Center of mass and a bit vector calculus

AI Thread Summary
The discussion revolves around understanding the center of mass for a rod with nonuniform density, as presented in Kleppner's Mechanics. A participant grapples with the integration of density functions and the implications of using the product rule in their calculations, mistakenly equating mass and density. Clarification is provided that the correct approach involves defining local density as the limit of mass over distance, leading to the integral form for total mass. Additionally, the conversation touches on the concept of vector operations, specifically questioning the meaning of raising a vector to a power and the distinction between scalars and vectors in such contexts. The thread concludes with a resolution of misunderstandings regarding integration and density calculations.
Shing
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Hi guys,
recently I am reading Kleppner's Mechanics
about center of mass,
well, I am always a fairly fast learner,
but I really got stick here.

you see,
in the page 119
example 3.3 (in short, a rod with nonuniform density )
let Q(x) be the function density of location vector

it said
M=\int{dm}
\int{dm}=\int{Qdx}


however, when I do the calculation on my own.

dm=Qdx+xdQ
then intergrate both sides

\int{dm}=\int{Qdx}+\int{xdQ}

so should there be Int(xdQ) ?

2.) I just out of the blue thinking of during my calculation, what if I come across a vector times itself two times?
as for just one time, then it is a dot product,
but, a scalar simply can't dot product with a vector!
so what is the meaning of {v}^3? (v is a vector)

Thanks for your reading!
 
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ohoh I see now
as Q will not change with time (but do change with distance)
so dQ=0!

well... but do dQ here really means dQ/dt...? or just mean a "change" merely?
 
Hmm..your error lies rather elsewhere, I'm afraid:

The quantity Qx doesn't equal "m" or M at all!

Let the rod start at x=0, and end at x=X.

Let m(x) be the mass of the piece of rod extending from 0 to x.
In particular, m(X)=M, where M is the total mass of the rod.

Now, we may DEFINE Q(x), the local density, as the limit of the expression:
\frac{m(x+\delta{x})-m(x)}{\delta{x}} as "delta x" goes to 0.

That is, Q(x)=dm/dx, by definition (and therefore, dm=Qdx).
 
I see what you did there. You applied the product rule of differentiation. However:

<br /> m \ne Q(x) \, x<br />

This would mean that you assume the density of the whole rod up to x is uniform and equal to Q(x). To see the absurdity of your assumption, consider some other y &gt; x. Then, according to your formula m(y) = Q(y) y. So, this means the density now chaged to Q(y) for all points between 0 and y, including the points from 0 to x.

When dealing with position dependent densities, you need to apply the "differential-integral" method (DI). What does that mean? Divide the whole rod into infinitesimally short segments of length dx You label each segment by choosing a particular point within that segment, let's say it's left side and using its coordinate x. Then, you assume nothing varies within the segment, but varies from segment to segment (expressed by the fact that all the quantities are coordinate dependent, like your density for example). Then, you apply the formula for constant density for each line segment. The mass of those segments is infinitesimally small:

<br /> dm(x) = Q(x) \, dx<br />

This is the differential part of the DI method.

To get the whole mass, you sum the masses of all the segments. However, when dealing with infinitesimally small quantities, summation means integration:

<br /> m = \sum{\Delta m(x)} \rightarrow \int{dm(x)} = \int{Q(x) \, dx}<br />

This is the integral part of the DI method.
 
Oh! thank you a lot! now I learned my mistakes! :D

sometimes, in my calculations, I intergrate around things,
for a example
\int{\overline{v}^2dv
then we get
{{1}\over{3}}\overline{v}^3+C
so ... what exactly \overline{v}^3 means here?
as a scalar (V^2) can't dot a vector...
 
So I know that electrons are fundamental, there's no 'material' that makes them up, it's like talking about a colour itself rather than a car or a flower. Now protons and neutrons and quarks and whatever other stuff is there fundamentally, I want someone to kind of teach me these, I have a lot of questions that books might not give the answer in the way I understand. Thanks
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