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Density and integrals

  1. Feb 16, 2007 #1
    1. Find the mass of a rod length 10 cm with density d(x) = e^-x gm/cm at a distance of x cm from the left.
    2. Find the center of mass of a system containing three point masses of 5gm, 3gm, and 1 gm located respectively at x = -10, x = 1, and x =2.

    for number two what I did is just this:

    (5(-10) + 3(1) + 1(2)) / (-10+1+(-2)) and I got 45/7
    Last edited: Feb 16, 2007
  2. jcsd
  3. Feb 16, 2007 #2
    Hint For the first: Imagine the rod to be made of tiny mass elements [itex]dm[/itex]. Each mass element is equal to [itex]\rho dx[/itex], where [itex]\rho[/itex] is the linear density within the differential [itex]dx[/itex]. If you sum up all such [itex]dm[/itex]'s you arrive at the mass of the rod.

    2) How do you define, mathematically, the centre of mass of a system of particles?
  4. Feb 16, 2007 #3
    so here's what I did for number one.

    I take the integral of 0 to 10 of e^-x dx, is that all?

    what's wrong with my number 2??
  5. Feb 16, 2007 #4
    Yes, that's right

    Again, I ask you, what's the definition of the centre of mass?
  6. Feb 16, 2007 #5
    the center of mass as of my understanding is the point/position where there's a balance/equilibrium.
  7. Feb 16, 2007 #6
    I was talking about the mathematical definition, which is: [tex]\vec{R}_{cm} = \frac{\sum_{i=1}^{n}m_i\vec{r}_i}{\sum_{i=1}^n m_i}[/tex]

    If you look at your answer in the first post, you may notice that it is not dimensionally correct.
    Last edited: Feb 16, 2007
  8. Feb 16, 2007 #7
    Oh clumsy mistake, so it should be:

    (5(-10) + 3(1) + 1(2)) / (5+3+1)) and results as -45/9 = -5 right??

    I have one more question and this is kind of hard:

    A cardboard figure has a region which is bounded on the left by the line x = a, on the right by the line x=b, above by f(x), and below by g(x). If the density d(x) gm/cm^2 varies only with x, find an expression for the total mass of the figure, in terms of f(x), g(x), and d(x)
  9. Feb 16, 2007 #8
    The principle's the same as in post #2, except that, here you have an area instead of a line. Use double integrals.
  10. Feb 17, 2007 #9

    what do you mean here as double integral?? so I still use the same formula as what I did in number 2?
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