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The Center of Mass of a Rod

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    A rod of length 30.0 cm has linear density (mass-per-length) given by

    λ = 50.0 g/m + 20.0x g/m2

    where x is the distance from one end, measured in meters. (a) What is the mass of the rod?

    (b) How far from the x = 0 end is the center of mass?

    2. Relevant equations

    Xcm = l/2

    3. The attempt at a solution

    λ = 50.0 g/m + (20.0)(.3m) g/m2

    λ = 56 g/m * .3 m

    λ = 16.8 g

    Xcm = .3/2 = .15

    The actual answers are 15.9 g and .159 m

    Am I not using the formulas correctly?
     
  2. jcsd
  3. Oct 4, 2009 #2
    [tex]m=\int\limits_0^L \lambda(x) {\rm d}x[/tex]

    [tex]x_G=\frac 1 m \int\limits_0^L x\lambda(x) {\rm d}x[/tex]
     
  4. Oct 4, 2009 #3
    Would I consider λ a constant.

    m = λ1/2x2

    m = (50.0 g/m + (20.0)(.3m) g/m2)(1/2)(.32)

    of integrate it as well. Both ways that doesn't seem to be the right answer either. I get 2.43 if I integrate λ, and 2.52 if I don't.
     
  5. Oct 4, 2009 #4
    [tex]\lambda[/tex](x) is a function of x that happens to be 50.0 + 20.0x, so you replace

    [tex]\lambda[/tex](x) by 50.0 + 20.0x and integrate the result.
     
  6. Oct 4, 2009 #5
    This is ridiculously when you put it that way. Thanks, I understand how to apply the formulas now. Hopefully when it gets rearranged for a test I will still see it.
     
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