Center of Mass of a Semicircle

Date: 06/14/99 at 18:43:50
From: Mayyada Abdou
Subject: Center of mass of a semicircle

I'm studying Mechanics (IGCSE - AS level), and need to get an equation 
or a standard formula for the center of mass of a semicircle. I'm sure 
that it will be on the symmetry line of the semicircle, but where?

Thanking you Dr. Math in advance for your help.

Date: 06/14/99 at 19:07:28
From: Doctor Pat
Subject: Re: Center of mass of a semicircle

I will assume the semi-circle in question is centered at the origin, 
located in the first two quadrants, and its circular periphery can be 
given by y = sqrt(r^2-x^2). If these things are not so, a translation 
and rotation will produce the given answer.

Draw the semi-circle in question with a radius of r...

The center of gravity, as you note, will be on the y-axis. To find out 
where, we need to find y-bar, the center of rotation about the x-axis. 
The value is given by Mx (the moment about x-axis) divided by the 
area.

The moment can be found by drawing a single representative segment and 
integrating this across all values of x. Draw a thin rectangular 
region reaching from the x-axis up to the curve. The area (weight, 
force) of this section is its height times its width. Its height is 
f(x) = sqrt(r^2-x^2) and its width will be the infinitesimal dx. Now 
we need to find the center of gravity of this thin section. Since it 
is infinitely thin, its center will be halfway up, or f(x)/2. Now the 
force x distance formula for the moment clears the square root for us, 
leaving a moment for each tiny section of (r^2-x^2)dx/2 (remember r is 
a constant, x is the variable). We can integrate this from -r to r to 
sum up the total moment due to ALL the little sections like the one we 
drew.

   (1/2) Int (-r,r) [r^2-x^2] dx

When I did this I got 2/3 r^3, but check my work. 

Now to find y-bar we divide this by the area of the semi-circle 
.5 pi r^2 and come up with (4r)/(3 pi). This should put the center of 
gravity about 42% of a radius out (or up) the y-axis.  

I hope this helps. Good luck.

- Doctor Pat, The Math Forum
  http://mathforum.org/dr.math/