Below, I have written how I think about rolling. I wish to know if my idea as to how it works is correct.

Consider a rigid ball on a plane which has considerable friction. If you push the ball the ball rolls because when you push it the part of the ball in contact with ground doesn't move due to the friction but the upper part of the ball (which is not in contact with the ground) tips over. Because the ball is rigid, the movement of the upper part of the ball gets the part of the ball in contact with the ground moving upwards. Meanwhile, the upper part moves downward. This, we perceive as rolling.

This also leads me into thinking that the ball will roll only on a frictionless plane.

If the way I think about rolling is right then is there a way a ball can be "induced" to roll even on a frictionless plane?

  • $\begingroup$ Wouldn't giving the ball angular momentum("rolling it" or "spinning") before dropping it a frictionless surface work? $\endgroup$ Commented Apr 4, 2016 at 6:12
  • $\begingroup$ Well, would it? And is the explanation for rolling right? $\endgroup$ Commented Apr 4, 2016 at 6:14
  • $\begingroup$ I mean if there is no friction (not even in the air), then there would be nothing to stop the ball from rolling. $\endgroup$ Commented Apr 4, 2016 at 6:21
  • $\begingroup$ So that would mean no force is acting on the ball whatsoever. Consider a point on the surface of the ball. If the ball is rolling then that point is changing the direction of its motion continuously or, in other words, it is accelerating. So there is a force acting on that point. Contradiction? $\endgroup$ Commented Apr 4, 2016 at 6:28
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    $\begingroup$ @QuantumMOCHACCINO It wouldn't work because the ball did not have forward velocity initially and there is no force on the ball that could give it forward velocity (remember friction is missing). $\endgroup$ Commented Apr 4, 2016 at 6:48

1 Answer 1


Your idea of how rolling works seems (and probably is) correct. But that is not enough to give a mathematical or physical description. There is another way to see that.

The motion of the individual particles of a rolling ball is pretty complicated, but that of its centre is very simple, motion with a speed $v$ in linear forward direction. So we think, how do the "companion particles" around it appear to this "central particle"?

So we hop onto a car (that is, in a frame) which moves with the same speed $v$ moving in the direction of the rolling ball. And voila! We see that the centre is stationary, and all other particles are moving in perfect circular motion around it!

This is what rolling motion is, superposition of uniform linear motion and rotational motion about an axis. This can be used to describe any kind of rolling motion, including skidding tyres on a road. So, we can see that to set a ball rolling on a frictionless plane, we have to simply give it both a forward motion (with some speed $v$), and a rotating motion (with some angular speed $\omega$). Note that here, both $v$ and $\omega$ can be independent of one another.

But friction helps us in setting up this rolling motion, because it removes the need of us setting up rotational motion! Suppose we throw a ball perfectly rectilinearly (without rotational motion or spinning), then after some time, the ball starts spinning because of friction! How does it do that? image here Well, if we again sit in that car moving with speed $v$, we see that the ball is at rest, and the floor is rubbing it in the opposite direction, and so it applies friction on the ball in the backward direction. This applies a backward force on the ball acting to reduce its speed, and a forward torque which acts to increase the forward rotation of the ball. Now, till when will this continue to happen? Till the speed of the ball is reduced to a speed $v_1$ and the angular speed increased to a value $\omega$, such that: $$v_1 = R\cdot \omega$$ where $R$ is the radius of the ball. This is the special situation which we call pure rolling. Now from the car (now moving with speed $v_1$), we see that the ground is moving backwards with speed $v_1$ and the particle of the ball touching the ground also moves backward with speed $v_1$. As there is no relative motion between them, so friction stops acting, and the ball continues to roll with this speed $v_1$.

One more thing, if you dropped the ball spinning with angular velocity $\omega$ (and zero linear velocity) on a rough surface, the ball still starts rolling, because this time, friction acts in the forward direction, and thus increases linear speed and reduces angular speed till again, $v=R\cdot\omega_1$.


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