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Center of Mass by integration

  1. Dec 17, 2003 #1
    Show that the center of mass of a uniform semicircular disk of radius R is at a point (4/(3*Pi))R from the center of the circle.

    well I know I am suppose to find this by integration. By this equation

    [tex]M\vec{r_{cm}}=\int\vec{rdm}[/tex]

    But, I am not sure how to find dm in this case...

    do I divide mass by area? or by circumference?

    and since it's a disk, and it was 2 variables, would I have to integrate for 2 variables? (x and y)?
     
  2. jcsd
  3. Dec 17, 2003 #2

    NateTG

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    In general, CM calculations end up being double or triple integrals, but this one is pretty straightforward since the disk has constant density.

    I'd be inclined to cut the half-disk into parralel strips and integrate along the CM of each strip.

    Since the density is uniform, you might as well assume that it's one.
     
  4. Dec 17, 2003 #3
    so I divide M by [tex]2R\sin{\theta}[/tex]?
     
  5. Dec 18, 2003 #4

    HallsofIvy

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    Since you are talking about a semicircular disk, I would be inclined to use polar (actually cylindrical coordinates).

    The differential of area in polar coordinates is r dr dθ so the differential of volume in cylindrical coordinates is r dr dθ dz.

    Taking the density to be ρ(r,θ,z), dm= ρ(r,θz) dr dθ dz.

    The mass, for semicircular disk of radius R and thickness h would be M= ∫(z=0 to h)∫(θ=0 to π)∫(r= 0 to R) ρ(r,θ,z)r dr dθ dz.

    If ρ is a constant, this is just ρπr2h.

    Since x= r cosθ, the formula for the xcm would be Mxcm= ∫(z=0 to h)∫(&theta= 0 to π)∫(r= 0 to R)(r cos θ)(ρ(r,θ,z)r dr dθ dz)=
    ∫(z=0 to h)∫(&theta= 0 to π)∫(r= 0 to R)(ρ(r,θ,z)r2cosθ dz dθ dz.

    Since y= r sinθ, the formula for ycm would be Mycm= ∫(z=0 to h)∫(θ= 0 to π)∫(r= 0 to R)(&rho(r,θ,z)r2sinθ dz dθ dz.

    The formula for zcm would be Mzcm= ∫(z=0 to h)∫(θ= 0 to π)∫(r= 0 to R)(&rho(r,θ,z)zr dr dθ dz.

    Of course, if ρ is a constant, from symmetry (cosine is an even function) xcm= 0 and zcm= h/2.
     
  6. Dec 20, 2003 #5
    Ok, I got that one.

    Moving on, the next problem is confusing me as well.

    A baseball bat of length L has a peculiar linear density (mass per unit length) given by [tex]\lambda=\lambda_0(1+x^2/L^2)[/tex]

    so what I've done is
    [tex]\int_{0}^{L}x\lambda_0(1+x^2/L^2)dx[/tex]

    which gives
    [tex]\frac{3\lambda_0L^2}{4}[/tex]

    so I know that M * x_cm = that but how do I find M so I can find x_cm?
     
  7. Dec 20, 2003 #6

    krab

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    C'mon. You know the mass per unit length. How can you not know the mass?

    [tex]M=\int_0^L\lambda dx[/tex]
     
  8. Dec 21, 2003 #7

    HallsofIvy

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    krab: Watch it now! You're starting to sound like me!
     
  9. Dec 21, 2003 #8

    HallsofIvy

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    By the way, why were these posted under the "classical physics" forum? The look like fairly standard calculus homework problems.
     
  10. Dec 21, 2003 #9
    well I got it from a "physics for scientists and engineers" from tipler. So I figured it would fit better under physics.
     
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