Centripetal force pointing outward

Question

What's going on physicists.

I would like some explanation here. I haven't been satisfied with the ones I gathered on the internet. Here is it:

If a particle moves in circular motion, there will be a force pushing it towards the center of the circle, which is the centripetal force. However, as most of us know, when you're in those kinds of roller coasters that move in circles, you feel a strong force pushing ou OUTWARDS, not inwards, and it goes up as $\omega^2$. The same is true for when, for example, you put a block of mass $m$ on a round table and start rotating it with an angular frequency $\omega$. The block will slide off and outside of the ball.

So why is it that the centripetal force points inwards in some cases and outwards in others?

Answer 1

What you feel in a rollercoaster as the carriage you are sitting in is pulling G's is that your body is compressed.

Let me elaborate on what I mean here with compressed.
Imagine a stack of, say, 10 weighing scales. The image below gives a schematic view

The weighing scale at the base of the stack is supporting the weight of the stack of weighing scales that is resting on top of it. So in the schematic view the weight indicators of the respective weighing scales are each pointing in a different direction, indicating a different force, ranging from top to bottom.

Of course we know what is compressing the stack: gravity is causing that compression.

In daily life:
In every position of our body we experience that compression.

Note that we are in a very real sense not experiencing gravity directly, there is something indirect to it.

If a rope is tied around your waist, and something is pulling the rope then you feel that tug at your waist.

But gravity acts equally on all the parts of your body; you don't feel gravity tugging at any specific part of your body.

However, when standing up, for example, then each part of your body is carrying the burden of all of the weight above that part of your body. That compressive effect of gravity is what you feel.

That compression of your body is never not there. And because that sensation is never not there you are rarely if ever consciously aware of it. That is how our cognitive systems work: you are aware of things that are different from moment to moment. Everything that is always there is duly internalized. Automated parts of your cognitive systems are dealing with that; no conscious awareness necessary.

So when you are taking a ride in a roller coaster, and the carriage you are sitting in is pulling G's then what you feel is a pattern of compression of your body, and that pattern of compression is identical to the pattern of compression when subjected to a gravitational force. Note that when the carriage is pulling G's the compression occurs from outside-to-inside

When the carriage is pulling G's then your cognitive system automatically interprets the physical sensation as you being subjected to a force in outward direction. You cannot stop that automatic interpretation, it's an internalized function of your cognitive system.

The fact that the two sensations are identical goes back to the observation that gravitational mass and inertial mass are equivalent.

Some additional remarks:

In the case of gravity and the stack of weighing scales:
Take the following case: the stack of weighing scales is set up in an elevator cabin; initially the elevator cabin is not moving. Then start the elevator cabin moving upwards. That change of velocity will then propagate along that stack of weighing scales in upwards direction. That is: the direction of causality is upwards.

(The direction of causality is always upwards: if the elevator cabin reverses from ascending to descending then that change of velocity will propagate along that stack of weighing scales in upwards direction.)

Of course we are correct in our automatic attribution of the direction of gravity; gravity is acting downward, towards the center of the Earth. But the compression that gravity causes is an effect in upwards direction; any change propagates in upwards direction.

With all of that in mind: when you are in a roller coaster carriage, and the carriage is pulling G's, then the causality of the compression is from outside-to-inside. You automatically attribute that compression to a force acting in outward direction.

Answer 2

There is no force pushing you outwards. What you are experiencing is the effect of your own inertia. If you are in the left hand seat of a car turning a fast corner to the right, say, the inertia of your body wants you to go straight on, so as the car turns, and your body doesn't want to, you end up pressing against the car door. It feels like a force pushing you against the car door, but that's an illusion-it's just the effect of the car door pushing against your body.

Answer 3

The issue is the choice of reference frame, and we can demonstrate it with any arbitrary force.

Suppose you are in a car which suddenly accelerates forwards. Clearly, the normal force of the car must point forwards on you to bring you up to speed; yet, as any passenger knows, you feel that you are pushed backwards. This is because your reference frame is accelerating, and hence you must take in account fictitious forces to apply Newton's laws. In the frame of the ground, the normal force $N=ma$ acts unopposed (neglecting friction, etc.) to accelerate you. In your frame, the normal force must oppose a fictitious backwards force $-ma$ to keep you at rest with respect to the car. These two analyses are perfectly consistent.

Now replace the uniform force with a centripetal one. In all cases, an object executing uniform circular motion with angular velocity $\omega$ experiences a net force $m\omega^2 R$ towards the center of the circle. But in the reference frame of that object, it is equivalent to say that a fictitious centrifugal force of strength $m\omega^2 R$ pointing outwards must be opposed by something (in the case of a roller coaster, the normal force from the coaster car) to keep the object moving at uniform speed.

