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Mass of 2 dimensional object - line integral

  1. Nov 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the mass and the coordinates for the center of mass of a thin wire formed like a quarter circle.

    2. Relevant equations
    Circle equation: x2+y2=r2
    Mass density: rho=x+y

    3. 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

    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:

    The line integral that I try to use: (C is the quarter circle wire)

    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.
  2. jcsd
  3. Nov 23, 2011 #2


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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].
  4. Nov 23, 2011 #3
    Ahh, noob error :-p Thanks :-) I'll try that
  5. Nov 23, 2011 #4
    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.



    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:


    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.
  6. Nov 23, 2011 #5


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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.
  7. Nov 23, 2011 #6


    When I factor out r2, it gets canceled by the mass. My error seems to be here someplace...
  8. Nov 24, 2011 #7
    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:



    Then the double integral becomes:


    I find: -[itex]\frac{7r^2}{24}[/itex]
  9. Nov 24, 2011 #8


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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.
  10. Nov 24, 2011 #9


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    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].
  11. Nov 25, 2011 #10
    Thanks for the replies :-)
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