# Weights on circular rotating circle

1. Aug 20, 2016

### artriant

circular cable i mean *
Hey all second time to use this forum, hope to get same awesome help.

Lets say that we have a cable in a shape of a closed perfect circle rotating in perfect conditions no air zero gravity etc around a center.

The cable is 100 m circumference, rotating at 1rpm

My problem is to find out a way to support how much weight i can distribute. But weights can be distributed in different ways(space in between, how many, theoretical or actual shape) . So im trying to go from a simplified version to a more complex one gradually.

So lets make the simplified version CASE 1:
We equally distribute 10 points of mass around.the cable.
Lets assume wire has zero mass no actuall volume and bend resistance is absent, but preserves other characteristics like it cannot brake easily(can hold 10.000 Kg tensile force with safety).

So since we add more mass on specific points the shape will change, and in equelibrium still we force it on 1 rpm, we are going to end up with a rotating Decagon! with 10 x 144deg interior angles and the r on each point of added mass will be now 16.1803m (previusly r was 15.92m while on perfect circle).

The question is how much mass is allowed on each point, to end up with a safe tensile force (10K kg).

2. Aug 20, 2016

### billy_joule

Your problem is a variation of the common tension in a string problem:
http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html#strmas
http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html

Draw a free body diagram of one of the weights and solve for mass when tension is at it's max allowable.
work symbolically and you can form an equation for max mass in terms of: Tmax, radius, RPM (or rad/s) and # of (equally spaced) masses.

See what happen to the allowable mass as the interior angle goes to 180 (eg we add many masses and the cable resembles a circle)

It'll get quite a bit more complicated when the weights &/or spacings are not equal as the radius for each mass will differ.

3. Aug 20, 2016

### artriant

Ok this is a portion of the decagon. Not sure if i do understand exactly what you proposed i expect the allowable weight to be more in total and more easy to calc as we get perfect distribution in countless parts (not sure tho ;p).

By using this calculator (http://www.calctool.org/CALC/phys/newtonian/centrifugal) i found out that if a single mass m1 was hanging only in the red imaginary string should be 552685Kg to provide 10000 Kg-force at 1 rpm. But i think i cannot just brake that in ten parts.

Back in the original setup no red cable.
If i add the 2 vectors of tensile forces at these angles (in image) i get a sum vector 0.61803 times smaller compared to them.
(So technically i wonder if i pull down a horizontal string and i get it at that angle with 61 newtons downwards, at the anchors it`ll be 100N each? Is that right?)
I assume is right for now.
So max tension allowed is 10000Kg-force within the black cable and im allowed to put 10000/10weights* 0.61803=618Kg-force on each point.
At that spinning rate and radius we need 34155.88 earthly Kilograms to get to that max tension within the black surrounding cable.

Can anybody confirm that im not doing nonsense? lol

4. Aug 21, 2016

### billy_joule

I don't quite follow your approach. It's much better if you show all your working and work symbolically, just plug the numbers in right at the end.

You free body diagram is right, lets label those forces TL (tension left), TR (tension right) & FC (centrifugal force) and leave the internal angle as θ.

Solving for forces in the x direction:
∑FX = 0
We know that FC has no x component and that TL and TR are of equal magnitude and symmetric about the Y direction so obviously:
TL,x + TR,x = 0

Now for the Y direction:
∑FY = 0
TL,y + TR,y +FC = 0

(we can see TL,y = TR,y, so lets just say T = TL,y = TR,y )

2Tcos(θ/2) + FC = 0
and we know:
FC = mv2 / r
so:
2Tcos(θ/2) + mv2 / r = 0
and solving for m:
m = 2rTcos(θ/2)) / v2

So as θ goes to 180 the mass goes to zero, which should be intuitive; the cable tension force needs some Y component to counteract a centripetal force.

5. Aug 21, 2016

### artriant

m = 2rTcos(θ/2)) /v^2
Max T allowed 98066N

m=2*16.1803m*98066N*cos(144deg/2)/(1.7^2) m/s

m=3173474,5996*cos 72/2,8706

m=341621,12~ Kg

split that in 10 and i get 34162,1 Kg for each point in the decagon

Quite close with my approach :D ,which is more empirical rather than proper physics.
So thats it? That is the max mass allowed per point? If so im so happy. Is this approach from inertial reference point right? Im trying to write an article you get honorable mention alrdy!

6. Aug 21, 2016

### billy_joule

7. Aug 21, 2016

### artriant

If you r right this is hell of impressive.

Dont we need to split 10 at any point?

This number is like 3.4 milion Kg all around! This is like the mass of 2 houses per point!!.

Could be true cause the generated centrifugal is just super small 0,018G. So each point pushes with 6112.25Kg outwards, is that ok? i get that this will produce a local tension of 10000. But arent tensions adding up to something like 10x? I think they dont now that im thinkin twice hah.

Thats a big finding right there for my stupid head. heh, confirm pls

Last edited: Aug 21, 2016
8. Aug 22, 2016