Answer 4

Any and all "forces" that are caused by a coordinate system that is not in free fall (or is "rotating", relative to what, idk$^1$), are fictitious forces.

For instance: consider a mass $m$, at rest on a table. It experiences a fictitious force:

$$ F = mg $$

that we call "gravity" (I'm not putting vector symbols on, but feel free to do so). Is this a real force because Newton, or a fictitious force because Einstein? I'll go with the latter, but the former is fine, if you don't think too deeply about it.

Now, consider the same mass on the same table, and make a coordinate system where it sits at a radius $R$.

Is it experiencing centrifugal force? centripetal force? Coriolis force?

I suspect y'all said "no", "no" and "no". But the correct answer is: well that depends on the coordinate system.

I told you where the origin is, but I didn't tell you if it was rotating (relative to the table). Let's spin it at $\Omega$.

Now we have a problem. The mass is now undergoing uniform circular motion at $-\Omega$, which means it needs a centripetal force:

$$ F_{real} = -m\Omega^2 R $$

to keep it revolving. (I used subscript "real" since we're always told centripetal forces aren't fictitious). Also: the minus sign indicates it must push inward.

Is there a real force causing it to revolve? Maybe friction? No, the table is frictionless, ofc.

In the rotating coordinate, there is also a centrifugal force pushing it outward:

$$ F_c = +m\Omega^2 R $$

While the magnitude matches $F_{real}$, it is in the wrong direction.

Let's add the Coriolis force:

$$ F_C = 2m\Omega v $$

with $v=-\Omega R$:

$$F_C = -2m\Omega^2 R $$

Well that's convenient. The total fictitious forces are:

$$ F_F = F_c + F_C = +m\Omega^2 R -2m\Omega^2 R = -m\Omega^2R = F_{real} $$

exactly what it takes to keep the mass revolving--even though the naive view is that nothing is even moving.

Now to tackle you question: let's turn friction back on, and spin the table at $\omega$

The centripetal force keeping it revolving is now:

$$ F_{real} = -m(\omega-\Omega)^2 R $$

The centrifugal force does not depend on velocity, only distance from the rotation axis, so it is unchanged:

$$ F_c = m\Omega^2 R $$

and the Coriolis force depends on the velocity ($v=(\omega-\Omega)R$, measured in the rotating frame):

$$ F_C = 2m\Omega v = 2m\Omega(\omega-\Omega)R $$

which satisfies:

$$ F_{real} = F_c + F_C $$

...but, if we want to know how much of that centripetal force is actualy "real", that is: attributable to friction, we need to set $\Omega=0$, so that:

$$ F_c = F_C = 0 $$

$$ F_{real} = m\omega^2 R $$

That is, in an inertial frame there are no fictitious forces and a real frictional force is required to keep the mass revolving in a circle.

If we imagine that the mass/fraction models a roller coaster car/track, then we set $\Omega=\omega$ and find that:

$$ F_C = 0 $$

There is no Coriolis force, because the roller coaster has no velocity in a roller coaster frame.

The centrifugal forces is:

$$ F_c = m\Omega^2 R $$

which means you, in the car (you're $m$) feel a gee-force pulling you outward.

When we calculate the "real" centripetal force we find:

$$ F_{real} = -m(\omega-\Omega)^2 R = 0 $$

Which seems problematic. There is no centripetal force because you are stationary--in the rotating frame.

The reason you don't fly off is because there still is a frictional force (or track force, irl, I'm still talking about the model of a mass on a spinning table) that counteracts the centrifugal force.

Now when this stuff is traditionally taught, we always say there is fictitious force pulling your outward and a centripetal force keeping you in circular motion, but as demonstrated: those statements are mixing reference frames, which leads to confusion.

The naming of all the forces depends on the choice of coordinates.

1 We traditionally say we rotate "relative to the fixed stars" (or maybe now, we can say the CMB or JWST's $Z=10$ ultra distance galactic catalogue), but there is a problem. What if my little though problem is set up near the fictional Gargantua, and a non-zero vortex line:

runs through my table? Well, then I have to rotate relative to the fixed stars in order to have zero centrifugal and Coriolis forces.

Answer 5

When you accelerate hard in a car you feel a strong force pushing you backwards into the back of the seat due to forward acceleration or force acting on the car. Similarly when moving in a circle on a fairground ride, the strong force you feel acting outwards on you is an indication you are being accelerated inwards. The force you feel is the opposite direction to the actual acceleration. Similarly if you attach a weight to a spring and accelerate forwards, the spring is stretched and the weight moves in the opposite direction to which you are accelerating. When moving in a circle the weight moves outwards, indicating you are accelerating inwards.