Surface Integrals: Finding Centroid & Inertia of Circle

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SUMMARY

This discussion focuses on calculating the centroid and moments of inertia for a circle with a surface density that varies as the nth power of the distance from a point O on the circumference. The centroid's x-coordinate is derived as x' = 2a(n-2)/(n+2), while the moment of inertia about the diameter through O involves integrating the expression -4(a×a)M. The calculations utilize polar coordinates, specifically the equation r = 2cos(θ), and the bounds for integration are set between -π/2 and π/2 for θ, and 0 to 2cos(θ) for r.

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CaptainX
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How to find the centroid of circle whose surface-density varies as the nth power of the distance from a point O on the circumference. Also it's moments of inertia about the diameter through O.

I'm getting x'=2a(n-2)/(n+2)
And about diameter
-4(a×a)M{something}
 
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Working in coordinates, we can assume that ##O## is the origin, and the circle is ##(x-1)^2+y^2=1##. In polar coordinates, this circle is ##r=2\cos(\theta)##. The ##y##-coordinate of the center of mass is ##0## by symmetry, and the ##x##-coordinate is the integral of ##x r^n dx dy=\cos\theta \ r^{n+2} dr d\theta ## over the disk, divided by the mass, which should be straightforward to calculate (bounds are ##-\pi/2\leq\theta\leq\pi/2## and ##0\leq r\leq 2\cos\theta##).

The moment of inertia calculations are similar.
 
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Infrared said:
Working in coordinates, we can assume that ##O## is the origin, and the circle is ##(x-1)^2+y^2=1##. In polar coordinates, this circle is ##r=2\cos(\theta)##. The ##y##-coordinate of the center of mass is ##0## by symmetry, and the ##x##-coordinate is the integral of ##x r^n dx dy=\cos\theta \ r^{n+2} dr d\theta ## over the disk, divided by the mass, which should be straightforward to calculate (bounds are ##-\pi/2\leq\theta\leq\pi/2## and ##0\leq r\leq 2\cos\theta##).

The moment of inertia calculations are similar.
Thanks a lot!
 

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