, forgot to quite MA, whatever, you see what direction I was pointing at.

What if it's just a simple positive bar graph? There is no need for negative factors?

Ooops, forgot to quite MA, whatever, you see what direction I was pointing at.

If it's a simple positive or even negative (as in it points downward) bar graph, then it's just as example 3 shows. However, if the axis is positioned in an arbitrary point between the top and bottom (or left and right depending which direction your axis runs from) then your limits of integration will be different. But looking back at example 3.. - We want to find the moment of inertia about axis BB - The axis BB runs horizontally, so we can disregard the position of our shape with respect to any vertical axis. - Axis BB is located at the bottom of the shape - Now lets see what information we're given: The shape is a rectangle The height = h Base = b - Now because our reference axis is horizontal, we find the y-distances of our shape to the reference axis (vice versa if our reference axis is vertical). We have to square it, so our equation becomes: ∫y^2dA but now note this: dA = the differential of area of our shape. Knowing that we're dealing with a rectangle, A = xy. However, because we're taking the differential of area for this rectangle with respect to an axis, one of our dimensions won't be constant. In this case, it is our y-value, or in other words h (height). x is constant, so x = b So dA = xdy = bdy (good way to remember this is that like DTBK said, you can call them "dee-ay" and such, but personally I like to call it "delta-ay", "delta-x", etc, because by definition, delta usually indicates a change in something. There probably is a better way to explain it, but for me this is how i like to remember it. So now that we have ∫by^2dy = b∫y^2dy (in calculus, we can move constants out of our integrals) just what are our limits of integration going to be? So let's see. the range from the top of the shape to our BB-axis... is h (simply because our axis is location at the bottom of our rectangle) the range from our BB-axis to the bottom of the shape.. is 0 (obviously because our BB-axis is location at the bottom of the shape.) Limits of integration: 0 to h (if you were to label this on the integral sign, h would be at the top and 0 at the bottom) so now our equation becomes: from 0 to h: b∫y^2dy = b[y^3/3] from 0 to h: plug in 'h' into y: b[(h)^3/3] then plug in '0' into y: [b[(0)^3/3] subtract the latter from the former: b[h^3/3] - b[0^3/3] you get: bh^3/3 Now suppose our BB-axis were not located at the bottom of the rectangle. Say it was location at a distance h/3 above the bottom of our rectangle. Then everything would be the same up till the limits of integration. So instead of it being from 0 to h, it would then be from -h/3 to 2h/3

If it's a simple positive or even negative (as in it points downward) bar graph, then it's just as example 3 shows.

However, if the axis is positioned in an arbitrary point between the top and bottom (or left and right depending which direction your axis runs from) then your limits of integration will be different.

But looking back at example 3..
- We want to find the moment of inertia about axis BB
- The axis BB runs horizontally, so we can disregard the position of our shape with respect to any vertical axis.
- Axis BB is located at the bottom of the shape
- Now lets see what information we're given:
The shape is a rectangle
The height = h
Base = b
- Now because our reference axis is horizontal, we find the y-distances of our shape to the reference axis (vice versa if our reference axis is vertical). We have to square it, so our equation becomes:
∫y^2dA
but now note this: dA = the differential of area of our shape. Knowing that we're dealing with a rectangle, A = xy. However, because we're taking the differential of area for this rectangle with respect to an axis, one of our dimensions won't be constant. In this case, it is our y-value, or in other words h (height). x is constant, so x = b
So dA = xdy = bdy (good way to remember this is that like DTBK said, you can call them "dee-ay" and such, but personally I like to call it "delta-ay", "delta-x", etc, because by definition, delta usually indicates a change in something. There probably is a better way to explain it, but for me this is how i like to remember it.

So now that we have ∫by^2dy = b∫y^2dy (in calculus, we can move constants out of our integrals) just what are our limits of integration going to be?

So let's see. the range from the top of the shape to our BB-axis...
is h
(simply because our axis is location at the bottom of our rectangle)

the range from our BB-axis to the bottom of the shape..
is 0
(obviously because our BB-axis is location at the bottom of the shape.)

Limits of integration: 0 to h (if you were to label this on the integral sign, h would be at the top and 0 at the bottom)

so now our equation becomes:
from 0 to h: b∫y^2dy = b[y^3/3] from 0 to h:
plug in 'h' into y: b[(h)^3/3]
then plug in '0' into y: [b[(0)^3/3]
subtract the latter from the former: b[h^3/3] - b[0^3/3]

you get: bh^3/3



Now suppose our BB-axis were not located at the bottom of the rectangle. Say it was location at a distance h/3 above the bottom of our rectangle.

Then everything would be the same up till the limits of integration.
So instead of it being from 0 to h, it would then be from -h/3 to 2h/3