The discussion revolves around finding the center of mass (CoM) of a semicircle, focusing on the mathematical integration methods involved in determining the centroid. Participants explore various approaches to calculate the CoM using integrals and question the assumptions behind their methods.
The discussion is ongoing, with participants providing various insights and corrections. Some have proposed calculations for the CoM, while others challenge the validity of those calculations and suggest alternative interpretations. There is no explicit consensus on the correct approach, but several productive lines of reasoning are being explored.
Participants note the complexity of the integration involved and the potential confusion between different moments and mass calculations. There is also mention of the constraints of the problem, such as the specific quadrants in which the semicircle lies and the assumptions about the axis of symmetry.
Panphobia said: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)?
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?Panphobia said:Ok so when I try to get the centroid of a semicircle using ∫rdm/∫dm,
Wrong by a factor of 2 throughout.Twich said:∫2*Pi*r/4 dr = ∫Pi*r/2 dr= area of semi circle = Pi*r^2 /4
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.Twich said:∫Pi*r/2*ρ(r)*r dr = moment of inertia of semicircle
haruspex said: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.
Yes, that sounds right.Panphobia said: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)
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.Twich said:It give the distance from origin to COM.
Correct, but that is not what you previously posted. You wrote r, not y.Twich said: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)) ...#