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Mass of 2 dimensional object - line integral

  • Thread starter Liferider
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  • #1
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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=∫[itex]_{C}[/itex](rho(x,y)*ds)=∫[itex]^{1/2*pi}_{0}[/itex](rho(x,y)*r'(t)*dt)
=r2∫[itex]^{1/2*pi}_{0}[/itex](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.
 

Answers and Replies

  • #2
HallsofIvy
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You appear to be arguing that [itex]r'= r(-sin(t)i+ cos(t)j)dt[/itex] 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, [itex]ds= \sqrt{(dx/dt)^2+ (dy/dt)^2}dt[/itex].
 
  • #3
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Ahh, noob error :-p Thanks :-) I'll try that
 
  • #4
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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.

[itex]\bar{x}[/itex]=[itex]\frac{1}{M}[/itex][itex]\int[/itex]x(t)*[itex]\rho[/itex](x(t),y(t))*dt

M=2r2
x=r*cos(t)
y=r*sin(t)
[itex]\rho[/itex]=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:

[itex]\bar{x}[/itex]=[itex]\frac{1}{8}[/itex]([itex]\pi[/itex]+1)

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

I fairly new to vector calculus, double and triple integrals +, so please bear with me.
 
  • #5
HallsofIvy
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Well, when I do the calculation, I get [itex](r/8)(\pi+2)[/itex]. Unless you show exactly what you did, we can't say exactly what you did wrong.
 
  • #6
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Hmm

[itex]\bar{x}[/itex]=[itex]\frac{1}{2r^2]}[/itex][itex]\int[/itex][itex]^{0.5pi}_{0}[/itex]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...
 
  • #7
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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:

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

I find: -[itex]\frac{7r^2}{24}[/itex]
 
  • #8
vela
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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
[tex]M = \int dm[/tex]where 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
[tex]\bar{x} = \frac{1}{M}\int x\,dm.[/tex]Compare what you wrote down when calculating M and your integral when calculating [itex]\bar{x}[/itex]. The only difference should be that the latter has a factor of r cos θ in it.
 
  • #9
HallsofIvy
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Hmm

[itex]\bar{x}[/itex]=[itex]\frac{1}{2r^2]}[/itex][itex]\int[/itex][itex]^{0.5pi}_{0}[/itex]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, [itex]ds= \sqrt{(x')^2+ (y')^2}dt= rdt[/itex]
The density is given by x+ y= r(cos(t)+ sin(t)) so the mass is given by the intergral
[tex]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[/tex] 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 [itex]r^3[/itex] multiplying the integral, not [itex]r^2[/itex].
 
  • #10
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Thanks for the replies :-)
 

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