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Homework Help: Center of Gravity slender rod

  1. Apr 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider a slender rod satellite in a circular orbit. Show that the position of the center of gravity, ##\rho_G##, can be written in terms of the position of the center of mass, ##\rho_C##, as: ##\rho_G = \rho_C (1+\frac{l^2}{4\rho_C^2})^\frac{1}{4} ##, where ## l## is the length of the rod.

    I'm not sure how to best describe this, but the rod is oriented like this, perpendicular to the position vector:

    2. Relevant equations
    I know the total force on the rod acts on the CoG:
    ## F = \frac{\mu mM}{\rho_G^2} ##

    m is the total mass of the bar, M is the mass of the attracting body, ##\rho## is a position vector.
    I need to set this force equal to an integral, which I believe goes something like this:

    ## F = \int_{0}^{m} \frac{\mu M dm}{\rho^2} ##

    3. The attempt at a solution
    I changed to a length integral (half of the bar multiplied by 2), and have this:
    ##F = 2 \int_{\rho_C}^{\sqrt{\rho_C^2+l^2/4}} \frac{\mu Mmd \rho }{l \rho^2} ##

    using the fact that ##dm = \frac{m}{l} d\rho##

    I'm not confident that this is correct, as I can't seem to get the right answer. Something seems off about this integral, but I'm not sure what.
  2. jcsd
  3. Apr 19, 2015 #2
    Can you define the terms?
    ρ = position vector = distance from the earth's center of mass M to any arbitrary point?
    ρC = distance from the earth's center of mass M to the cylinder's center of mass m?
    ρG = distance from the earth's center of mass M to the cylinder's center of gravity?

    Are you sure the cylinder is oriented perpendicular to the position vector?
    Last edited: Apr 19, 2015
  4. Apr 19, 2015 #3
    You are correct on all three accounts:
    ρ is the position vector from the earth's center (mass M), to the infinitesimal mass element dm.
    ρc is the distance from earth's center to the rod's (mass m) center of mass
    ρG is the distance from earth's center to the rod's center of gravity.

    And yes, the rod is perpendicular to the position vector. The first part of the problem had the rod aligned "vertically" with respect to the surface of the earth. In this case, it is perpendicular.
  5. Apr 19, 2015 #4
    OK, then it looks like what you are doing is taking ρ from ρC to one end of the rod in your integral and then using symmetry to double it.

    I am not sure that you can do this since this would give you the same center of gravity as a parallel rod with length = √(ρC2+(L/2)2) -ρC. Maybe it works out to be the same.

    Could you instead take the angle θ and create an integral from θ=0 to L/(2ρC) ?
  6. Apr 19, 2015 #5
    Hm I hadn't thought about using the angle. I will give that a try tomorrow and report back. Thanks for the responses by the way, I really appreciate it.
  7. Apr 19, 2015 #6


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    Doesn't look right to me. Working with the angle should make it clearer.
    I think that's ok.
  8. Apr 19, 2015 #7
    dm=m/l dρ
    Seems OK to me. m/l is the density of the rod.
  9. Apr 19, 2015 #8


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    Sure, but ρ is the distance from Earth's centre, not a distance along the rod.
  10. Apr 19, 2015 #9
    But dρ is along the length of the rod.
  11. Apr 19, 2015 #10


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    Eh?! dρ is a change in distance from Earth's centre as we move along an element of the rod. That will not be the same as the length of the element.
  12. Apr 19, 2015 #11
    Yes, of course, you're right. I was thinking for a moment the rod was parallel.

    That's probably why he was getting the wrong answer.
  13. Apr 20, 2015 #12


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    Hello sun18,

    It's an interesting problem, for sure! Anyway I worked through it and there are two mistakes in your general approach and perhaps another quibble about your notational semantics. I'll start with notation part.

    Not bad so far. But let me comment on the notation.

