# Is accelerated rolling with slipping possible?

I have a conceptual question regarding the following tasks. Two cylinders with different coefficients of friction are rolling down a inclined plane, thus accelerating. According to the task, due to the different coefficients of friction, they will eventually collide. I understand, that rolling with slipping is possible if the center of mass moves with a constant velocity, however I do not see how it is possible how the cylinders can accelerate with rolling and slipping. In my opinion the rolling condition can either be fulfilled or not. If the contact point with the ground moves, there cannot be any torque that increases the angular velocity and it would just be slipping without rolling.

However under these assumptions the question would not make sense. Please let me know if I understand things wrong and if accelerated rolling with kinetic friction (slipping) is possible.

Accelerated rolling with slipping is indeed possible. Consider a car traveling down a hill in snowy conditions. It is possible for the car to lock the brakes and slide down the hill. Depending on the steepness of the hill and the friction of the snow, it is possible that the car accelerates down the hill.

• But isn't it the case with your example, that for a short time the car slips and then again goes to a rolling motion (back and forth)? How would you quantify this motion. Of course the accelerating force is mg*sin(φ) if φ is the angle of the plane, but what would be the effective acceleration of the center of mass if we take into account both rolling and slipping? Of course the rolling condition omega = v * r is not fulfilled. Commented Dec 3, 2022 at 15:54
• I meant v = omega * r, sorry Commented Dec 3, 2022 at 16:11
• @EliasK. Yes, it could be accelerated rolling with slipping for only a finite time. Nevertheless, you would need to model that situation because it can indeed happen, even if only briefly.
– Dale
Commented Dec 3, 2022 at 17:09

I will try and give you an answer without providing a solution to the problem!

The no slipping condition is that $$v_{\rm com} = r \omega$$ which when differentiated with respect to time gives $$a_{\rm com} = r \alpha$$ where $$a_{\rm com}$$ is the linear acceleration of the centre of mass and $$\alpha$$ is the angular acceleration.
So it is a "balance" between two different accelerations.

What determines the linear acceleration of the centre of mass, $$a$$?
$$ma = W \sin \theta-F$$
$$mg\sin \theta$$ is the component of the weight down the slope, $$F$$ the frictional force up the slope, $$m$$, the mass of the cylinder, and $$\theta$$ is the angle of the incline.

What determines the angular acceleration, $$\alpha$$?
$$\tau = F\,r=I_{\rm com} \alpha$$
$$\tau$$ is the torque on the cylinder, $$r$$ the radius of the cylinder and $$I_{\rm com}$$ the moment of inertia of the cylinder about its centre of mass.

There is a limitation on the magnitude of the frictional force $$F\le\mu N$$ where $$N$$ is the normal force, $$N=m\,g\cos \theta$$, ie $$F\le\mu m\,g\cos \theta$$.

This means that the maximum value of $$\alpha$$ is $$\dfrac{\mu\,N\,r}{I_{\rm com}} = \dfrac{\mu\,m\,g\cos \theta\,r}{I_{\rm com}}$$

Consider the cylinder rolling down the slope without slipping.
Now consider what happens if the incline angle $$\theta$$ is increased.
The normal force decreases and so the maximum value of the angular acceleration,$$\alpha$$, decreases.
There will come a time when $$a> r\,\alpha$$ because of the limiting value of $$\alpha$$ and the cylinder will start to slip, ie its increase in angular velocity is not sufficient to keep up with the increase in linear velocity.

Without giving too much away, you can show that for a cylinder to roll down an incline without slipping the coefficient of static friction $$\mu \ge \frac 13 \tan \theta$$.