If some object is kept in a radial groove made on a disk and the disk is rotated with say constant angular velocity.

Now as we know that the object moves outwards ie away from the centre of the disk. I want to know which force is it which throws the object outwards.

As centrifugal force is a pseudo force and hence when the observer is in ground frame it shouldn't be present so which force exactly pulls it out.

  • 3
    $\begingroup$ What direction is the groove? Radial? Circumferential? Other? $\endgroup$ – S. McGrew Aug 28 '18 at 17:19
  • $\begingroup$ A careful viewing of archive.org/details/frames_of_reference# would help clarify our thinking... $\endgroup$ – DJohnM Aug 28 '18 at 19:09
  • 2
    $\begingroup$ Obligatory XKCD: xkcd.com/123 $\endgroup$ – Mark Aug 28 '18 at 21:31
  • $\begingroup$ What direction is the groove you're describing oriented in? Radial or circular? $\endgroup$ – Beanluc Aug 28 '18 at 22:50
  • $\begingroup$ @Aaron Are you sure about radial? That means a straight groove from the centre to the rim, like the spokes on a bike wheel. Circumferential (or circular) would be like the groove on an old-timey record. $\endgroup$ – Oscar Bravo Nov 8 '19 at 13:22

No force throws it outwards.

There is also no force pulling it inwards. Therefore it doesn't follow the circular motion. Such a force would have been friction for example. Instead it just continues straight ahead, away from the disk.

All in all, when an object has a speed, then it keeps going at that speed until a force pulls in it. Like a spaceship drifting forever at constant speed effortless. When the object in the disk has an initial speed, and the disk's angular speed is suddenly set too high for the object to follow, then the object will tend to continue straight ahead with this speed. And that means away from the disk.

The fact that it is stuck in a groove may push on it from the sides. But the fact that it moves outwards at all is due to the above describe tendency to continuestraight ahead with constant speed.

"Centrifugal force" is indeed non-existing.

  • $\begingroup$ I think the OP is wondering what has made (or is making) the object move away from the disk if we are looking at in in an inertial frame. $\endgroup$ – Aaron Stevens Aug 28 '18 at 15:08
  • $\begingroup$ Yes exactly @Aaron Stevens $\endgroup$ – Arnav_05 Aug 28 '18 at 15:57
  • 2
    $\begingroup$ But of course, the direction called radial is changing every instant.... $\endgroup$ – DJohnM Aug 28 '18 at 19:11
  • 1
    $\begingroup$ @AaronStevens. What do you mean with the radial speed? Why is that important? $\endgroup$ – Steeven Aug 28 '18 at 19:18
  • 2
    $\begingroup$ Sure thing @AaronStevens, thanks for the comment. I have nowhere claimed a constant radial speed and I disagree with it being important. So I will leave this as is. But I appreciate the comment. $\endgroup$ – Steeven Aug 28 '18 at 20:21

For the following discussion, let's assume there is negligible friction present in the system.

You are correct in saying that when we are looking at the object from an inertial frame (let's say we are the ones who started spinning the apparatus) then there is no centrifugal force. Correct me if I am wrong, but I think the essence of your question lies at how do we explain in our inertial frame why the object moves outwards when there is no radial force pushing it outwards.

One thing to keep in mind that there is a force acting on the object. This is the force supplied by the groove in order for the object to keep moving along with the groove at the same angular velocity. This force always acts perpendicular to the groove, so this force is always changing directions in our inertial frame.

Now, let's consider some small change in time. The direction of the groove changes, but our object has inertia. The velocity cannot change to be perfectly perpendicular to the new groove position due to inertia (mass) of the object. Therefore the velocity picks up a component that is parallel to the groove. This is the "centrifugal force" in the inertial frame. It is the inertia of the object along with it being constrained to rotate along with the groove There is no force that can make this component go to $0$ in this system, so the object cannot travel in a circle.

So to summarize, the centrifugal force itself is just from the inertia of the object. The object wants to keep moving tangent to the circle, which would mean moving away from the center of rotation. At the end of the day, even though some might dismiss the force acting on the object supplied by the groove, this is actually the only force acting on the object in our inertial frame, so it is this force that must be responsible for the entire motion of the object.


Something that has been bothering me for a while is that this does not explain why the object's speed outward from the center of rotation increases so drastically. After trying to simulate this system from just using the force applied by the groove, I have realized that the tremendous increase in velocity outwards comes from the fact that the force applied to the object by the groove is not constant in magnitude. This is because as the object moves outward, the torque the object applies to the disk increases (the force is the normal force, and the distance from the disk center is increasing, hence $\tau=Fr$ is increasing. If you specifically do the work out the torque must be $\tau=2m\omega r \dot r$).

