Thread: Physics question

Your not alone, i do design engineering; but I forgot what exactly it is, so had to open up me old motion & energy book.

It's a relation between W=mg, where W = weight (in newtons) M = mass (kg) & g = gravitational force.

(this is only so you know that whatever you have sitting on it, won't fall off)!

&

Fc (centripedal force) = (mv^2)/ r - where R = the radius. M = mass (kg), V= Velocity (m/s)

Mass of 1 g is 9.81N , Which reaches that at roughly 6400 km radius ove the 24 hour time frame.

(Edit: keep in mind that everything is in per unit time. I.E meters per second)

Originally Posted by SnowLeopard

Is it as easy as that?

I would think the density of the body's material would have something to do with it.......

'
I've got my pen & paper out now, but i don't have a damned calculator I may end up scrubbing that! But it rings a bell, Im just running it through as Fc=mrw^2
Fc=mv^2/r
therefor
mrw^2=mv^2/r

should be a little bit of catch up, I refuse to look for the solution online!

Edit, hashing that through, I get r^2 x w^2 = V^2.

So by that logic, The Square of the Radius, times the Square of the angular displacement (radians) = the Square of the velocity. Therefore if you know the angular displacement, & velocity, you should get the radius required.

If Fc = Fc the mass is cancelled re-arranging the equation yes? the units come out as above and below as 9.81. Which either tells me that the Mass doesn't actually matter, or I've got my wires crossed in the equation im using!

Edit - I think i've got my wires crossed here, as if I use earths stats it comes out wrong by a margin of almost 2.5

Pari, Maybe we should keep this thread going. I could assist in finding the answer if needed but I do certainly like the topic of the thread. Good luck with your answer. And hopefully this thread stays alive.


*Besides it will give the college attending members a life line of reference if needed. Just imagine the physics that are known here between members of different backgrounds and vocations.

P.S. Better Physics then doing Logirythms.

Just to clarify, the object isn't the earth, and the 1 g is supplied by the centrifugal force provided by the spin of the circle. Don't worry about the mass of the rotating object for the moment

Originally Posted by Parihaka

Just to clarify, the object isn't the earth, and the 1 g is supplied by the centrifugal force provided by the spin of the circle. Don't worry about the mass of the rotating object for the moment

It does not matter if the large object is earth, mars or venus. It provides inward gravitational 'g' because of its large mass, not because the planet rotates. A larger planet have a larger 'g'.

There must be an mg force pulling in, and a cenrtipetal force mv2/r pulling out to balance it. Here, m is the mass of the orbiting satellite, not the planet.

How many years ago did you last learnt physics? I last learnt it 51 years ago.

Originally Posted by Merlin

It does not matter if the large object is earth, mars or venus. It provides inward gravitational 'g' because of its large mass, not because the planet rotates. A larger planet have a larger 'g'.

There must be an mg force pulling in, and a cenrtipetal force mv2/r pulling out to balance it. Here, m is the mass of the orbiting satellite, not the planet.

How many years ago did you last learnt physics? I last learnt it 51 years ago.

You're not understanding what I'm saying. This has nothing to do with a planet. It is a man made space station. It does not orbit a planet, though it will orbit a sun just as our planet does.
A wheel spinning in space.
To achieve comfort, it needs to achieve 1g by centrifugal/centripedal/whatever force.
What I am asking is what is the radius of that circle to achieve 1g if the rotation is 24 hours, the reason being it's a natural human cycle. If you're on the side closest to the sun it's night time, if you on the side furthest from the sun it's day time.

Originally Posted by dalem

Neat little Java app.

-dale

Nice, I'll have a play with it tonight when hopefully my head cold clears up. Still looking for better brains than myself to hand me the answer on a plate though

Originally Posted by Merlin

It does not matter if the large object is earth, mars or venus. It provides inward gravitational 'g' because of its large mass, not because the planet rotates. A larger planet have a larger 'g'.

There must be an mg force pulling in, and a cenrtipetal force mv2/r pulling out to balance it. Here, m is the mass of the orbiting satellite, not the planet.

How many years ago did you last learnt physics? I last learnt it 51 years ago.

I think Dalems applet makes sense of what you're trying to tell me as well..

Originally Posted by Merlin

It does not matter if the large object is earth, mars or venus. It provides inward gravitational 'g' because of its large mass, not because the planet rotates. A larger planet have a larger 'g'.

There must be an mg force pulling in, and a cenrtipetal force mv2/r pulling out to balance it. Here, m is the mass of the orbiting satellite, not the planet.

How many years ago did you last learnt physics? I last learnt it 51 years ago.

The problem as presented treats the mass of the "ring" Pari asked about as negligible. For purposes of the question imagine the ring he's spinning for gravity on the inner rim is made of cotton candy or dandelion tufts.

-dale

p.s. Pari, here's a (yuk) Wiki that seems to address the issue pretty well, but I only skimmed it.

Wiki (yuk).

Originally Posted by dalem

The problem as presented treats the mass of the "ring" Pari asked about as negligible. For purposes of the question imagine the ring he's spinning for gravity on the inner rim is made of cotton candy or dandelion tufts.

-dale

p.s. Pari, here's a (yuk) Wiki that seems to address the issue pretty well, but I only skimmed it.

Wiki (yuk).

And there it is, thanks Dale.

R = 9.81g/(((pi x rpm)/30)^2)