- #1

kostoglotov

- 234

- 6

## Homework Statement

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]]

## Homework 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]

## The Attempt at a Solution

I made the following diagram to help illustrate my process

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?