# I need a quike answer

1. Jul 5, 2007

### toutou

hi i need to know how can we find the gravitational force between two uniform rods of mass m and length l???

2. Jul 5, 2007

### mgb_phys

F=Gm1m2/r2
Where r is the distance between them and m1,m2 ae the masses of the two rods. G is the gravitaional constat = look it up.

3. Jul 5, 2007

### ice109

lol that's not right, that formula is for to spheres, which act like point masses. you have to do an integral for each dl over l for his question, though i don't know how to do it

4. Jul 5, 2007

### K.J.Healey

Assuming some constant densidy p,
I'd think you could do a dual integral.
Your "R" would be something like sqrt((y-x)^2+d^2). Draw some triangles and you will see why.
Your small mass dm = p*dx
And you have to integrated each point's (dm's) influence on each other dm on the second bar. So dual integral, over dx and dy=0..L
p = M/L (just easier a little to work with density).

Seems to end up being a log and arctan... Maybe I did that wrong.

5. Jul 5, 2007

### mgb_phys

Sorry was thinking a bit too much 'introductory physics'.
Or just assume the two rods are a long distance apart!

6. Jul 5, 2007

### K.J.Healey

Eh, and maybe I didn't. Seems any sum over some points seperated by a distance of 1/r^2, which becomes any 1/((x-y)^2+d^2) requires a trig sub leading to an arctan ((1/d)arctan((y-x)/d) if I remember it right, but dont trust me on that. Then the integral of that arctan becomes a log and another arctan something id guess would be (y-x)*(1/d)arctan((y-x)/d) again plus the log term, something like -(1/2)log((y-x)^2/d^2+1). Once again, I'm just doing this real fast, and its something like that.

haha, I just realized i did this all without the limits, hold on...
Yeah, it ends up a arctan and log but its simpler.
basically
G*P*[(2L/d)ArcTan(L/D) - ln(L^2/D^2 +1)]
where G = gravity const
P = density
L = length
D=Distance from eachother

Now the only problem is I did this for 2 parallel bars. If they were at an angle to eachother, thats a whole nother story. Youd have to recalculate R as a function of the angle of rotation between the two bars. Can't use simple right-triangle law of cosines.

7. Jul 5, 2007

### Schrodinger's Dog

hehe that's my bad I directed toutou here, I too thought it was a simple matter of the square inch law oops :)

Perhaps it could be moved to advanced physics then?

8. Jul 5, 2007

### K.J.Healey

I could be wrong in my way of thinking, but if one part of the bar is farther away than another, gravity will act differently and you cannot just use center of masses, right?

Two bars like this : | |
Will behave differently than : | --
And differently than: | /

But I'm sure its a function of their relative rotation, their lengths and the distance from their centers of mass.

9. Jul 5, 2007

### mgb_phys

Actually that makes me think my simple answer is correct!
Imagine each of the rods is surrounded by a light sphere - now imagine the calculation where you don't know the orientation of the rods.

10. Jul 5, 2007

### Schrodinger's Dog

Interesting point, but there is no information given to conclude they are not uniform in their length, mass and orientation¦¦ damn my keyboard doesn't do single unbroken bar?

I guess we'll have to wait to see what Tatou means, English isn't his first language though bear that in mind.

I'm kinda with Mgb_phys but the rod thing is throwing me towards a more complicated answer. Anyway I thought the same, he wanted a simple equation, but definitely could be wrong.

Last edited: Jul 5, 2007
11. Jul 5, 2007

### Kurdt

Staff Emeritus
One can use calculus to help them obtain the answer.

$$dF = \frac{Gdm_1dm_2}{r^2}$$

Integrate over the masses.

PS: If you want to go more in depth find a text on potential theory.

Last edited: Jul 5, 2007
12. Jul 6, 2007

### toutou

we should use the integral for both as we did with a point mass and uniform rod..the two rods lying on the x-axis with seperation a ___ ___ they are equal in mass and length

13. Jul 6, 2007

### Staff: Mentor

The separation distance, r, is the distance between the centers of mass (CM), with mass m applied at CM. That is what the double integral would provide.

Use F=Gmm/r2

14. Jul 6, 2007

### Staff: Mentor

Integrate!

I don't think so. Newton's law of gravity applies to point masses or special mass configurations (those with spherical symmetry). Of course, if the two rods are very far apart, you can approximate the answer by treating them as point masses with their mass concentrated at their centers. I presume that's not the case here.

For other mass configurations--like this one--you'll need to integrate.

15. Jul 6, 2007

### K.J.Healey

Their orientations would have to affect their movement. If you're talking about the apparent force on the center of mass, that may be one thing, but for two bars oriented at : \ / then you'll have to worry about moments as well.
And yes, I think dF=Gm1m2/r^2 is the way to go, which I did. The problem is that "r". When you integrate over the bar the r is a function of their rotation.

Though he did just say they were like this : ___ ___

16. Jul 6, 2007

### Staff: Mentor

Ah, so that's supposed to be the relative orientation of the rods. (D'oh!) In that case, the integration is easier.

17. Jul 6, 2007

### toutou

i just got hint from my teacher that i should use the double integral

18. Jul 6, 2007

### toutou

and by the way am a girl

19. Jul 6, 2007

### Kurdt

Staff Emeritus
Hello toutou,

For this problem you do indeed have to perform a vector sum over the two bars. This means its important to get your picture of what is happening correct. I've never done this myself so forgive me if I get part of this wrong.

For a point particle and an extended object we have:

$$\mathbf{F} = -Gm \int \frac{dM}{r^2} \mathbf{\hat{r}}$$

For two extended objects you'll probably have to have something of the form:

$$\mathbf{F} = -G \int \frac{dM_1dM_2}{\mid\mathbf{r}-\mathbf{r\prime}\mid^2}\mathbf{\hat{r}}$$

You'll have to find M1 and M2 in terms of r and r'.

I'm not 100% sure of this though because as I said i've never done it personally but perhaps someone older and wiser can confirm or deny this.

20. Jul 6, 2007

### Dick

Kurdt is correct. Where the dM_i is given by rho*dr_i (rho is the linear density) and the denominator is (r_1-r_2)^2. You can always take the rods symmetrical around the origin (eg [-b,-a] and [a,b]). The integration isn't particularly hard but you do have to be careful with signs. When you are all done it's a good exercise to take say, a=n and b=n+1 and make sure you get nearly the same answer when n is large as you do when taking the force between point masses at the center of mass of each of the rods.