## Sunday, November 25, 2012

### Centroid (Center of Mass)

This is a topic covered in Calculus II. The textbook explanation is inadequate, and I found nothing good online. So I wrote my own explanation. I now understand it better than I ever did before. (Not surprising, huh? If you're a student, this is an important principle of learning. After you think you understand something, try to write an explanation of it and see how much deeper your understanding can get.)

I'd love to improve this, so please let me know where it's unclear.

Imagine a thin sheet of metal cut in an artistic shape. Is there always a spot where you could hold it balanced on your finger? If there is, can we find that spot? We’ll assume the metal has uniform density. This allows us to treat area as equivalent to weight.

The 1-dimensional Case
To think about this, we first imagine a teeter-totter. We know that two people of the same weight must sit the same distance from the fulcrum (balance point) to balance. We also know that a heavier person must move inward if they want to balance with a lighter person. Experiments show that the weights times the distances from the fulcrum must be equal on the two sides for the teeter-totter to balance. (I wonder if there’s a thought experiment we could do that would convince us this must be true, without the actual experimental evidence.)

Example 1: I weigh 170 pounds and sit 5 feet from the center. My son weighs 75 pounds and sits in front of me, 4 feet from the center. Weights times distances = 170*5 + 75*4 = 1150 feet-pounds. We need someone who weighs 230 pounds to sit 5 feet from the center on the other side. We could write this as:  w1*d1 + w2*d2 = w3*d3

If we change our perspective to a number line below the teeter-totter, with 0 at the fulcrum, then the values on the left will be negative. We won’t have the same equality – we’ll have
w1*p1 + w2*p2 = -(w3*p3),
where each d (for distance, always positive) was replaced with a p (for position). This becomes
w1*p1 + w2*p2 + w3*p3 = 0,
given that the fulcrum is at 0. But suppose we don’t know where the fulcrum is? Let’s just put our 0 at the left end, and let the former proper place for the 0 - at the fulcrum - be f. Then the equation becomes
w1*(p1 - f) + w2*(p2 - f) + w3*(p3 - f) = 0,
or  w1*p1 + w2*p2 + w3*p3w1*f + w2*f + w3*f
or w1*p1 + w2*p2 + w3*p3 =  f(w1 + w2 + w3)
Let W = the sum of all the weights, then we have
f = (w1*p1 + w2*p2 + w3*p3)/W,
which of course extends from 3 weights and positions to n weights and positions.

On to 2 Dimensions
If we use areas instead of weights, we can look for the fulcrum of the x-values (written as an x with a bar over it) and the fulcrum of the y-values (written as a y with a bar over it). For a finite number of small areas, we would get (the same as above)
xbar = (a1*x1 + a2*x2 + ... + an*xn)/A.

If we imagine a shape formed by the area under a function f (where f has positive y values), between x=a and x=b, sliced into infinitely many vertical strips, with the area of each vertical strip given by height times width = f(x). Δ x, then taking the limit as Δ x goes to 0 gives us

For the y value, we need to notice that the vertical center of each vertical strip is at 1/2 *f(x), and we use this instead of x for the position. So we get
or

Many times the area we're interested in will not be touching the x-axis, and so we need area between a top function, f(x), and a bottom function, g(x). The height of each slice will be (f (x)- g(x)), making the area  (f(x) - g(x)). Δ x. The vertical center is now given by averaging f and g. We get:
or

Now if only I could describe this bird shape with functions...

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