Recent content by rzyn

Well you took the time to calculate the slowing of the Earth's rotation and a gain in weight around the Earth so small it doesn't matter. Don't you think at orders of magnitude so small you can say an effect doesn't matter, tidal effects would be of comparable magnitude or larger? You took the...

Well the thing is I think the weight loss is there but can't be found algebraically. Plus I think because angular momentum is conserved centripetal force toward the center of the system must remain constant. I think everyone for their input though. It is helpful, although I think it...

You went through all these calculations and eventually were able to contradict me by a tiny tiny amount [but not enough to matter]. You'll go through efforts to find a tiny tiny effect as long as you're satisfied with it being there. But you didn't go back to see if the effect you assumed a...

Hmm maybe the teacher's wording is sloppy. Judging from follow-up question it seems like he's trying to get us to determine whether its angular momentum in a wheel that could cause weight loss or east-west oscillation alone, or neither. According to the account holder you referred to, it's...

Yes I understand that centripetal forces about the axel are zero-sum and need only be considered for determining tensile forces developed in the spokes. The part I'm having trouble visualizing is how topside east rotation of the Ferris wheel could not exert upward tension through a spoke which...

Hmm I thought you could do a calculation for whatever particle happened to be at the very top of the wheel. Sum up the Earth's linear velocity with the wheel's linear velocity to get v. Then add acceleration due to gravity. Then subtract acceleration due to rotation around the earth. The add...

F=mv^2/R As R increases on a rotating reference frame v increases linearly however v^2 increases parabolically therefore F also increases parabolically for a corresponding point-mass. A stationary Ferris wheel, simply by being vertical rather than horizontal would have the effective gravity at...

Hmm. I've thought it over. It appears a craft could not lose weight just by oscillating east-west within. Now I'm wondering, suppose there were a Ferris wheel instead. If it were stationary relative to inertial space, it would be heavier? As it started rotating with the Earth it would be...

Ah, I mean little g. The g has to be adjusted for the eastward mass and westward mass. But when you do that, because of v^2 the eastward mass will lose more acceleration than the westward mass will gain. It has to fall at the same g though whether it contains oscillation or not, right?

Does the object fall at the same rate of acceleration as an identical object not containing oscillation?

*effective gravity

Yeah but only looks half that from earth

Force due to effective gravity in its non-rotating state.

The geometry of the masses themselves in relation to the Earth given that R is large and v's are high, is negligible. Only 3 significant digits.

The same frame of reference used to arrive at the G (the earth) because the G is used in the problem.