# Centripetal vs Centrifugal confusion crisis

#1
Okay sorry guys I'm having a mix-up and it needs quashing!

So in 2001: A Space Odyssey, on Discovery 1, the astronauts inside the centrifuge "feel" Earth weight why exactly?

Had you asked me this a year ago I'd have said "Centrifugal forces" but I know that's not a force but merely conversation of momentum. I don't get then how they are able to walk about?

The two situations I can understand circular motion are orbits and a tennis ball around a thingy. I can see how the force acts hence I can understand how it's circular motion. In both cases they're being pulled in towards the centre.

But after that it starts to feel tenuous. The one I can't seem to understand how being on a loop-de-loop works (as in pretend there are no rails and so it's just circular motion).

I see the maths in the textbook but it doesn't seem to mean much - sure I can use it in equations but it doesn't make sense.

Can someone just say some stuff and hopefully I can rid myself of this confusion-gremlin?
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4 years ago
#2
(Original post by DrSebWilkes)
...
In all ref frames, circ motion => inward centripetal force (net inward force), balanced by a reactive centrifugal force by N3 (action/reaction pair).

In rotating frames, person/object feels an outward (centrifugal) force; pseudo/inertial. If they try walk to/from axis of rotation, they experience a coriolis force too (pseudo/inertial).

To work in rotating frames, must account for these in terms of maths. But outside observers in inertial frames, don't detect such forces on person/object.

At top of looped rollercoaster, should know for circ motion: Fc = mv^2/r = N + mg, so N = m(v^2/r - g). Require N > 0 to remain on tracks => v > sqrt(g/r).
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#3
So just to make sure I totally get it and there is no more confusion I drew this diagram for the rollercoaster where the centripetal force = mg
And does that mean in virtually all cases of unresolved force diagrams, there is a force vector acting outwards which is opposite to the net vector right to conserve N3?

Presumably for any values >mg then it'd just be a mg*(1+/−sin(theta)) graph (where you take the mid elevation to be the axis from which the angle is taken - like an argand diagrams) , right?

And then if there was no gravity (or the FOR of the circle is experiencing any non-local acceleration), then R would be constant because and the "weight" (like in 2001) is just the reaction to the centripetal motion on the floor?

Assuming that's true, what I just said, why is the net force towards the centre? For the case of the people aboard Discovery 1, they are clearly undergoing CM but surely their "weight" is equal to the force experienced which is itself equal to the force acting inwards meaning there is no force - they cancel each other out? But if there is no force ... then how are they accelerating all the time?

Yeah as I say, really sorry but I just want to make sure I get rid of this mental-glitch!
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4 years ago
#4
(Original post by DrSebWilkes)
Assuming that's true, what I just said, why is the net force towards the centre? .......... then how are they accelerating all the time?

Yeah as I say, really sorry but I just want to make sure I get rid of this mental-glitch!
Acceleration requires a force n'est ce pas?

Consider the motion of a point (person) on the rotating circumference w.r.t. x and y co-ordinates. Now resolve the velocity components of the circular motion into those perpendicular axes.

Start with the point at top dead centre. At that instant, the motion of the point is a straight line tangent to the circumference, moving in the +x direction, with no motion in the y direction.

Now do the same for the point at the clockwise 90 degree position. At that instant, the motion of the point is still a straight line tangent to the circumference, but now the motion is resolved with no x direction component, but there is now motion on the -ve y direction.

From the starting position, the point decelerated to stop moving in the x direction, and accelerated in the -y direction.

To do that, the resolved average force acting must have acted in the -x direction and also in the -y direction.

Those resolved vectors are the result of the centripetal force always acting towards the centre of rotation.

Go ahead and do the same for resolved forces at any two angular positions. (easier if you do it in the same quadrant.

The acceleration came from the centripetal reaction force of the space station rim acting on the astronauts feet.
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