How Do You Calculate the Center of Mass with Non-Constant Density Using Vectors?

[Ataman]

I am looking for a way to find the center mass of an object whose area is enclosed by $$x^{2}$$ and $$\sqrt{x}$$ without computing the x and y seperately (a great deal of paperwork).

So...

$$M\overrightarrow{R_{cm}} = \int \overrightarrow{r} dm$$

$$\sigma = \frac{M}{A} = \frac{dm}{dA}$$

$$\sigma A \overrightarrow{R_{cm}} = \int \sigma \overrightarrow{r} dA$$

$$\overrightarrow{R_{cm}} = \frac{\int\int \sigma \overrightarrow{r} dy dx } {\int \sigma (f(x)-g(x))dx}$$

Because they are constants, the sigmas cancel and I eventually end up with...

$$\overrightarrow{R_{cm}} = \frac{\int^1_0\int^{x^{2}}_{\sqrt{x}} (xi+yj) dydx}{\int^1_0 x^{2} - \sqrt{x} dx}$$

(The answer is $$\frac{9}{20}i + \frac{9}{20}j$$)

But what happens when sigma/density is not constant, but is given a value say... xi or something like that? Obviously taking the dot product will not work, and I am unsure about the cross product (I haven't done a lot of vectors).

-Ataman

 

[John Creighto]

What do you mean that the dot product won't work? I don't see any dot product in the above derivation.

 

[Ataman]

In the above derivation, the density is constant, so it is not defined by a vector.

What I am looking for is a case where there is a varying density within the region.

-Ataman

 

[John Creighto]

In the above derivation, the density is constant, so it is not defined by a vector.

What I am looking for is a case where there is a varying density within the region.

-Ataman

Why define it as a vector? Why not make it as a function of the position vector?

 

[Ataman]

That's what I meant. Excuse me.

-Ataman

 

[John Creighto]

That's what I meant. Excuse me.

-Ataman

Then don't you know the answer?

 

[HallsofIvy]

Then it is still not a vector. "Density" is a numeric value, not a vector function. You don't need a "dot product", you just have a scalar product- multiply each component of the position vector by the density function.

You should notice that you aren't really doing less work that if you did x and y as separate integrals. Since $\int u\vec{i}+ v\vec{j} dx= \int u\vec{i}dx+ \int v\vec{j}dx$ you are just writing two integrals as if they were one.