1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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 :-)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Mass of 2 dimensional object - line integral
Loading...