I'm a graduate student in mechatronics, but I give myself headaches on this astonishingly simple 2D problem.

Imagine a ball (perfectly rigid) on an horizontal plane (perfectly rigid), initially at rest. The only forces acting on it are its weight $P$ and the normal reaction of the plane.

Now, at time $t = 0$, I apply an additional horizontal force to the ball $F$, big enough ($F > \mu \times P$) in its centre of mass. An opposite tangential friction force appears ($T = \mu \times P$). And the problem begins! Since the horizontal forces aren't equal, the ball will accelerate linearly. Moreover, the tangential force create a torque, and the ball will also rotate.

But because we are still at time $t = 0_+$, both the linear and the angular velocity are nil. So the cinematic rolling-without-slipping condition $v = R \times \omega$ is fulfilled, so in fact, there is no friction and the ball is only slipping… That's pretty paradoxical.

It's easy to know if the ball will move or stay immobile. But how can I determine if the ball will slipping and rolling, or just rolling without slipping?

  • $\begingroup$ In your argument: " Since the ball is at rest, there is no friction ..." is wrong. $\endgroup$
    – velut luna
    Jun 15, 2014 at 14:21
  • $\begingroup$ @user139981 You're right. Edited. $\endgroup$
    – Blackhole
    Jun 15, 2014 at 14:32
  • $\begingroup$ I don't think rolling without slipping always implies zero friction. $\endgroup$
    – velut luna
    Jun 15, 2014 at 14:48

1 Answer 1


I'd disagree with the statement that there is no friction when the ball is at rest, or rolling without slipping. You have to distinguish between static friction and kinetic friction. Static friction is what keeps the ball rolling without slipping when it is doing so. When the ball is slipping, kinetic friction comes into play, and it can exert a torque on the ball that will cause it to spin. Also note the coefficient of static friction is generally greater than the coefficient of kinetic friction for most materials.

An analysis of the forces and torques on the ball using its moment of inertia (assuming it's of uniform density) ,will reveal that if it is hit at a height

$h= {\dfrac{7r}{5}}$

where $r$ is the radius of the ball, it will always roll without slipping, no matter how hard it is hit.

Hitting it at any other height, including at the center of mass, the magnitude of the force and the coefficient of friction determines whether it will slip initially or not.

In your case, hitting it at the center of mass, since the only torque on the ball is due to the static friction at the point of contact, the ball will slip if

$F_{hit} > \mu_{static}mg$

There's a pretty good analysis of this problem and the conditions under which the ball will roll without slipping, roll with slipping, or spin, at this site :

Physics of Billiards


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