This means that whatever agent is causing the disk to rotate at a constant angular velocity must be increasing its applied torque so that the disk remains rotating at a constant angular momentum. Therefore, the object experiences a force that is increasing in magnitude while changing direction around the disk. This is where the extra energy comes from that allows the object to drastically increase in speed. In the rotating frame, this force is exactly canceled by the Coriolis force. However in our inertial frame it is this increasing normal force that causes the object to pick up speed as it moves out. In fact, if the disk starts at some initial angular velocity and is left on its own, the angular velocity decays towards $0$ and the object will move away from the center of rotation at a constant velocity due to the lack of the "centrifugal force".


There is indeed a force acting on the object in the groove. That force comes from the rotation of the disk - it pushes the object in a direction perpendicular to the radius the object lies on (the direction is tangential to the "expected" circular path).

From there, the Coriolis effect takes over. The force acts in a linear direction. The fact that that direction is continually changing doesn't erase the fact that it only has one direction at any given instant. That direction is not "around the circle", it's pointed at the outside edge of the disk. The object moves "outward" because any force vector which is perpendicular to the radius which the object is lying on is necessarily tangential to the circle which the object would move in if it was attached to the disk. Tangent = away from the center, and away from that circular track. This is the Coriolis effect - objects given a linear impetus across a rotating surface appear to follow a curving path toward the outside of the rotation.

Coriolis is an effect, not a force, but it only exists where forces exist. The effect is that forces appear to act in unintuitive or surprising directions, though instantaneous momentum is always strictly linear.

This might be easier to imagine if you picture that groove as being in the radial direction (like the rays of a sunburst) rather than the circular direction (like the record). When rotation starts, the groove acts like a lever arm pushing the object in the tangential direction.

If we're talking about circular grooves, and everything is perfectly frictionless, then the rotational force isn't transferred to the body resting in the groove. The disk would turn under it but none of that motion would be transferred to the body because its own inertia would keep it at rest in its original spot. If no force causes it to move at all, then, it won't have any tendency to move toward the outside of the disk.

  • $\begingroup$ Very nice answer. I might need to edit mine accordingly. So does the increase in radial speed come from both the centrifugal force as well as the Coriolis effect in the rotating frame? $\endgroup$ – Aaron Stevens Aug 29 '18 at 2:47
  • $\begingroup$ On second thought, I am not sure if this is true here. The equation for the Coriolis force in the rotating frame is given by $2m\vec v \times \vec \omega$ where $\vec v$ is the velocity of the object in the rotating frame. $\vec v$ is along the rod, so the Coriolis force will not responsible for pushing the object outwards. This is only due to the centrifugal force in the rotating frame. $\endgroup$ – Aaron Stevens Aug 29 '18 at 10:22
  • $\begingroup$ The increase in radial speed comes from the continued application of force from the disk to the body. If the disk doesn't slow down and stop turning, then, the body will continue to be influenced by its motion due to the friction which exists between them and the fact that it's not possible for the body to travel along a path on the disk's surface which is completely inertial. There's going to be more pushing unless the friction disappears or the disk stops turning. $\endgroup$ – Beanluc Aug 30 '18 at 0:11
  • $\begingroup$ Keep in mind that what's pushing the body "outwards" is the fact that the push happens in a direction perpendicular to the radius the body lies on. That direction is necessarily toward the edge of the disk and not toward its center. It's not DIRECTLY toward the edge of the disk, straight out along the radius, but, it's tangential to a circle whose radius equals the current, instantaneous distance of the body from the disk's center. The body moves along that tangent, gets farther away from the center, therefore it moves "outward". $\endgroup$ – Beanluc Aug 30 '18 at 0:14
  • $\begingroup$ That force isn't "Coriolis force" (which is a directional effect, not a distinct force) and it's a bit useless to talk about whether it's a "centripetal force" or a "centrifugal force", it's just a sideways impulse from the disk's rotation which transfers momentum to the body due to friction. That sideways direction is necessarily "outward" at all points in time and trajectory. It can't ever be inward nor can it ever be of a constant radius from the disk's center - UNLESS the body's motion is bounded by something else. Like, perhaps, the circular groove which the OP may have been describing. $\endgroup$ – Beanluc Aug 30 '18 at 0:17

There is no force throwing it outward, as if there was the object would accelerate away from the disc instead of moving around it. The centripetal force is the actual force on the object, and the centripetal force is an inward force, and if the object circles the disk at a constant rate in a perfect circle the centripetal force is perpendicular to the objects tangental velocity. An object moving with constant velocity will tend to maintain that velocity unless acted on by an outside force, and in this case moving with a constant velocity will mean getting further from the center of the disk, so if the centripetal force is not great enough the object will escape and move away from the disk from its own inertia, and not from an outside force.

