Mass of 2 dimensional object - line integral

[Liferider]

Homework Statement

Find the mass and the coordinates for the center of mass of a thin wire formed like a quarter circle.

Homework Equations

Circle equation: x²+y²=r²
Mass density: rho=x+y

The Attempt at a Solution

I know that: x²+y²=cos²(t)+sin²(t)=1
This leads to: x²+y²=r²(cos²(t)+sin²(t))=r²
x=r*cos(t)
y=r*sin(t)

The quarter circle is described by the vector equation:
r(t)=r*cos(t)i+r*sin(t)j; 0<t<1/2*pi
The density function becomes:
rho=x+y=r*cos(t)+r*sin(t)=r(cos(t)+sin(t))

The line integral that I try to use: (C is the quarter circle wire)
M=∫_{C}(rho(x,y)*ds)=∫^{1/2*pi}_{0}(rho(x,y)*r'(t)*dt)
=r²∫^{1/2*pi}_{0}(cos²(t)-sin²(t))*dt

This ends up in M=0... FFFUUUU! The right answer is supposed to be 2r²

I can't seem to find the right solution to this... please help.

 

[HallsofIvy]

You appear to be arguing that r&#039;= r(-sin(t)i+ cos(t)j)dt and then taking the dot product with r cos(t)i+ r sin(t)j. But x+ y is NOT a vector. Instead you need to use the scalar differential, ds= \sqrt{(dx/dt)^2+ (dy/dt)^2}dt.

 

[Liferider]

Ahh, noob error Thanks :-) I'll try that

 

[Liferider]

Ok, I found the correct solution for the mass, but now I keep hitting the wall with the center of mass...

The way I have tried to find it, is by two integrals, one for the x coordinate and one for the y coordinate.

\bar{x}=\frac{1}{M}\intx(t)*\rho(x(t),y(t))*dt

M=2r²
x=r*cos(t)
y=r*sin(t)
\rho=x+y

I regarded r as constant, so its a single integral. Should this be a double integral instead, with r as a variable? The answer I found with this integral was:

\bar{x}=\frac{1}{8}(\pi+1)

It should be \bar{x}=\frac{r}{8}(\pi+2)

I fairly new to vector calculus, double and triple integrals +, so please bear with me.

 

[HallsofIvy]

Well, when I do the calculation, I get (r/8)(\pi+2). Unless you show exactly what you did, we can't say exactly what you did wrong.

 

[Liferider]

Hmm

\bar{x}=\frac{1}{2r^2]}\int^{0.5pi}_{0}r²(cos²(t)+cos(t)sin(t)*dt

When I factor out r², it gets canceled by the mass. My error seems to be here someplace...

 

[Liferider]

I wonder why I can't make a double integral with x and y as variables. Just insert the limits of t in the x and y equations to get the limits for x and y, like this:

x₀=r*cos(0)=r
x₁=r*cos(0.5pi)=0

y₀=r*sin(0)=0
y₁=r*sin(0.5pi)=r

Then the double integral becomes:

\bar{x}=\frac{1}{M}\int^{0}_{r}\int^{r}_{0}(x*(x+y))dydx=\frac{1}{2r^2}\int^{0}_{r}\int^{r}_{0}(x²+xy)dydx

I find: -\frac{7r^2}{24}

 

[vela]

The latter method doesn't work because x and y aren't independent. You're integrating over the rectangle [0,r]x[0,r] whereas you only want to integrate along the wire.

When you calculated the mass, you did some integral which amounts to
M = \int dmwhere dm corresponds to whatever expression you had in your integral. For the center of mass, you do a similar integral with a factor of x or y thrown in
\bar{x} = \frac{1}{M}\int x\,dm.Compare what you wrote down when calculating M and your integral when calculating \bar{x}. The only difference should be that the latter has a factor of r cos θ in it.

 

[HallsofIvy]

Hmm

\bar{x}=\frac{1}{2r^2]}\int^{0.5pi}_{0}r²(cos²(t)+cos(t)sin(t)*dt

When I factor out r², it gets canceled by the mass. My error seems to be here someplace...

As I said before, ds= \sqrt{(x&#039;)^2+ (y&#039;)^2}dt= rdt
The density is given by x+ y= r(cos(t)+ sin(t)) so the mass is given by the intergral
r^2\int_{t=0}^{\pi/2} (cos(t)+ sin)(t))dt= r^2(sin(t)- cos(t))_0^{\pi/2}= r^2((1- 0)- (0- 1))= 2r^2 so you have that right.

Your error is in the "r^2" you have with the integral on the right. There is an "r" in ds= r dt, there is an "r" in x+ y= r(cos(t)+ sin(t)), and there is an "r" in x= r cos(t). You should have r^3 multiplying the integral, not r^2.

 

[Liferider]

Thanks for the replies :-)