The discussion focuses on calculating the center of mass (CoM) of a semicircle using integral calculus. The correct formula for the CoM is derived as (0, 4r/(3π)), where r is the radius of the semicircle. Participants clarify the distinction between first moments and moment of inertia, emphasizing that the mass of the semicircle is calculated using the integral of density times area. The integration process involves using the equations for the semicircle and requires knowledge of calculus techniques, particularly integration of square roots.
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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)) ...#