# Density and centre of mass

1. Oct 18, 2008

### d_b

Hi. I'm not sure how to picture it on the x-y-axis.

I found this in the text book:
If i have a plate bounce by 0<= x <=2 and 0<= y <= 3. and the density is constant. i'm looking for two integrals that corresponding to stirps in the x-direction and y-direction.

let say the density is 5gm/cm^2 then to integral should be finding the centre of mass right?? and using the center of mass how do i find the total area if i dont know the radius?? correct me if i'm wrong, isn't the centre of mass will be close to the x-axis then y-axis due to its momentum?

sorry i'm not really good in physic its hard to get my head around combining physic and cal together.

2. Oct 18, 2008

### HallsofIvy

If the density is constant, then the "center of gravity" is just the center which, for this rectangle, would be the point (1, 3/2).

Integrating WHAT? The integral of the density function over a figure is the total density of the figure, not the centre of mass. The integral of 1 over a figure is the area of the figure. Since the centre of mass of a two-dimensional figure involves two values (the x and y coordinates) there is no single integral that will give the centre of mass.

Typically, one finds the center of mass by by first finding the area rather tnan the other way around! And what radius are you talking about? The example you gave, a rectangle, doesn't have a "radius".

Perhaps it would be better to post this in a physics thread. And please post the actual problem itself! Are you talking about "Pappus' theorem" that states that the volume (not) area, of a 3 dimensional figure, formed by rotating a two-dimensional figure around an axis is the area of the figure times the circumference of the circle the centre of the figure describes during the rotation. That has nothing to do with "centre of mass" strictly speaking but it is easy to see that the centre of mass of an object with constant density is the same as its geometric centre.

Suppose the rectangle, 0<= x<= 2, 0<= y<=3, is rotated around the y-axis. As I said before its centre is just (1, 3/2) and the distance from that to the y-axis is just the x-coordinate of the centre, 1. The circle the centre describes in rotating around the y-axis has that radius, 1, and so circumference $2\pi$. The area of the rectangle is, of course, 2(3)= 6 so Pappus' theorem says that the volume of the figure formes is $(6)(2\pi)= 12\pi$.

Of course, that figure is just a cylinder with base radius 2 and height 3: its volume is the area of the base, $\pi (2)^2= 4\pi$ times the height 3, giving volume $12\pi$.

I doubt that is the question you are asking but I really have no idea what problem you are talking about.

Last edited by a moderator: Oct 20, 2008
3. Oct 24, 2008

### d_b

thanks for helping...i found out how to find it. Also at first I got confuse between the length of a strip in two dimension and the area of a slice in 3 dimension because when you times length with delta x then you get an area, area times delta x you get volume....
and i just have to use the definition to find the centre of mass which is fairly easy....