1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gravity of a disk acting on a mass on the z axis

  1. May 22, 2015 #1
    1. The problem statement, all variables and given/known data

    A lamina has constant density [itex]\rho[/itex] and takes the shape of a disk with center the origin and radius R. Use Newton's Law of Gravitation to show that the magnitude of the force of attraction that the lamina exerts on a body of mass m located at the point (0,0,d) on the positive z axis is

    [tex]F = 2 \pi Gm \rho d \left(\frac{1}{d}-\frac{1}{\sqrt{R^2+d^2}}\right)[/tex]

    [Hint: Divide the disk and first compute the vertical component of the force exerted by the polar subrectangle [itex]R_{ij}[/itex]]

    2. Relevant equations

    p = diagonal distance between mass m and a polar subrectangle of the disk.

    [tex]F = G\frac{Mm}{p^2}[/tex]

    [itex]F_h[/itex] = horizontal component of the force of attraction

    [itex]F_v[/itex] = vertical component of the force of attraction

    [tex]F^2 = F_v^2+F_h^2[/tex]

    3. The attempt at a solution

    I made the following diagram to help illustrate my process

    lP0o2HU.jpg

    First, all the horizontal components of the force F will cancel out due to symmetry. So I just need to find a Riemann Sum for all the vertical components.

    [tex]F_v^2 = F^2 - F_h^2[/tex]

    [tex]F_h = F \sin{\phi} = F \frac{r}{r^2+d^2}[/tex]

    [tex]F_v^2 = F^2 - \left(F \frac{r}{r^2+d^2}\right)^2 = F^2\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)[/tex]

    [tex]F_v = F \sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)}[/tex]

    Now, to just consider the polar subrectangle [itex]R_{ij}[/itex]

    [tex]F_{v_{ij}} = G\frac{mM_{ij}}{r^2+d^2}\sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)}[/tex]

    and I figure that [itex]M_{ij} = \rho \Delta A[/itex]

    so [itex]dA = rdrd\theta[/itex] when changing to polar form.

    I get the double integral

    [tex]F_v = Gm\rho \int_0^{2\pi}\int_0^R \frac{r}{r^2+d^2}\sqrt{\left(1 - \left(\frac{r}{r^2+d^2}\right)^2\right)} drd\theta[/tex]

    Is this right? Because I can't get Matlab to evaluate it analytically. I've tried trig. sub and u sub on it. It looks like there should be a way through via some sort of fairly standard method, but the change of the infinitesmals always gives me some hideous substitute that just makes the integral more complicated.

    Do I need to revisit my model, or do I have the correct double integral and I just need to figure out how to solve it?
     
  2. jcsd
  3. May 22, 2015 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Check your expression for ##F_h##. You are missing a square root in the value for ##sin(\theta)##. And while you are at it, why don't you just compute ##F_h## using the cosine of that angle?
     
  4. May 22, 2015 #3

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    If you're interested in only the vertical component of the gravitational force, why are you wasting time calculating the horizontal component?

    After all, you can use Fv = F cos φ just as easily, instead of going thru the complication of the Pythagorean relation to find cosine.

    [EDIT: I think Dick means why don't you just calculate Fv = F cos φ?).
     
  5. May 22, 2015 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, I did mean that. Thanks for the correction!
     
  6. May 22, 2015 #5

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You made this more complicated than you needed to.

    You recognized that the horizontal components cancel, so you only needed to consider the vertical components, and then you proceeded to calculate the individual vertical components by using the individual horizontal components.

    This would all have worked if you had used the correct expression for ##\ \sin\phi\ .##

    ##\displaystyle \ {F_v}^2 = F^2 - \left(F \frac{r}{\sqrt{r^2+d^2\,}}\right)^2 =F^2\left(1 - \left(\frac{r}{\sqrt{r^2+d^2\,}}\right)^2\right)\ ##​

    However, you could directly use ##\displaystyle \ F_v = F \cos{\phi}=F\frac{d}{\sqrt{r^2+d^2\ }}\ .##

    Added in Edit:

    While I was messing around formatting things, the SteamKing and Dick beat me to this. -- but after all that effort, I'm leaving this up.
     
    Last edited: May 22, 2015
  7. May 22, 2015 #6
    Because I like to punish myself. :P
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Gravity of a disk acting on a mass on the z axis
Loading...