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Center of Mass of a Disk with a Hole

Date: 02/19/2001 at 01:27:18
From: CK
Subject: Center of Mass of a Doughnut

A planar circle of radius R with center (0,0) is of uniform density. 
A circle of diameter R, symmetrical about the x-axis, is cut out of 
the right side of the first circle and removed. What is the location 
of the center of mass of the remaining piece?

Date: 02/19/2001 at 01:54:47
From: Doctor Pat
Subject: Re: Center of Mass of a Doughnut

I will refer to the radius of the smaller circle as r, and the larger 
as R.  

There are three figures involved in this problem: the small circle 
(A), the large circle with a small circle cut out of it (B), and the 
large circle formed by putting the first two together (C). It is easy 
to find the center of mass of the two complete circles A and C, but 
what about the third?

Remember that if two masses with known centers are joined as a 
system, the center of mass of the whole must be on the centerline 
joining the first two centers at a distance determined by the 
relative masses. All of this is better handled if we reduce each 
shape to a point mass at its center.

Assuming a uniform density of 1: Shape A is a small circle with mass 
pi*r^2 at some point (x,0) which is the center of the little circle. 
Shape C has a mass of pi*R^2 and a center at (0,0). Since C is a 
combination of B and A, it must be true that B has a center on the 
x-axis to the left of the origin. We'll call it (-p,0).  

For (0,0) to be a true center of mass, the following must be true: 
The mass of A times the distance from (0,0) to A's center is equal to 
the mass of C times the distance from (0,0) to C's center. This gives 
us the equation

     p(pi)*(R^2-r^2) = x(pi)r^2

We can simplify the pi away by dividing both sides and distributing:

     pR^2 - pr^2 = xr^2

Since p is what we want (everything else is a constant in the 
problem), we get:

     p = (xr^2)/(R^2-r^2)  

In your case, r = x = R/2 and p is equal to R/6.

If the center of the small circle is not on the x-axis, but at some 
other point (x,y), then use the distance from the origin to (x,y) in 
place of x in this equation, and the distance in the opposite 
direction from the origin to the unknown center in place of p.  

Nice problem! Hope that is all clear. 

- Doctor Pat, The Math Forum   
Associated Topics:
College Physics
High School Physics/Chemistry

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