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Centre of mass of semi circle

  1. Nov 17, 2013 #1
    Ok so when I try to get the centroid of a semicircle using ∫rdm/∫dm, ∫dm = ρ∫dxdy for x^2, but now that the area is (0.5)∏(x^2 + y^2) what would ∫dm be? If I did use ρ∫dxdy I have to do some weird integration of sqrt(30^2 - x^2) which we haven't learned yet. If I do have to ρ∫dxdy how would I go about integrating sqrt(30^2 - x^2)?
     
    Last edited: Nov 17, 2013
  2. jcsd
  3. Nov 17, 2013 #2
    ∫2*Pi*r/4 dr = ∫Pi*r/2 dr= area of semi circle = Pi*r^2 /4
    ∫Pi*r/2*ρ(r) dr = mass of semi circle
    ∫Pi*r/2*ρ(r)*r dr = moment of inertia of semicircle
    Distantce from origin to Centre of mass of semi circle = moment of inertia / mass of semi circle
     
  4. Nov 17, 2013 #3

    haruspex

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    What makes you think that will give you the location of the CoM? If the semicircle lies in the first two quadrants, centre of curvature at the origin, you just need to compute the y coordinate of the CoM. That's ∫ρydxdy/∫ρdxdy. For the mass, isn't that obvious? What's the mass of a circle radius r?
    Wrong by a factor of 2 throughout.
    Don't confuse moment (1st moment) with moment of inertia (2nd moment). Moment of inertia (about an axis perpendicular to the semicircle and through the centre of arc) would have r2 in there.
    Neither is ∫Pi*r/2*ρ(r)*r dr the first moment. See my response to the OP.
     
  5. Nov 17, 2013 #4
    By the way I think I figured out the centre of mass of the semicircle with ∫rdm/∫dm and I used ρ∫dxdy, the integration got a little tricky but I did it, the centre of mass I calculated was (0, 40/∏), is that right?(if the axis of symmetry is on the y axis)
     
  6. Nov 17, 2013 #5
    Sorry for using wrong words.

    It give the distance from origin to COM.

    if semi circle lie in Q1 and k is the distance from origin to COM then you could find COM at (kcos45°, ksin45°)

    if Y-axis is the axis of semretical then COM is (0,√2 /3) since k = √2 /3
     
    Last edited: Nov 17, 2013
  7. Nov 17, 2013 #6

    haruspex

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    Yes, that sounds right.
     
  8. Nov 17, 2013 #7

    haruspex

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    Wrong. It would give the CoM of a triangle with one corner at the origin and its base along y = r. Try it, you'll get 2r/3. The correct answer is 4r/3π. It must obviously be < r/2.
     
  9. Nov 17, 2013 #8
    Area of the semi circle is (0.5)∏(x^2 + y^2)
    The circle equation should be x^2+y^2 = r^2
    So the semi circle equation in Q1Q2 should be y = √(r^2-x^2 )
    |x|=√(r^2-y^2 )
    Distance from origin to COM is ∫y dm/ ∫dm
    dm = ρ 2|x|dy = 2 ρ|x|dy = 2ρ√(r^2-y^2 ) dy ;from y= 0 to r
    ∫dm = 2 ρ ∫√(r^2-y^2 ) dy

    ∫dm = ρ(y√(r^2-y^2) + r^2 arctan (y/√(r^2-y^2))) = ρ r^2*π/2

    ∫y dm = -(2/3) ρ (r^2-y^2)^(3/2) = (2/3) ρ r^3

    ∫y dm/ ∫dm = 4r/(3π)

    COM of semi circle is (0, 4r/(3π)) ...#
     
    Last edited: Nov 17, 2013
  10. Nov 17, 2013 #9

    haruspex

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    Correct, but that is not what you previously posted. You wrote r, not y.
     
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