Physics question. Complicated. Centrifugal force.
 

Ok, here's a complicated physics question ya.

If I am in space, and I have a one pound weight, and I tie the weight to a rope that is three feet long and swing or rotate the weight, how fast must the weight rotate at the end of the rope in order to produce a one pound pull on the rope?

It will be very hard to rotate because there's an elephant in the way.;)

I believe you're looking for F=mass x w (angular speed in rads/sec)^2 x radius

couple that with F = mass x acc (centripetal) and you get

acc (centr) (looking for 32f/s^2 here = w^2 x r

so square root of 32/3

or 3.266 rads / sec which is about 1/2 rev per sec.

I think that's correct, but not positive. Will have to check tomorrow after some sleep.

Dredging into the archaeological recesses of memory.....

As I recall, it will depend on your mass. In space you will operate as a system rotating around your mutual center of mass; therefore, the "orbital" radius of the 1lbm object will be larger as your mass is larger. Iow, if your mass is also 1lbm, the radius of the orbit is 1.5ft. If your mass is infinite, the radius is 3ft.

The force on the rope is:
F = ( m * v^2) / r

Combine with masraum's eq'ns.

Not enough information to solve, though with a bit of time you could probably have a result that is expressed as a function of M2.

It has been a long time since I last did that stuff.
Jim

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OK, that's hilarious. :D

Of course, if you're in space (and I'm assuming by "space" you mean away from the influence of any gravitational fields) then your "one pound weight" doesn't weigh one pound. Weight is a function of mass acted upon by gravity. No gravity = no weight.

I worked on the Spacelab program for about a decade and we used to do "microgravity" experiments in the shuttle on orbit. The interesting thing is that gravity on the shuttle in orbit is about the same as it is for people standing on the Earth's surface. The shuttle merely simulates zero-g by inducing a state of constant freefall when in orbit. It's not exactly zero-g, so they called it "micro" g. Sort of like the world's longest vomit comet ride.

Probably common knowledge, but I always thought that was interesting.

so if the shuttle stopped moving at it's orbit distance gravity would be same as on earth?

I don't think the gravitational attraction of the two bodies will have an appreciable effect on this situation. (assuming we are someplace with no gravity from a nearby planet or celestial body). And, yes, the 1 lb weight won't weigh anything, but it will still have the same mass in "slugs" (English unit of mass).

Either way, we are looking for the component of the angular acceleration perpendicular to the tanget to the circle that's described by the rotating mass to be 32ft/sec^2. I could definitely be wrong, it's been a long time, but I think you guys are making this much harder than it should be.

Basically, yes. Gravity's effect on you (or any mass) is a function of your distance from the center of the mass. Assuming the diameter of the Earth is about 8000 miles (I don't know what it is exactly), if you're standing on the surface of the Earth, you're about 4000 miles from the center. If you're on the shuttle in orbit (typical orbit is around 200 miles), the you're only about 5% farther away from the center, so the effect of gravity is only slightly different. You'd be hard pressed to notice.

If the shuttle stopped moving, it would drop like a rock. In fact, that's how they change the orbit - speed up and they climb to higher orbit, slow down and they fall to a lower orbit. To de-orbit they slow down a bunch.

I think I was correct last night. Another way to do the problem that I think comes out with basically the same answer.

http://hyperphysics.phy-astr.gsu.edu...gmec/cfmag.gif
again, set that equal to F=ma which if we are interested in weight would be
weight equals mass times acceleration due to gravity

1lb = mass x 32ft/sec^2

so

mass x 32 = mass x velocity ^2 / radius

because the mass is the same you can cancel the masses from both sides. You end up with

v^2 / 3 =32

v^2 = 96

v = √96 = 9.8 fps

the circumference of the circle is 2πr which is 6π or about 18.85 feet so again, we are at just a tad more than half a rev per second to match accel due to gravity.

If I'm wrong, please, someone let me know. Really. It's been a long time. I'm reasonable certain that my thinking is correct, but I may be wrong.

You are correct, masraum. I got into bed and had the "Oh Crap!!" moment when I realized I had typed the basic gravitational equation. Then it became the "do I get up and start the computer?" moment.

Note to Self: "Self, don't do equations when tired"

Sorry. Updated the equation.