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I don't understand how the centripetal force, which always points to the center of our circular motion can cause this scenario:

We have a big stone which spins very fast, so fast that a part breaks down, because of the centrifugal force (this is at least how my text books describes it).

My problem: the centrifugal force does not really exist (we only use it in accelerated frames of reference, so the newton laws still work there), so if we're in a laboratory frame of reference, which force would "pull" that piece of the stone to the outside of the circle, when we only have the centripetal force (as mentioned pointing to the center of the circular motion...)?

(Please don't try to explain it in an accelerated frame of reference, because there I understand it, but I don't understand it in a laboratory frame of reference)

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    $\begingroup$ Perhaps it would be useful for someone to expand on @valerio92 comment. In general relativity the effect handled by 'the force of gravity' in Newtonian physics is an inertial pseudo-force and has exactly the same status as centrifugal force. This difference in point of view is a stumbling block for many people on first encountering GR. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jul 26, 2017 at 18:56
  • $\begingroup$ I've deleted some comments that didn't seem to serve the purpose of improving the question. $\endgroup$
    – David Z
    Commented Jul 26, 2017 at 19:36

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In the lab frame of reference, you need to reverse the question - don't ask yourself what pulls the particles apart but what keeps them together.

By Newton's laws, everything on which no force acts keeps travelling in a straight line. So what requires explanation is not that a collection of moving particles - such as a rotating flywheel - flies apart but what keeps them together. The force that keeps them together is a centripetal force, in this case exerted by the bonds that keep the material together. When you reach a velocity where this force is not enough anymore to keep the particles on a circular trajectory/bound orbit, they fly apart.

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  • $\begingroup$ but how do I express this mathematically? I can write down the case where we have an equilibrium: the sum of forces is equal to the gravity (or the centripetal force is equal to gravity). This implies that if the sum of forces is bigger then gravity, then the piece is going to fly away. Don't we get those other forces from the circular motion, implying that ther is an acceleration pointing to the inside, implying that we have a force point to the inside? Still, I don't really see what is making the stone fly away if evey force is pointing to the insid of the circel. $\endgroup$
    – Yalom
    Commented Jul 25, 2017 at 10:08
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    $\begingroup$ @Yalom Every piece of the object just wants to fly straight. You have to supply a centripetal force to cause it to turn. If you are trying to follow a specific path (like a circle), you need to apply the right force. Say for example you have a spinning mass on a string, the centripetal force is the tension in the string. This string isn't a very strong string. If at some point you spin it fast enough, the tension required for circular motion will be less than the strength of the string. If the string snaps, the object would continue to move forward. $\endgroup$
    – JMac
    Commented Jul 25, 2017 at 10:15
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    $\begingroup$ @Yalom Maybe you didn't understand my first sentence: There is nothing "making the stone fly away", instead there is nothing sufficient making it not fly away. E.g. if you solve the Kepler problem (things orbiting around a central body under the force of gravity), you find two types of solutions: Orbits and "escape" trajectories where the object just gets farther from the body (e.g. the Earth, sun, whatever). The only difference between these solutions are the initial conditions. There's nothing particular "making" it orbit in one case and not the other. $\endgroup$
    – ACuriousMind ♦
    Commented Jul 25, 2017 at 10:22
  • $\begingroup$ @Yalom You don't have equilibrium in that case. The stone is traveling in a circle which means that it is accelerating, so you should have $\sum_i \vec{F}_i = m \vec{a}_c$ not $\sum_i \vec{F}_i = 0$. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jul 25, 2017 at 15:25
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    $\begingroup$ @Yakk I don't think so, since neither the question nor my answer rely on the particular nature of that force. That the force in the flywheel example is atomic and molecular bonds and not gravity or macroscopic electric attraction is completely irrelevant - I could as well have taken a planet orbiting a star as the example. $\endgroup$
    – ACuriousMind ♦
    Commented Jul 25, 2017 at 17:39
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Try to imagine, instead of a big stone, a big plate. On top, you fill the plate with sand. Now you start spinning the plate.