  • $\begingroup$ But the velocity is not constant here. Also, what are you saying is the weak centripetal force here? $\endgroup$ – Aaron Stevens Aug 28 '18 at 17:21

Everything with velocity wants to go in its straight line, anything that pushes it offline or in a circular motion will feel a resistance or inertia and technically it's not a force but that's where the term came from.


This depends on what reference frame you are looking at.

From the viewpoint of someone sitting on the disk and riding along with it - i.e. like a kid on a merry-go-round (though sadly it seems such have been removed as a result of creeping mullycuddle but that's neither here nor there) - then yes, indeed the increase in distance of the particle from the disk's center is accompanied by a measurable acceleration and thus by Newton's second law reflects the presence of a force.

From the viewpoint of someone sitting on the ground or at least not rotating with the disk, the body moves in a straight line, with constant speed (i.e. ignoring other forces like gravity) after it is released. In this frame, there is no force after the body is released because in straight-line uniform motion there are no forces acting pretty much by definition, instead its distance increases from the center simply because it is moving away. But before it was released, and when it was circling on the disk there was a force - namely that which is accelerating it on its curvilinear path - and then when the ball was released, that force stopped.

In fact, the latter force is present in both scenarios. In the first scenario, the person sitting on the disk, were sie to be holding the object, would have attributed it to having to "try to pull it tight to keep it from 'wanting to fly away'". Since this is an actual exertion by hir own muscle power, both observers must agree it's happening since such a thing cannot be just made to go away by a simple coordinate transform, requiring a complex rearrangement of molecules in muscles, brains, etc. . Thus it must be a real force, and that force is the same one which in the second scenario is being exerted - in the same direction, i.e. toward the center - to keep deflecting its velocity around the circle.

The only difference is that due to the different vantage point, one interprets it as trying to "keep the ball from fleeing" while the other thinks it is exerted to "deflect it from its straight line motion".

Since in only one observation the outward force is noted, while in both the inward force is noted, we like to say the inward force is the "real" force since it represents an actual interaction, while the other represents the artifact of a coordinate transform.

But nonetheless, both are equally valid ways of looking at the same scenario, and both can be reconciled by noting that for the person sitting on the disk hir situation of forces must be different because to hir sie is stationary, while in the case of the external observer sie is moving. Nonetheless, the same motion, same exertions and other invariants must be preserved between both. I suppose there is some philosophic issues involved here, as there always will be when you dig into it, but this is basically the reasoning behind one common philosophic view and at least behind the apparently differing observations and their relation which you will still have to accommodate no matter your philosophy of "reality" (i.e. ontology).

And yes, for the question, this outward force seen by the observer standing on the disk is centrifugal force. Whether it's a "real force" or not is not relevant to that - that is the name which refers to that particular thing.


Rotational motion is just the movement of a points which have velocity they works bonds (centripetal) force. Because you have chosen the disk will be very easy. Disk is flat object with a lot symmetrical axes and it does not matter for calculations which symetri axis we choose. I have simplified the rigid body to two-dimensional model on the surface XY with five masses, two symmetrical on each axis and one in the center of the mass. All masses are equal $m_1=m_2=m_3=m_4=m_5$. The angular velocity is perpendicular to this body $\vec ω (0, 0, z)$ . enter image description here Because it is freely rotating rigid body, every point (mas) velocity must be compatible relationship $$ \vec v = \vec ω \times \vec r $$ where $\vec r$ is a position vector relative to the axis of rotation (point A). The points are affected by centripetal accelerations consistent with the formula $$ \vec a_c = \vec ω \times \vec v $$ Now we can calculate bonds force which work on points $$ \vec F_b = m \vec a_c $$ For each point work bonds force along the arm, directed to the axis of rotation. It gives us the opposite force of reaction working on point located on the axis of rotation. Resultant force acting on the axis of rotation is the sum of bond forces acting on points and opposite directed. $$ \vec F_r = -\sum_i \vec F_{bi} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.