I've learned that when a disc or any similar geometry rolls the instantaneous radius of curvature which is the point of contact of the rolling object and the surface has an acceleration upward.

My question is why does it? The point does not even have to be on the object how can centripetal forces act on it?

  • $\begingroup$ The acceleration in inward. What is in the upward direction could be angular acceleration, because angular volocity is upward, and so is the angular acceleration. $\endgroup$
    – ytlu
    Jan 27, 2021 at 5:27
  • $\begingroup$ physics.stackexchange.com/q/69345 $\endgroup$ Jan 27, 2021 at 6:07
  • $\begingroup$ If it wasn't clear I'm asking why the instantaneous axis of rotation accelerates not why the points accelerate upwards $\endgroup$ Feb 9, 2021 at 15:50
  • $\begingroup$ @Glowingbluejuicebox As explained in each answer so far, the geometric point in space aligned with IAOR of a rolling object does not accelerate upward. If anything, it can accelerate sideways if the object starts to roll faster or slower. $\endgroup$ Feb 10, 2021 at 14:10

3 Answers 3


The center of rotation (point for planar cases, an axis for spatial cases) is a mathematical construct and does not correspond to any physical particle of the body.

By definition, the rotation axis is the locus in space where the particles under it have velocity parallel to the axis. Or in 2D the locus where the particles under it have zero velocity.

The fact that it is not a physical particle, means that the location of the rotation center isn't subject to any physical laws and can instantaneously jump around from one place to another. Corollary to that is that the particles under the rotation center are subject to physical laws and do exhibit acceleration per the kinematics of the rigid body.

The two concepts are separate, one being particles with mass that move with the body and the other being a geometric location in space where something special happens.

The short answer is that the center of rotation does not belong to any particle of a body, but it is a property of the vector space described by a rotating frame.


A point on the perimeter of a rolling disk traces out a cycloid. Check out the animation at the top of this Wiki article. It certainly seems reasonable that the acceleration of that point, at the moment it touches down, is straight up.

EDIT: I did not previously understand the question, I think. My understanding of it now is: “Why is the acceleration of any point in a rolling disk equal to the centripetal acceleration toward the instantaneous axis of rotation in the lab frame (which is the point of contact with the ground) $plus$ an upward acceleration?”

One way to think about this is to imagine a gear with many sharply pointed teeth rolling along instead of a disk. Every point on the gear rotates around the tooth that’s touching the ground, meaning that the acceleration is toward that point. But during this time, the “average” position of the gear (the center of mass) changes, starting with a small upward component and ending with a small downward one. When the next tooth hits down, this vertical motion abruptly switches back to upward, which constitutes a sudden spike of upward acceleration. As you increase the number of teeth to the point where the gear becomes a disk, these intermittent bursts of acceleration decrease in amplitude and increase in frequency, smoothing out into a constant acceleration, meaning that any given spot on the disk is accelerating both toward the point which is located perpendicular to the velocity vectors of every point on the disk (the point of contact) and also upward.

  • $\begingroup$ Sir but why does the IAOR accelerate? $\endgroup$ Feb 9, 2021 at 15:51
  • $\begingroup$ I think I now understand your question. See my attempt at an intuitive explanation above. $\endgroup$
    – Ben51
    Feb 9, 2021 at 17:43
  • $\begingroup$ I've played around with gears before they move quite smoothly ,why do you say that they move up and down ? $\endgroup$ Feb 10, 2021 at 11:18
  • $\begingroup$ You have actually answered one of my previous unanswered question but I'm still find it hard to understand allow me to think about it $\endgroup$ Feb 10, 2021 at 11:22
  • $\begingroup$ The COM traces out a circular arc around each tooth while it’s in contact with the ground. The path of the COM is a series of circular arcs. $\endgroup$
    – Ben51
    Feb 10, 2021 at 13:20

EDIT: The geometrical point in space associated with the instantaneous axis of rotation does not accelerate, if the linear velocity $v$ of the center of the rolling disc is constant (same as the angular velocity being constant, as the disc has fixed radius). It simply follows the disc at velocity $v$, always tracking the point contacting the ground.

However, the fixed, physical point on the disc passing through the IAOR is accelerating, as the below answer explains.

As you zoom in closer and closer to the part of the circle that touches the ground, it looks flatter and flatter, and more and more like this intuitive picture:

Left side moving up,
|    ^
|    | 
                |     |
                v     |
          and right side moving down.

(All points are accelerating upward.)

The part touching the ground is in the center. It is clear that as a point on the circle progresses from being one represented on the right side of this picture to being one represented on the left, it is mostly just accelerating upward, with purely upward acceleration at the center of this picture.

The point touching the ground is being pushed up by the ground$^1$, and hence the acceleration is straight up. This doesn't mean the entire disc accelerates off of the floor though, because the rest of the disc is being pulled down by gravity. Why does this point accelerate up, then, and instead of being balanced out by the rest of the disc pushing it down?

Well, I imagine that's because it takes time for the compression forces to propagate through the disc; but by that that time, the disc has already rolled on to another point. Hence the cycle continues.

$^1$ In this case, it is useful to think of Einstein's principle of equivalence: we should be able to reproduce the exact same physics if we accelerate a rocket ship, and watch a disc being accelerated upward on the floor. It is no surprise that the acceleration of the point contacting the rocket floor is straight upward, event when the disc is rolling at a constant velocity. (This also clarifies why the disc doesn't randomly accelerate upward any more than the ground pushes it up: it has inertia, and follows Newton's first law.) If we imagine the floor of the rocket is a bit tilted relative to its acceleration, you will reproduce the physics for a ball rolling down a hill.

  • $\begingroup$ Awesome answer but it doesn't explain why the IAOR accelerates $\endgroup$ Feb 9, 2021 at 15:51
  • $\begingroup$ Why shouldn't the point located at the IAOR accelerate? It accelerates and then stops being the IAOR. Otherwise, essentially you'd just be rotating the disc around a wire. $\endgroup$ Feb 9, 2021 at 16:03
  • $\begingroup$ There's a reason why it is called the instantaneous axis of rotation. If we track this point in space over time, the IAOR actually does not accelerate, but moves rightward with the velocity of the center of the disc. $\endgroup$ Feb 9, 2021 at 16:09
  • $\begingroup$ Sir if it was just a geometrical point in space then why do we include it's acceleration In questions like this $\endgroup$ Feb 10, 2021 at 11:25
  • $\begingroup$ @Glowingblurjuicrbox That question was talking about the angular acceleration of a point about the IAOR, and you achieve twice the angle found from looking at the center because the path length subtended by an angle $\theta$ placed on the edge of a circle is $2\theta R$. $\endgroup$ Feb 10, 2021 at 13:08

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