    Newton's universal law of gravitation for point mass particles is often expressed as, using [itex] \vec \rho [/itex] as the displacement vector,
    [tex] \vec F = G \frac{Mm}{\rho^2} \ \hat \rho [/tex]

    That said, it's often expressed using [itex] \mu = GM [/itex] as
    [tex] \vec F = \frac{\mu m}{\rho^2} \hat \rho [/tex]

    where [itex] \mu [/itex] is called the "standard gravitational parameter" and has the mass of the gravitating body (often Earth when dealing with things orbiting Earth) folded into it along with Newton's gravitational constant.

    In the notation you were using, you've simply replaced [itex] G [/itex] with [itex] \mu [/itex], but didn't fold [itex] M [/itex] into it. The notation that I've seen most often has [itex] M [/itex] folded into the [itex] \mu [/itex], along with the [itex] G [/itex].

    Of course, this is all just notational semantics, and isn't important on the grand scheme of things. But for what it's worth, I thought I should mention it.

    There's one problem right there.

    The change in mass does not vary linearly with the change in [itex] \rho [/itex] in this case. It does vary linearly in the case where the rod is parallel to the displacement vector; in that case it would work fine. But here, with the rod perpendicular to the displacement vector, it does not hold true.

    Let's define [itex] x [/itex] as the distance from [itex] \rho_c [/itex] to some point along the rod. So you can say easily,
    [tex] dm = \frac{m}{\ell} dx [/tex]

    You should be able to form a relationship between [itex] x [/itex] and [itex] \rho [/itex] knowing that they form a right triangle. [itex] \rho^2 = \rho_c^2 + x^2 [/itex].

    Now for the second problem: you are forgetting about the pesky unit vector [itex] \hat \rho [/itex].

    In the case where the rod was parallel to the position vector, the direction of the force wasn't important; the force on every individual [itex] dm [/itex] all pointed in the same direction, so the unit direction vector didn't really matter. But in this case, where the rod is perpendicular, the each of the individual force directions are different.

    You need to break up the [itex] \hat \rho [/itex] into its x- and y-components. It's in the form, [itex] \hat \rho = a_x \hat x + a_y \hat y [/itex]. Due to symmetry you know that the horizontal (x-) components will all ultimately cancel. But the y-components will not. You need to put that into your expression: The way you express the unit vector for a given [itex] dm [/itex] should be should be a function of [itex] \rho_c [/itex] and [itex] x [/itex]. [Edit: or alternatively, express things in polar coordinates involving an angle. Either way will work, as long as you take directionality into account one way or the other.]

    Good luck! :wink:

    [Edit: Looks like @haruspex and others beat me to the punch. I suppose I should either type faster or write shorter. :smile:]
    Last edited: Apr 20, 2015
  14. Apr 20, 2015 #13
    Oh right of course, my mistake. I was forgetting that μ absorbs G and M. Luckily it should end up cancelling out anyways.

    I had a feeling that the directionality was non-trivial. The first part of the problem had the rod colinear with the displacement vector ρ, so it didn't matter then.

    So I had the integral ## \vec{F} = 2 \int_{\rho_C}^{\sqrt{\rho_C^2+l^2/4}} \frac{\mu md\rho}{l\rho^2}\hat{\rho} ##

    Using ## \rho^2 = \rho_C^2 + x^2 ##, ## \vec{\rho} = \rho_C \hat{y} + x\hat{x}##, ## \hat{\rho} = \vec{\rho}/\rho ##, and changing the bounds of my integral, I have
    ## \vec{F} = \int_{-l/2}^{l/2} \frac{\mu mdx}{l\rho^2}\hat{\rho} ## (I didn't replace the ρ's yet for brevity)

    I evaluated the integral, and as you said, the ##\hat{x}## components cancel, and I'm left with this integral:
    ## \frac{\mu m \rho_C \hat{y}}{l} \int_{-l/2}^{l/2} \frac{dx}{(\rho_C^2+x^2)^{\frac{3}{2}}} ##

    **EDIT: Yep! This got me the right answer. It's a tricky integral to do, I wonder if it's easier to do in polar coordinates?

    Thanks again for all the responses, I appreciate it!
    Last edited: Apr 20, 2015
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