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Gravitational Force Due to Infinitely Long Rod

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Infinitely long rod with the z axis at its center. The rod has a uniform mass per unit length [itex]\mu[/itex]. Find the gravitational force vector F on a mass m, at a distance [itex]\rho[/itex] from the z axis.


    2. Relevant equations
    F=-G*(M*m)/R^2 (times radial unit vector rhat for the vector form)


    3. The attempt at a solution
    I believe I can treat the rod as being very thin, with a center of mass along the z axis. Then I labeled the masses position as being on the y axis. I believe that the force exerted on m in the z direction cancel because of symmetry. I believe I need to work in cylindrical polar coordinates because of the problems use of [itex]\rho[/itex] and z.


    I do not see how to construct an integral (from - infinity to + infinity) in polar coordinates. I know I need to vary z. I tried to construct an equivalent integral in Cartesian coordinates as follows.

    -G[itex]\mu[/itex]m[itex]\int[/itex]dz/(y^2+z^2)


    G is the gravitational constant.
    integral from -infinity to infinity
    M= mass of 2nd object in Newton's law of gravitation was replaced by [itex]\mu[/itex]*z
    r= distance from z axis= (y^2+z^2)^1/2
    y is a constant because the mass is always a the same y position.

    But I think this is wrong. Can someone help?
     
    Last edited: Feb 15, 2012
  2. jcsd
  3. Feb 15, 2012 #2
    You're overthinking it. A cylindrical Problem like this doesn't require integration on the Z axis. What you need is the 2-dimensional form of the gravitation law. The rest of your reasoning about concentrating the mass on the axis is sound.
     
  4. Feb 15, 2012 #3
    But don't the elements of the rod exert nonzero bits of force on m in the negative radial direction (towards the rod)? Which don't cancel, as those in the z direction do.
    And how to I account for the mass of the rod? I.e. the "M" in F=-G*(M*m)/R^2 .
     
    Last edited: Feb 15, 2012
  5. Feb 16, 2012 #4

    vela

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    That's pretty close. You need to pick off only the radial component of the force. As you noted, the component in the z-direction will cancel out due to symmetry.

    The small piece of the rod between z and z+dz has a mass ##\mu\,dz##, so the force dF exerted on m is given by
    $$d\vec{F} = -\frac{Gm (\mu dz)}{y^2+z^2} \hat{r}$$ where ##\hat{r}## is the unit vector in the direction from the point on the rod to the location of mass m. So you got it right in your original post, except that you need to only sum the radial component of the forces.
     
  6. Feb 16, 2012 #5
    And I would sum the radial components by doing the integral (from z= -infinity to z= infinity) of the dF, as in the original post. But, now with the r-hat unit vector?

    Doing this improper integral, I get

    F= -G[itex]\mu[/itex]m[itex]\pi[/itex]/y [itex]\widehat{r}[/itex]

    If I orient my y and x axes, I can put the mass on the y axis at a Cartesian point (0,y,0). Then I could call rho from cylindrical coordinates y in Cartesian.
     
    Last edited: Feb 16, 2012
  7. Feb 16, 2012 #6

    vela

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    The integral sums the individual forces. It's a vector sum so you need to sum the individual components. It turns out if you integrate the z-component, it comes out to be 0, which you expect from the symmetry arguments, but you still need to sum the radial components. You haven't done that yet.
     
  8. Feb 16, 2012 #7
    Ok, so new approach. I'll call y, rho, and define an angle α such that [itex]\hat{r}[/itex]=cos(α)[itex]\hat{\rho}[/itex]+sin[itex]\hat{z}[/itex]. Then the radial components are the formula for dF above*cos(α), and the z components are dF*sin(α). As I let α go to -pi/2 and +pi/2, the integrals kill the z components, and the radial ones sum to

    F_rho= -2Gm[itex]\mu[/itex]/[itex]\rho[/itex]

    Is this a sound approach? So I was not correct before because I was not working with the unit vector to get the individual components?
     
  9. Feb 16, 2012 #8

    vela

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    Right, you needed to multiply by the cosine so that you summed only the radial component. If you just sum the magnitudes like you were doing before, you don't really get anything because that's not how vectors add.
     
  10. Feb 16, 2012 #9
    Thanks.
    So in cylindrical polars, my force vector is (-Gm[itex]\mu[/itex]2/[itex]\rho[/itex],0,0) where the last two are the phi-hat and z components.

    Not to drag this out, but I am then asked to show that the curl X F is zero, in (x,y,z) coordinates. The rho in the denominator is (x^2+y^2)^(-1/2) and

    rho-hat= cos[itex]\theta[/itex][itex]\widehat{x}[/itex]+sin[itex]\theta[/itex][itex]\widehat{y}[/itex]
    where [itex]\theta[/itex] is from the positive x axis.

    This puts a cos[itex]\theta[/itex] on the F_x and the sin[itex]\theta[/itex] on the F_y.

    And I can't get the curl to be zero.
     
  11. Feb 16, 2012 #10

    vela

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    So you have
    $$\vec{F} = -2Gm\mu \left(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2},0\right),$$ right?
     
  12. Feb 16, 2012 #11
    Why not just use Gauss's law applied to gravitation?
     
  13. Feb 16, 2012 #12
    No.

    -Gmμ2/ρ*[itex]\widehat{\rho}[/itex]

    then in x and y (with rho-hat= cosθ[itex]\widehat{x}[/itex]+sinθ[itex]\widehat{y}[/itex])

    and rho =(x^2+y^2)^(1/2)

    F=-Gmμ2(cosθ/(x^2+y^2)^(1/2) , sinθ/(x^2+y^2)^(1/2))

    It must be in the unit vectors that I am messing up.
     
  14. Feb 17, 2012 #13

    vela

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    You need to express the sine and cosine in terms of x and y.
     
  15. Feb 17, 2012 #14
    I now have the expression for F you mentioned. The curl of which is zero. Thank you.
     
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