Imaginary Center of Mass? What is that?

In summary: It is interesting to note that Mathematica is aware that ln(x) is real for x > 0, but it is not aware that ln(x) is real for -1 < x < 1... but that's another story.)Overall, it seems that Maple is not as reliable as Mathematica when it comes to handling complex numbers in integrals. It is important to pay attention to the results and to double check when necessary.
  • #1
flyingpig
2,579
1
Suppose I give you a curve

[tex]f(x) = \sin^2 (x) + ln(x)[/tex]

And suppose I tell you to rotate this curve about the x axis, we get disks. Now, I ask you, what is the center of mass of this object?

Now immediately, you could say that [tex]\bar{y} = 0[/tex] because it is symmetric about the x-axis. I don't argue, and I say you are right. But what about [tex]\bar{x}[/tex]?

It is actually NOT the same as the 2D- lamina. In fact

[tex]\bar{x} = \frac{\int_{a}^{b} \pi x[f(x)]^2 dx}{\int_{a}^{b} \pi [f(x)]^2 dx}[/tex]

Okay, so what am I trying get here?

Evaluate the integral of the curve from x = 1 to x = 10

[tex]\int_{1}^{10} \pi x (\sin^2 (x) + ln(x))^2 dx[/tex]

Do this (do it on a computer, trust me on this) and you get some nonreal numbers. I asked my TA about it and he said when I expand [tex]f(x) = \sin^2 (x) + ln(x)[/tex], I get a term that I won't know how touch until I get to analysis.

On Maple I got

[PLAIN]http://img143.imageshack.us/img143/9805/comh.jpg [Broken]

I asked my TA what it means physically and he said that he wasn't sure and maybe the complex/imaginary part meant it is significantly 0

So what does having an imaginary part mean?
 
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  • #2
You don't need Maple to carry out the integrals, you just apply integration by parts and obtain:

int(ln(x) dx)=x ln(x)-x
int(sin^2(x) dx)= (-sin(x) cos(x)+1)/2

Now apply Barrow's rule

My mistake. I didn't pay proper attention to the integrand
 
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  • #3
how about this?
 

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  • #4
flyingpig said:
I asked my TA what it means physically and he said that he wasn't sure and maybe the complex/imaginary part meant it is significantly 0

So what does having an imaginary part mean?
That is just 0 to within your numerical precision.
 
  • #5
flyingpig said:
So what does having an imaginary part mean?
It means Maple is stupid sometimes. The function is real everywhere in this interval, so there was no reason to introduce imaginary numbers. Yet it did.

It's obvious how Maple solved that integral: They integrated a truncated infinite series. Had they used a infinite rather than truncated series the imaginary part would have gone to zero. They use truncated series because they want to give an answer in finite time. This, combined with its penchant to introduce complex numbers at the drop of a hat, can lead to complex results with tiny imaginary part (that should be zero).
 
  • #6
If you think about it, the quoted answer in post #1 is in an unhelpful form. To illustrate this, you could write it as
Code:
 6.549237340?
-0.0000000004308080094 i
where "?" denotes an unknown digit. If the first "?" is small enough to be ignored, the whole of the imaginary part can be ignored too. (Bear in mind that any numerical solution like this can only ever be an approximation to the exact answer.)
 
  • #7
never mind...

@gnel

That is odd, but Wolframalpha takes log(x) as ln(x) right?
 
  • #8
flyingpig said:
That is odd, but Wolframalpha takes log(x) as ln(x) right?
Correct. Mathematica does things a bit better than does Maple in this regard. Maple is too quick to go to complex numbers. Mathematica tries to keep things real if it knows the function to be integrated is real throughout the integration interval. That, or it is "smart" enough to recognize that a non-zero imaginary part is bogus if it does use a complex formulation to calculate a result that is real.
 

What is the "Imaginary Center of Mass"?

The Imaginary Center of Mass, also referred to as the Center of Gravity, is a theoretical point in an object or system where the mass is evenly distributed in all directions. It is a useful concept in physics and engineering for understanding how objects move and interact with each other.

How is the "Imaginary Center of Mass" calculated?

The Imaginary Center of Mass is typically calculated using the formula:
xcm = Σmixi / Σmi
where xcm is the position of the center of mass, mi is the mass of each individual component, and xi is the position of each component relative to an arbitrary reference point.

Why is the "Imaginary Center of Mass" important?

The concept of the Imaginary Center of Mass is important because it helps to simplify complex systems and objects into a single point. This point can then be used to analyze the motion and dynamics of the entire system.

What is the difference between "Imaginary Center of Mass" and "Real Center of Mass"?

The Imaginary Center of Mass is a theoretical point that assumes uniform distribution of mass, while the Real Center of Mass takes into account the actual distribution of mass in an object or system. The Real Center of Mass can be calculated using more precise methods such as integration, while the Imaginary Center of Mass is a simplification used for theoretical and practical purposes.

How does the "Imaginary Center of Mass" affect an object's stability?

The location of an object's Imaginary Center of Mass can greatly affect its stability. If the center of mass is located within the base of support, the object is stable and less likely to tip over. However, if the center of mass is located outside of the base of support, the object becomes less stable and more likely to tip over. This is why it is important to consider the center of mass when designing structures and objects to ensure safety and stability.

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