I am confused about this whole subject. Let's suppose we have a disk standing still upon the ground. Now let's say someone inside the disk shoots a projectile to the left. From momentum conservation on the horizontal axis, the disk's center of mass will now have a velocity to the right. Suppose the projectile wasn't shot from the center of mass, but below it. That means the disk will also rotate clockwise. If there is static friction between the ground and the disk, it should exert a torque, thus changing the angular velocity of the disk overtime. Suppose also that the disk is rolling without slipping after the shot.

  1. My first question is, why doesn't the static friction exert a torque on the disk when it is rolling without slipping? and how can I see why that happens only when there is rolling without slipping? Is it because rolling without slipping means zero velocity at the point of contact, meaning no force is exerted?

  2. My second question is about the impulse exerted by the friction in this instance. I have seen that the change in momentum of the system I've described after the shot is equal to the impulse of the frictional force. Why does the frictional force exert an impulse at the moment of the shot of the projectile?

  • 1
    $\begingroup$ If a wheel is rolling without slipping on a flat surface - even a rough, flat surface - then there is no static friction acting. So no resultant force, and no torque either. In real life, we have rolling resistance which does provide a torque (search for a diagram). Thirdly, for a wheel rolling + accelerating down an incline there is now a force of static friction pointing up the slope, and for a wheel powered by a motor there is instead now a force of static friction pointing in the direction of acceleration! This makes rolling a very confusing topic! $\endgroup$
    – 13509
    Feb 22, 2020 at 16:22
  • $\begingroup$ it might help to have a sketch or edit - I am not sure from your description if the disk is vertical - like a car wheel so it can roll or lying flat on the ground. $\endgroup$
    – tom
    Feb 22, 2020 at 16:35

2 Answers 2


Here is my answer to both your questions:

  1. Neglecting the rolling resistance mentioned by @James Wirth and other deviations from the ideal case, the answer is yes; your presumption about it being because the relative velocity being zero at the contact is true: no force appears because both the linear and angular momenta of the wheel are being conserved by the uniform motion after the initial impulse.

  2. This is because in the absence of this initial impulse, the wheel would both spin and translate after the shot, leading to nonzero relative velocity at the point of contact with respect to the ground. A nonzero impulse appears at the point of contact precisely to force the motion to follow a different course, rotating about this point.

  • $\begingroup$ Is it right to say then that static friction doesn't do anything to the disk, considering it is a rolling without slipping situation? It IS there, but it doesn't exert any force? I don't quite get it. $\endgroup$
    – Darkenin
    Feb 22, 2020 at 18:48
  • $\begingroup$ No, it is there, but only during the acceleration phase. If the acceleration happens instantaneously, then the friction force must be infinite to supply the required impulse. That is obviously impossible, but so is the instantaneous acceleration. In any case, once the wheel has reached its final velocity, friction is not there anymore. If it were, it would, as you point out, decelerate the wheel. $\endgroup$ Feb 22, 2020 at 21:22

Consider first a slippy surface on which a wheel is rolling. Because the surface is frictionless the wheel just rotates in place, there is no rolling.

Now assume a rough surface, so there is friction and so the wheel rolls forward and it never slips.

Here, the translational velocity is the velocity of the axle, as the axle is above the point of contact then this point also moves at this velocity.

However, the resultant velocity at the point of contact is zero since the wheel is rotating with a rim speed at exactly this speed in the opposite direction.

Why is the rim moving at this speed? Well when the wheel rotates one whole revolution it covers a distance equal to the circumference.


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