Question

Find the centre of gravity of a mass in the shape of a semicircular disc of radius 4, if the density at (x, y) is 2y/x² + y²

Answer 1

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Let the diameter of the disc be along x axis with its center at Origin O(0,0). 
The formula for density is symmetric wrt x. ie., d(x,y) = d(-x,y). So clearly the COM lies on the y axis. ie., COMx = 0.

First find the mass M of semicircular disc. Let the radius be R.

$dm = d(x,y) dA = d(x,y) dx dy = d(r, \theta) (dr\ r\ d\theta)\\\\x=rcos\theta,\ \ y=rSin\theta,\ dx=-rsin\theta\ d\theta,\ dy=rcos\theta\ d\theta\\\\d(x,y)=\frac{2y}{x^2+y^2}=\frac{2r\ Sin\theta}{r^2}=2 sin\theta/r\\\\M=\int \limits_{r=0}^{R} \int \limits_{\theta=0}^{\pi} {d(r,\theta)*r} \, dr\ d\theta\\\\=\int \limits_{r=0}^{R} \int \limits_{\theta=0}^{\pi} {2sin\theta} \, dr\ d\theta=\int \limits_{r=0}^{R} {2} \, dr \ \int \limits_{\theta=0}^{\pi} {sin\theta} \, d\theta\\\\M=2R*[ -cos\theta ]_0^\pi=4R$

Now find the COMy.

$COM_y=\frac{1}{M} \int \limits_{r=0}^{R} \int \limits_{\theta=0}^{\pi} {r\sin\theta*d(r,\theta)*r} \, dr\ d\theta\\\\=\frac{1}{M} \int \limits_{r=0}^{R} \int \limits_{\theta=0}^{\pi} {r\sin\theta*2sin\theta} \, dr\ d\theta\\\\=\frac{1}{M} \int \limits_{r=0}^{R} \int \limits_{\theta=0}^{\pi} {r\sin\theta*2sin\theta} \, dr\ d\theta\\\\=\frac{1}{M} \int \limits_{r=0}^{R} {r} \, dr \ \int \limits_{\theta=0}^{\pi} {2sin^2\theta} \, d\theta\\\\=\frac{1}{4R}\frac{R^2}{2} [ \theta-\frac{sin2\theta}{2} ]_0^\pi$

Thus the answer comes to :    π R / 8
     = π/2   for R = 4 cm

COM =  (0, π/2)