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Calculating Mass and Center of Mass

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data
    Let D be the upper half of the disc x2+y2 ≤1(that is, the part of the disc with y ≥ 0). Suppose the lamina Ω fills the region D, and has density given by ρ(x, y) = |x|.

    (a) Calculate the mass of the lamina.
    (b) Find the coordinates of the center of mass of the lamina.

    2. Relevant equations



    3. The attempt at a solution
    So, first, I'm going to rewrite this |x| as y=-x when for when x≤0 and y=x for when x≥0. Therefore, the quarter-circle given by the boundary in the first quadrant allows 0≤x≤1 and 0≤y≤1.

    I integrate ∫(0≤x≤1)∫(0≤y≤1) x dydx and get 1/2. Observing that everything is symmetrical, I simply multiply by 2 to find the total mass, which I find to be 1.

    Again, by symmetry, I reason that the center of mass must lie on the y-axis, so I will let the x-coordinate of the center of mass equal 0. To find the y-coordinate, I integrate

    ∫(0≤x≤1)∫(0≤y≤1) xy dydx + ∫(-1≤x≤0)∫(0≤y≤1) -xy dxdy (???)

    and get 1/2. Therefore, the center of mass exists at (0,1/2).
     
  2. jcsd
  3. Nov 12, 2011 #2
    It looks like you're integrating over a rectangular region in your solution, but the region given is the upper half of a disk. Also, polar coordinates seem like they would be more friendly here.
     
  4. Nov 12, 2011 #3
    I was thinking polar would be easier, too. But I didn't know how to convert y = |x| into polar coordinates. But, yes, clearly my bounds are not correct.
     
  5. Nov 12, 2011 #4
    well your bounds should be pretty easy, theta goes from 0 to pi and r goes from 0 to 1. Your question of rho = |x| comes down to knowing x = r cos(theta) and |r cos(theta)| = r |cos(theta)|, which is just a question of knowing where cos(theta) is positive and where it is negative.
     
  6. Nov 12, 2011 #5
    Okay. So, first, why have you replaced y with rho in y = |x|? Does rho = rsin(theta)?

    But anyway, let's see if I can integrate this properly. I think I should have the sum of two integrals so that I can deal with the bounds independently. So, cos(theta) is non-negative in 0≤theta≤pi/2. I think one of my integrals should then be

    ∫(0≤theta≤pi/2)∫(0≤r≤1) r^2 * cos(theta) dr dtheta.

    This gives me 1/3. My other integral is

    ∫(pi/2≤theta≤pi)∫(0≤r≤1) r^2 * -1 * cos(theta) dr dtheta.

    This, unexpectedly, also gives 1/3. Therefore mass is 2/3 (?)
     
  7. Nov 12, 2011 #6
    i did this because in the problem statement you said rho = |x|. But yes, this looks to be correct.
     
  8. Nov 12, 2011 #7
    Are you referring to Ω? Isn't that omega?
     
  9. Nov 12, 2011 #8
    [tex] \rho (x,y) = |x| [/tex]
     
  10. Nov 12, 2011 #9
    Oh, right. I'm just being stupid.
     
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