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Center of mass and a bit vector calculus

  1. Jul 5, 2010 #1
    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

    however, when I do the calculation on my own.

    then intergrate both sides


    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 [itex]{v}^3[/itex]? (v is a vector)

    Thanks for your reading!!
    Last edited: Jul 5, 2010
  2. jcsd
  3. Jul 5, 2010 #2
    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?
  4. Jul 5, 2010 #3


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    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:
    [tex]\frac{m(x+\delta{x})-m(x)}{\delta{x}}[/tex] as "delta x" goes to 0.

    That is, Q(x)=dm/dx, by definition (and therefore, dm=Qdx).
  5. Jul 5, 2010 #4
    I see what you did there. You applied the product rule of differentiation. However:

    m \ne Q(x) \, x

    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 [itex]y > x[/itex]. Then, according to your formula [itex]m(y) = Q(y) y[/itex]. 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 [itex]dx[/itex] 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:

    dm(x) = Q(x) \, dx

    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:

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

    This is the integral part of the DI method.
  6. Jul 6, 2010 #5
    Oh! thank you a lot! now I learned my mistakes! :D

    sometimes, in my calculations, I intergrate around things,
    for a example
    then we get
    so ....... what exactly [itex]\overline{v}^3[/itex] means here?
    as a scalar (V^2) can't dot a vector....
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