What's going to happen to the sand? The sand is going to leave the plate very fast, and spill in all directions. This is the basic, natural state of things, and it's from here that you should start questioning.

How do we keep the sand from leaving the plate? The answer is the centripetal force. If you tie each sand grain to the center of the plate with strings, when rotating each string will pull on its grain of sand, and prevent it from leaving the plate.

Using a big rock is the same thing; instead of using strings you're just relying on the intrinsic cohesivity of the rock with itself. Here each "grain of sand", or piece of the big rock, is attached to the pieces adjacent to it. They in turn transfer all forces to the ones adjacent to themselves, and this is how the centripetal force is propagated from the outside to the center of the big rock.

Once any piece of the rock fails to sustain this force, the bonds break. Once again you have sand, and as we said above, sand is going to shoot in all directions.

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    $\begingroup$ "The sand is going to leave the plate very fast, and spill in all directions" -- or to be more specific, the sand will rotate with the plate for as long as friction between the plate and the sand is able to exert sufficient force to cause the sand to move in a circular path. Once static friction would have to be exceeded to make that happen, we know that's impossible, so the sand is going to move with respect to the plate. All that's left to decide is which direction it moves, and the answer is "all directions" because a rotating plate of sand has grains going in any direction you choose. $\endgroup$
    – Steve Jessop
    Commented Jul 26, 2017 at 15:54
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    $\begingroup$ The case of frictionless plate and sand is uninteresting, because then if you spin the plate it just rotates under the sand, which sits perfectly still in the lab frame of reference :-) $\endgroup$
    – Steve Jessop
    Commented Jul 26, 2017 at 15:57
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You don't have to explain this by centrifugal force, or any fictitious force at all. All what centrifugal force is about is inertia.

As your stone is spinning, it has some velocity. But since initially there's a centripetal force, this velocity constantly changes towards the center of rotation. When part of the stone breaks off, it's no longer held by centripetal force, so it just flies due to inertia with constant velocity.

It's just when you go to the frame of reference associated with the stone, only there you get centrifugal force — as a device to make Newton's laws look unchanged.

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You seem to ask two different questions:

  1. Why does the stone break apart? For the smaller parts of the stone to travel along circular paths, a centripetal force must be exerted on each of them. These forces are exerted through the forces holding the parts together. When these forces become too large for the particular material, the stone breaks into smaller pieces.
  2. What makes the parts fly outwards after the stone has broken apart? Since there is no longer any centripetal force making the parts travel along circular paths, they simply continue traveling straight in the direction they were traveling before the stone broke apart. This means that they will fly outwards, although not straight radially.
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The centrifugal effect is not a force, true, but it does indeed exist.

The effect happens, not because any force pulls the piece outwards in the circle, but rather because of the already existing centripetal force that pulls all the rest of the material inwards in the circle. The piece wants to continue its straight line motion, as everything does when not experiencing forces, which brings it away from the circle.

If you sit in a car and feel squeezed to the side as the car turns, it is not you being squeezed into the side, but rather the car squeezing into you; turning and trying to take you with it.

This tendency of wanting to continue in a straight line while being pulled around in a circle, is what we can call the centrifugal effect (I deliberately don't call if "force" to avoid this confusion).

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It might help to note that using polar coordinates, at the instant a part breaks away, the part only has a tangential velocity and zero radial velocity. The break away part is not moving "outwards", but instead "forwards" (absent gravity and drag, there would be zero net force on the part). After some amount of time, that forwards velocity does result in the part moving away from the center of the spinning stone.

A better example of centrifugal force in a non-rotating frame is the reaction force in response to a centripetal force, a Newton third law pair of forces, but each part of the pair exerts a force on the other object. In the case of a string exerting a centripetal force on a stone, the stone exerts a centrifugal reaction force onto the string. Wiki has an article about this:

https://en.wikipedia.org/wiki/Reactive_centrifugal_force

There is an example where the only forces are centripetal, a "two body" system where two objects are orbiting each other about a common center. Each object experiences a centripetal force towards the common center (which is also towards the "other" object), but since the force is gravity, there are no reactive forces. In the case of gravity, the Newton third law pair is the gravitational force each object experiences.

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