# Rolling without slipping violates $F=ma$?

Suppose you have a ball at rest on a floor with friction. After applying a force $$F$$, the ball rolls without slipping. Therefore, friction from the floor should equal $$-F$$. Then $$F_{net}=\frac{dp}{dt}=F-F=0$$, but obviously there is linear momentum (towards the right) because the ball is rolling forward.

What is the contradiction here? Is my understanding of rolling without slipping correct? Additionally, if there is no friction, will the ball roll?

• I feel your question is similar to this one: physics.stackexchange.com/questions/45653/… Do you agree? Commented Jan 6, 2023 at 20:17
• @Semoi. I don't think so. The box is different than the ball Commented Jan 6, 2023 at 20:21
• Why should the force at the top of the ball be equal to the force at the bottom of the ball, when the top of the ball is observed to accelerate at a greater rate than the bottom of the ball?
– g s
Commented Jan 6, 2023 at 20:22
• In addition to the comment by @gs, two forces applied at different points can generate a torque… Commented Jan 6, 2023 at 20:32
• @gs sorry, will clarify. the ball is at rest initially and then rolls without slipping after the force is applied, so an equal force must be applied to the bottom, correct? Commented Jan 6, 2023 at 20:36

The friction on the floor does not equal $$-F$$. There are three unknowns in this problem: the linear acceleration $$a$$, the angular acceleration $$\alpha$$, and the friction force $$F_f$$. We can produce three equations: Newton's 2nd law, Newton's 2nd law for rotation, and the no-slip condition. Writing these three equations and solving for the three unknowns will give you a $$F_f \ne -F$$.

$$F-F_f = m a$$ $$r_F F + r F_f= I \alpha$$ $$a=r\alpha$$

• Comments are not for extended discussion; this conversation has been moved to chat.
– rob
Commented Jan 7, 2023 at 4:52

Just to add to @Dale 's answer (which is excellent and concise), the third equation is key to the no slipping requirement. The angular acceleration and linear acceleration of the center of mass must match for the given radius for no slipping to occur.

Hope this helps.

• Yes, and with no slip the force of friction does no work. (We are restricting evaluation to a rigid body and using the purely mechanics definition of work as change in kinetic energy, not the more general thermodynamics definition of work. No "heating" effects.) Commented Jan 6, 2023 at 22:07
• @JohnDarby Correct Commented Jan 6, 2023 at 22:08

Assume a ball rolls without slipping in a straight line on a horizontal surface. If the center of mass (CM) of the ball is moving at constant velocity, there is no net force in the horizontal direction and the force of friction is zero. If an external force is applied to the CM of the ball, the force of friction acts such as to impede acceleration of the CM and to increase the rotational velocity of the ball; the net horizontal force (applied minus friction) is positive and the CM of the ball accelerates. For pure rolling (no slip), the force of friction does no work: the decrease in translational kinetic energy of the CM due to friction is compensated exactly by the increase in rotational KE of the ball due to friction with respect to its CM. Once the external force is removed, the ball continues to move at constant velocity and the force of rolling friction is again zero.

If the ball slips when the external force is applied, the force of friction does work since the decrease in translational kinetic energy of the CM due to friction is less in magnitude than the increase in rotational KE of the ball due to friction with respect to its CM.

An automobile on a level surface has no external horizontal force except friction. The internal engine increases the rotational energy of the drive wheels, and the force of friction opposes that increase thus providing a force to accelerate the car forwards.

See Is work done by torque due to friction in pure rolling? and other related discussions of friction on this exchange.

With no friction, the ball will not roll if an external horizontal force is applied to its CM; the ball slides forward on the frictionless surface. Without friction the car using an internal engine will not move forward; the wheels will rotate faster and faster.

Therefore, friction from the floor should equal $$-F$$.

The static friction force would equal $$-F$$ in the case of something like a block on a floor with friction. In that case, where the only potential motion is sliding (or tipping) and no rolling, the static friction force matches and opposes the applied force up until the maximum possible static friction force is reached at which point the block slides. However, in the case of rolling motion the role of static friction is to contribute to, where needed, the necessary net torque to satisfy no slip conditions (discussed below)

If you simultaneously solve the equations given in the answer by @Dale for the no slip condition, you will find the magnitude of $$F_F$$ will always be less than the magnitude of $$F$$ if there is linear acceleration $$a$$, thus not violating Newton's 2nd law. It is also important to keep in mind that the maximum applied force $$F$$ is limited to that corresponding to the maximum possible static friction force of $$F_{F}=\mu_{s}mg$$, which would result in slipping while rolling.

Additionally, if there is no friction, will the ball roll?

I assume you mean roll without slipping. The ball can roll without slipping in the absence of static friction if the no slip condition of $$a=r\alpha$$ can be satisfied with $$F_{F}=0$$ in the figure below.

Then, per Newton's 2nd law for linear motion we where $$a$$ is the linear acceleration of the ball, we have

$$F=ma\tag{1}$$

Per Newton's 2nd law for rotational motion we have

$$\tau=r_{F}F=I_{com}\alpha\tag{2}$$

Where $$\tau$$ = the torque applied by $$F$$, $$I_{com}$$ = the moment of inertia of ball about its center of mass and $$\alpha$$= the angular acceleration of ball.

The no slip condition is given by

$$a=r\alpha\tag{3}$$

Combining equations (1), (2) and (3) we obtain

$$r_{F}=\frac{I}{mr}\tag{4}$$

For a solid sphere

$$I_{com}=\frac{2}{5}mr^2\tag{5}$$

Substituting (5) into (4)

$$r_{F}=\frac{2}{5}r$$

So if the force $$F$$ is applied a distance of exactly $$\frac{2}{5}r$$ from the COM friction is not required for rolling without slipping.

However, I would argue since any deviation in the location of the applied force from the no static friction "sweet spot" would result in slipping while rolling in the absence of static friction, that the no friction condition represents an inherently unstable condition.

Hope this helps.

To get the ball rolling on a horizontal surface by the application of a horizontal force whose line of action passes through the centre of mass of the ball there must be a frictional force but once the ball is rolling with the centre of mass moving with a constant linear velocity and there is no slipping, the frictional force is zero.

If the ball starts from rest then there are two horizontal forces acting on it.
The external force $$F$$ and the frictional force $$f$$.

The system will try and move towards the no slipping condition, $$v=R\omega$$, and to that end you can think of the frictional force as trying to reduce the linear acceleration by opposing the external force, $$F-f=ma$$, whilst at the same time providing a torque about the centre of mass to increase the angular velocity, $$fR=I_{\rm com}\alpha$$.

If the external force is removed and the no slip condition is satisfied there os no frictional force on the ball and the net horizontal force on the ball is zero as is the torque about the centre of mass of the ball.

• Suppose the applied force acts above the COM and would, by itself, provide a torque greater than $I_{com}\alpha$? It seems to me that would necessitate the static friction force act forward to increase the linear acceleration and decrease the angular acceleration to satisfy the no slip condition. Your thoughts? Commented Jan 9, 2023 at 16:31
• I purposely chose the line of action of the external force to pass through the centre of mass so it does not produce a torque about the centre of mass. This is not true if the force does not act through the centre of mass and then you get situations which are often described via a Yoyo. For example, Yoyo Pulled Along the Ground for the force to act below the CoM and you can formulate a situation with the force acting above the CoM Commented Jan 9, 2023 at 23:18