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Centroid of an Arc

Date: 03/20/2002 at 07:18:00
From: derrick weedon
Subject: Centroid of an arc

How do I find the centroid (or centre of gravity) of an arc of a 

Date: 03/20/2002 at 08:07:07
From: Doctor Jerry
Subject: Re: Centroid of an arc

Hi Derrick,

Suppose a curve C is described parametrically by 

   x = x(t)
   y = y(t)

where t varies over the interval [a,b]; also suppose that the density 
of the curve at a point (x,y) is g(x,y).  The total mass m of the 
curve is given by the integral

m = int(t=a,t=b,g(x(t),y(t))*sqrt(x'(t)^2+y'(t)^2)*dt).

If (X,Y) is the center of gravity, then

X = (1/m)*int(t=a,t=b,x(t)*g(x(t),y(t))*sqrt(x'(t)^2+y'(t)^2)*dt)

Y = (1/m)*int(t=a,t=b,y(t)*g(x(t),y(t))*sqrt(x'(t)^2+y'(t)^2)*dt).

If C is the arc circle of radius a described by

   x = a*cos(t)
   y = a*sin(t)

where t varies over the interval [0,T] (radian measure), and the 
density is constant, say, g(x,y) = k, then 

   m = k*a*T,

   X = a*sin(T)/T
   Y = a*(1-cos(T))/T

If a = 1 and T = pi/2, then

   X = 2/pi
   Y = 2/pi.

- Doctor Jerry, The Math Forum   
Associated Topics:
High School Calculus

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