This thought experiment is in a way related to the (in)famous airplane on a treadmill problem.

If you take a ball and place it on a treadmill, will the ball:

  1. Move backwards relative to the ground at the same speed as the treadmill (as if placing any other non-circular object on the treadmill)?
  2. Roll in place without moving relative to the ground (the speed of the treadmill is converted directly into rolling motion of the ball)?
  3. Exhibit some other behavior such as rolling while also moving backwards?

For this problem assume that there is no slippage between the treadmill and the ball (sufficient friction to make full contact at all times), and assume that the ball has mass.

I know the answer is not #1. I am not sure if the answer is #2 or #3. If the answer is #3, what factors affect the movement of the ball? Is it the mass of the ball, the speed or acceleration of the treadmill, and/or other factors?

  • 1
    $\begingroup$ There had to be a net force backwards on the ball from the treadmill surface, assuming a non-zero coefficient of friction began the ball and the treadmill. The ball, since it does not slip, has to begin rotating. So the frictional force is both providing linear acceleration backwards and rotational acceleration, and the work done goes into both linear and rotational kinetic energy. So the ball had to move backwards relative to the ground but not as fast as the treadmill. Just how fast it moves depends on the moment of inertia and mass of the ball. $\endgroup$
    – PhillS
    Mar 30, 2016 at 16:50
  • $\begingroup$ @PhillS that should be an answer, not a comment $\endgroup$
    – David Z
    Mar 30, 2016 at 17:07
  • $\begingroup$ I wanted to do the actual maths before doing a proper answer, but was writing on my tablet while cooking dinner which isn't very good for writing full answers. .. $\endgroup$
    – PhillS
    Mar 30, 2016 at 17:30
  • $\begingroup$ Hello, and welcome to Physics SE! Look around, and take the tour. As it stands, this is very close to a homework-like question, which we tend to frown on (we don't ask you to do our homework, don't ask us to do yours...). What have you done to approach the problem, and what really is at the heart of it? $\endgroup$
    – Jon Custer
    Mar 30, 2016 at 18:34
  • 2
    $\begingroup$ Jon Custer: I can assure you that this is not a homework question as I completed college a long time ago. Retrospectively I can see how it would be mistaken as one given the way I worded it. The reason for including choice #1 was to highlight for others a commonly incorrect intuition when thinking about the airplane on a treadmill problem (that I admittedly originally had myself). $\endgroup$
    – Dave M
    Mar 30, 2016 at 19:09

5 Answers 5


CASE I : No friction
If the surface is friction less then, the ball wont move neither rotate, it will be motion less

CASE II : Friction
We know that $$ \tau= I\alpha $$ $$\tau=\mu mgr $$ and $$ I \alpha K$$ K is radius of gyration

Above statements provide the equations related to ball's rotational motion
Regarding transnational motion $$ F=frictional force $$ $$ F=ma$$ $$ a= f/m = \mu mg/m=\mu g $$

Thus above equations clarify the transnational motion, and the motion will be backwards :)
So I will go with Option 3

This is my first activity on physics.stackexchange.com so I hope I answered your question well :)

  • $\begingroup$ Your answer seems to suggest that the ball has a constant acceleration, independent of the relative velocities of the treadmill and the ball! so if we say the treadmill has velocity $u$ then according to your model it will take $$t=\sqrt{\dfrac{2u}{\mu g}}\ \text{seconds}$$ for the ball to be moving as fast as the treadmill and then it will overtake! I believe the mistake is taking the frictional force to always be at it's theoretical maximum: $f_s = \mu_s N$. The answer to this question is certainly #3 but the actual mechanics will depend on how the treadmill accelerates. $\endgroup$
    – Angus Leck
    Mar 30, 2016 at 23:30
  • $\begingroup$ You made me confused, bdw once pure rolling starts $$f=0$$ so it wont overtake $\endgroup$
    – Jimmy Kudo
    Mar 31, 2016 at 20:58
  • $\begingroup$ If we consider ideal condition I dont think ball will go backwards after slipping stops. in case of real condition it will move back $\endgroup$
    – Jimmy Kudo
    Mar 31, 2016 at 21:08
  • $\begingroup$ The mechanics are more complicated than this. So long as the ball is not rotating at $$\omega = \dfrac{v_\text{tread}}{r}$$ then there is a net force on the ball and it will move backwards. Note though this condition needn't be met for the ball to move without slipping. $\endgroup$
    – Angus Leck
    Mar 31, 2016 at 22:34

The treadmill will apply both a non-zero torque and a net force to the ball. Thus the ball will both move backwards and roll. To know precisely how much of each the ball does we need to know more about the initial setting:

Does the treadmill start from rest and then accelerate? How does it accelerate?

Then using the Principle of Least Action I believe we can model exactly the dynamics of this scenario.


I've thought about the steady state scenario, for a ball rolling without slipping on a horizontal treadmill surface.

The rotational velocity of the ball will be given by $\omega = v/r$, where $v$ is the velocity of the treadmill. However, the motion of the ball's centre of mass due to its rotation will be $u = \omega r = v$, but in the opposite direction. So the ball should be rotating without moving its centre of mass from the external observer's frame of reference.

On an experimental note, there are some videos of a ball on a treadmill available (eg) that show it rolling slowly along with the motion. I think that this is due to slipping of the motion.

