Find Center of Mass of Thin Plate w/ y^2=4Ax & x=A

[Punkyc7]

Suppose that A is a positive constant. Find the center of mass of a thin plate covering the region bounded by y^2= 4Ax and the line x=A. Assume that mass density at (x,y) is proportional to x.

y=sqrt(4Ax)

K is what I am saying is proportional to x


xbar=My/m
ybar=MX/m


Im not concerned with the answers of xbar and ybar, just the concept of setting up the integrals.

For M we integrate ksqrt(4Ax) dx from -A to A

for My we intgrate k(xtilda)*(ksqrt(4Ax)) dx from -A to A, xtilda=x

for Mx we integrate k(ytilda)*(ksqrt(4Ax) dx from -A to A, ytilda= (ksqrt(4Ax)-ksqrt(4Ax))/2

I just want to make sure I am doing this right. Also I am not sure if the second k term should be included in My and Mx

 

[vela]

Both the integrand and the limits on your integrals are incorrect. Did you sketch the region?

Explain how you derived your integral for M so we can see where your mistakes are.

 

[Punkyc7]

The limits should be from 0 to A, multiplied by two because of symmetry

M is the integral of the dA * rho

 

[vela]

Right. So you now have

$$M = \int_0^A \rho 2\sqrt{4Ax}\,dx$$

What do you want to plug in for the density ρ?

 

[Punkyc7]

so since rho is proportional to x you would let rho =kx
and another thing, how do you do the math symbols?

 

[vela]

Check the following thread for links about LaTeX.

https://www.physicsforums.com/showthread.php?t=386951

You can also click on a typeset formula to see the code used to generate it.So you're pretty much set for M and $\bar{x}$. The one for $\bar{y}$ is a little more complicated, though you may be able to just say what the answer should be.