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

[kostoglotov]

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?

 

[Dick]

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?

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?

 

[SteamKing]

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?

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?).

 

[Dick]

[EDIT: I think Dick means why don't you just calculate Fv = F cos φ?).

Yes, I did mean that. Thanks for the correction!

 

[SammyS]

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

[ IMG]http://i.imgur.com/lP0o2HU.jpg[/PLAIN]

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}\ \quad\ \ \color{red}{\text{Take note: }\sin\phi=\frac{r}{\sqrt{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 infinitesimals 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?

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.

 

[kostoglotov]

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

Because I like to punish myself. :P