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Gravitation force caused by uniform rod

  1. May 9, 2007 #1
    1. The problem statement, all variables and given/known data
    What is the gravitational force caused by a thin uniform rod of length L on a point mass located perpendicular to the rod at it's center? Assume the point mass is a distance R perpendicular to the rod.

    2. Relevant equations
    [tex]g = - \frac{GM}{r^2}\hat{r} = - G\rho \int \frac{dV}{r^2}\hat{r}[/tex]

    3. The attempt at a solution

    [tex]g = -G\rho \int \frac{Rdx}{\sqrt{x^2+R^2}^3} [/tex]
    [tex]x = Rtan(\phi) [/tex]
    [tex]g = -\frac{G\rho}{R} * ( \frac{x}{\sqrt{x^2+R^2}} ) [/tex]
    Using limits of integration from -L/2 to +L/2 to yield:
    [tex]g=-\frac{2G\rho}{R}*(L/\sqrt{L^2+4d^2})[/tex] after removing a factor of 4 from the (L/2)^2 from the square root.

    Since the y components cancel, I took theta to be the angle nearest the point mass, and called rhat cos(theta) which I evaluated to R/sqrt(x^2+R^2) which could combine with the r^2 in the denominator from the gravity equation.

    If this is wrong, I'll be glad to provide more steps to see where I went wrong. I ended up with a 1/(x^2+R^2)^(3/2) in the integral, so I did a trig substition to integrate it, and came up with the formula wikipedia has in its irrational integrals table, so I'm pretty sure that's alright, but wasn't sure if I set it up right.
    Last edited: May 9, 2007
  2. jcsd
  3. May 10, 2007 #2
    Looks okay to me. This is how I did it (I get the same integrand form)

    [tex]d\vec{f} = \frac{Gdm}{R^2+x^2}\left(-\cos\theta \hat{i} + \sin\theta \hat{j}\right)[/tex]


    [tex]\cos\theta = \frac{R}{\sqrt{R^2+x^2}}[/tex]
    [tex]\sin\theta = \frac{x}{\sqrt{R^2+x^2}}[/tex]

    The integrand over [itex]\hat{i}[/itex] is an odd, so it integrates to zero ([itex]x[/itex] ranges from [itex]-L/2[/itex] to [itex]L/2[/itex]). You get

    [tex]\vec{f} = -\hat{j}G\lambda R\int_{-L/2}^{L/2}\frac{1}{(x^2+R^2)^{3/2}}dx[/tex]
    Last edited: May 10, 2007
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