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Homework Help: 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
    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.
     
  2. jcsd
  3. Nov 23, 2011 #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].
     
  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.

    [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.
     
  6. Nov 23, 2011 #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.
     
  7. Nov 23, 2011 #6
    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...
     
  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:

    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]
     
  9. Nov 24, 2011 #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.
     
  10. Nov 24, 2011 #9

    HallsofIvy

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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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