# Center of mass of parabolic plate

1. Oct 20, 2006

### hbomb

I'm having a hard time with these center of mass problems. The lecture that my professor gave on this wasn't very descriptive. Could someone show me the whole process of finding the center of mass?

A uniform plate of height 0.720 m is cut in the form of a parabolic section.
http://img174.imageshack.us/my.php?image=prob03para2cz9.gif
The lower boundary of the plate is defined by: y=1.600x2. Find the distance from the rounded tip of the plate to the center of mass.
The fact that it is uniform means that it has the same density throughout. Since it is two-dimensional, it is a surface density, mass divided by area.

2. Oct 20, 2006

### OlderDan

By symmetry you know the x-coordinate of the CM, right? For the yCM coordinate, you need to divide the plate into pieces that have the same y coordinate. Those would be horizontal strips having area dA = 2x(y)dy where x(y) is the x value at the edge of the plate corresponding to the value y. The mass of such a strip would be dm = σdA where σ is the area density of the plate. yCM is the value obtained by integrating ydm for y ranging from the bottom to the top of the plate and dividing that result by the total mass of the plate.

3. Oct 20, 2006

### hbomb

Could you set up the integral for me because I still don't understand.

4. Oct 20, 2006

Look up the definition of the center of mass, and you'll know how to set up the integral.

5. Oct 20, 2006

### OlderDan

I could, but it will be better for you to give it a try yourself. I will tell you that from

y = 1.600x^2

it follows that

x(y) = sqrt(y/1.600)

The limits of integration over y are from 0 to the maximum height of .720 m.

6. Oct 20, 2006

### hbomb

Tha't what I did before and it didn't work for me. I got .527046y^(3/2).
I plugged in .72 to evaulate it and my homework site said it's wrong.

7. Oct 20, 2006

### OlderDan

I have not seen your work, so I do not know how you arrived at that expression. It is not correct. You should have two integrals to evaluate. One for the integral of ydm and one for the integral of just dm the find the mass. When you take the ratio of the two, constants like σ and sqrt(1.600) divide out leaving you with a simple ratio of two integrals

yCM = [Int(y^(3/2)dy]/[Int(y^(1/2)dy]

where the limits of integration on both integrals are from 0 to .72 m

See if you can set up the problem and reduce it to that equation before doing the integrals, then evaluate.

Last edited: Oct 20, 2006
8. Oct 21, 2006

### hbomb

That's what I don't understand, I don't know how to use the integration setup properly. This is what I should be using:

R=1/M*integral of rp(r)dV

9. Oct 22, 2006

### OlderDan

ρ is constant throughout the plate. So for this problem it can be takenout of the integral. Let Int[] be "the integral of"

R=ρ/M*Int[rdV]

R and r are vectors in this equation. The equality holds for each component. So you actually have three equations

X=ρ/M*Int[xdV]
Y=ρ/M*Int[ydV]
Z=ρ/M*Int[zdV]

In all of these the mass is found by integrating over the volume of the plate

M=ρ*Int[dV]=ρ*Int[dxdydz]

Since there are no functions of position in the integrand, the integral is partially separable. The limits of the z integral are independent of x and y coordinates so we can write

M=ρ*Int[dxdydz]=ρ*Int[dz]*Int[dxdy]

The origin is at the vertex of the plate and lets say it is in the middle of the thickness of the plate, which I will call T. The z integral is easy; it is the thickness of the plate.

M=ρ*T*Int[dxdy]

It is customary to define the product ρ*T to be the surface mass density σ, so this becomes

M=σ*Int[dxdy]

It is in fact so easy to do the z-integral that when approaching a problem for a uniform plate it is customary to sart with this expression instead of the integral over volume.

The limits of integration for one of the remaining integrals is a function of the other variable, so these cannot be separated. Either integral can be done first as long as the limits are handled correctly. Let me use Int[<L,U>ƒ(ξ)dξ] to show the limits, where L is lower, U is upper, and ξ is whatever the integration variable is. Then

M=σ*Int[<0,H>Int[<-w(y),+w(y)>dx]dy]

where w(y) is found by solving the boundary equation of the parabola that was given and H is the given height of the parabola. Doing the x integral gives

M=σ*Int[<0,H>2w(y)dy]

From y = 1.600*x^2 we get w(y) = sqrt(y/1.600). So

M=(2σ/sqrt1.600)*Int[<0,H>sqrt(y)dy]

Now you can follow the exact procedure I used here for the three remaing integrals to find X, Y, and Z. Or you can simplify things by taking advantage of symmetry. The plate is a mirror image of itself in the z-y plane. So Z must lie in the x-y plane. Hence Z = 0. The plate is a mirror image of itself in the y-z plane. Hence X = 0. That just leaves X. You should be able to follow what I did for M to set up the integral for X. When you do, the division by M will cancel some constants and you will be left with a ratio of two integrals.