  • $\begingroup$ A ball can certainly be made to remain stationary on a treadmill if launched correctly (although it will eventually start to lag for the same reason that a ball rolling along the floor in a vacuum would - tiny compressions of the floor and ball as it rolls will slowly convert the energy to heat and stop the rotation). But you won't get there from a scenario where the ball is initially non-rotating and stationary, and only the treadmill force (plus gravity) acts on it. $\endgroup$
    – PhillS
    Mar 31, 2016 at 10:51
  • $\begingroup$ Maybe you will never get there, but you will asymptotically approach it, which as the famous quote goes is "good enough for all practical purposes" $\endgroup$
    – Gremlin
    Mar 31, 2016 at 10:55
  • $\begingroup$ I think this question is more concerned with the initial behaviour than the asymptotic. $\endgroup$
    – Angus Leck
    Mar 31, 2016 at 22:38

The motor which accelerates the treadmill must deliver a certain amount of power when a ball is not on the treadmill.
When a ball of radius $R$ is put on the treadmill the ball must gain kinetic energy.
If the acceleration of the treadmill $a_T$ stays the same the motor must deliver more power and this will mean that the force the motor exerts on the treadmill and the ball must be greater than it was before.
So the frictional force on the ball $f$ is in the direction of the acceleration of the treadmill.

Assuming that the ball and the treadmill start from rest the no slipping condition requires that at all times the velocity of the part of the ball in contact with the treadmill must be the same as the velocity of the treadmill.

After a time $t $ the no slipping condition gives

$v_B + R \omega_B = a_T t $

where $v_B$ is the velocity of the centre of mass of the ball, $\omega_B$ is the angular velocity of the ball.

Differentiating this equation with respect to time gives a relationship between the accelerations.

$a_B + R \alpha_B = a_T $

where $a_B$ is the linear acceleration of the centre of mass of the ball and $\alpha_B$ is the angular acceleration of the ball.

For the ball's linear centre of mass motion $f = M a_B$
and for its rotational motion about its centre of mass $fR = I_B \alpha_B$
where $M$ is the mass of the ball and $I_B$ is the moment of inertia of the ball about its centre of mass.

Some algebra gives $f = \dfrac{2Ma_T}{7}$ and hence $a_B = \dfrac {2a_T}{7}$.

So relative to the ground the ball moves in the same direction as the treadmill but moves backwards relative to the treadmill.


I think that this is one of those problems that becomes immediately more comprehensible when you find the right reference frame to look at it. The question as posed has a ball on a treadmill. To satisfy the non-slipping criteria, the ball and treadmill must both be at rest at first. The treadmill accelerates to a speed $u$ backwards, so what does the ball do?

Now look at the question in a reference frame moving with the final velocity of the treadmill, namely at speed $u$ relative to the original reference frame. In this frame, we have a ball and treadmill moving forwards with speed $u$. The treadmill then decelerates to come to rest. I think most people will spot quite easily that the ball initially non-rotating and moving forwards with speed $u$, will finish up rotating and moving forwards with a lower speed, and the details of the acceleration don't make any difference. This can be solved by a) conservation of energy and b) the requirement that the ball doesn't slip.

The initial energy of the ball mass $m$ is just its kinetic energy:


The final energy of the ball is its new KE plus its rotational energy


where for a uniform sphere of radius $r$, $I=\frac{2}{5}mr^2$.

The non-slipping requirement means that the surface of the ball in contact with the treadmill must be stationary, from which we get $\omega=\frac{v}{r}$.

Put all that together and you get $\frac{1}{2}I\omega^2=\frac{1}{5}mv^2$ and invoking conservation of energy ($E_i=E_f$) we get:


which can be trivially solved to give $v=u\sqrt{\frac{5}{7}}\approx0.845u$.

Transferring this back to the original frame of reference, it means that the ball that was originally at rest is now moving backwards at $\left(1-\sqrt{\frac{5}{7}}\right)u\approx 0.155u$ and rotating with an angular speed depending on its radius $\omega=\frac{u}{r}\sqrt{\frac{5}{7}}$

As an aside, it doesn't matter if the ball slips during the acceleration or not (or indeed if the ball is dropped onto an already moving treadmill). The only condition is that it is not slipping in the final state. Which makes the ball on a treadmill problem functionally equivalent to ten pin bowling if the ball is not initially rotating (as viewed in a reference frame moving along with the initial velocity of the bowling ball).

To apply this to the plane on a treadmill problem (with the plane providing no thrust), you just have to account for the much larger mass of the plane in the KR term versus the (relatively) much smaller moment of inertia of the wheels in the rotational energy term. The net result is that $u$ is only very slightly lower than $v$ - ignoring the usual suspects of air resistance and friction in the wheel bearings. Obviously in the long term in the real world, you are going to end up in some equilibrium between bearing friction trying to make the plane move along with the treadmill and air resistance trying to keep the plane at rest relative to the air, but that is complicated and depends on drag behaviour, and details of the plane etc.

  • 1
    $\begingroup$ There is no reason to expect that $E_i = E_f$, because work is being done on the ball by the treadmill. $\endgroup$
    – Gremlin
    Mar 31, 2016 at 10:54

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