# How to determine the direction of the tendency for a rolling object to slip?

I know that the direction of static friction in a rolling motion has been asked about before, but I am specifically asking how to determine the direction of the tendency to slide. In this picture, the particle P has a tendency to slip to the left as a wheel rotates in a clockwise manner:

On a slope, the direction of tendency to slip is down the slope, so in this picture, the rolling object's tendency to slip is downwards:

What confuses me is this: what indicates the direction of the tendency to slip? In the first picture the tendency to slip is to the left because it's angularly accelerating clockwise but as soon as that rolling object is placed on a slope such as in the second picture it seems like that no longer applies and now the direction of tendency only has to do with it being on an incline. Does the direction of the tendency to slip depend only on the direction of the acceleration of the COM of a rolling object? Sorry if this is a bad question! Trying to understand.

Thanks to the OP for posting this great question. The figures are of great help in explaining this interesting aspect of friction related phenomena. In what follows, this answer provides a detailed analysis of the rolling without slipping phenomenon which underlies the physics addressed in the OP. $$\underline{\textit{Qualitative analysis}}$$

It is easily observed that the contact force referred to as kinetic friction is applied by one surface on another in the direction opposite to that of the relative velocity of the latter surface.

Let us conduct the following static friction related thought experiments. These empirical thought-proofs show that this type of contact force is indeed, applied in the opposite direction of 'impending' relative velocity. Let $$f$$, $$\mu_s$$, $$W$$, $$f\leq F$$ and $$\tau$$ denote the relevant component of force of friction, relevant component of static friction coefficient, weight of the object and the externally applied force and torque on the body.

$$\textit{Friction block thought experiment:}$$ In the figure below, we know that in the static case, $$f=F$$ and that if $$F=f\leq\mu_sW$$, then the block has vanishing acceleration. Clearly, the 'impending' relative velocity of the block is in the direction of $$F$$ towards the right of the figure and the force of static friction is acted by the ground surface on the block surface in the opposite direction of this relative velocity.

$$\textit{Rolling without slipping thought experiments:}$$ In the case of the circular body rolling without slipping in the figures below, we know that the translational velocity (in the direction of $$F$$ or towards the right of the figure), $$v$$, of the center of the circle is given as $$v=-\omega R$$ due to the assumption of rolling without slipping, where $$\omega$$ is the non-vanishing $$Y$$ component of the rotational velocity (with the other components necessarily being vanishing due to the assumption of planar motion) and $$R$$ is the radius of the circle. Therefore, assuming that the center of the circle is also the center of mass (COM) of the body, we obtain the translational acceleration of the COM as $$a = -\alpha R$$, where $$\alpha$$ is the non-vanishing $$Z$$ component of the rotational acceleration (with the other components necessarily being vanishing due to the assumption of planar motion) of the body. Further, the analysis of the angular momentum implies that $$\tau_\text{ext}=I\alpha$$ where $$\tau_\text{ext}$$ is the externally applied torque on the body and $$I$$ is the moment of inertia about the axis passing through the center of the circle. Further, the rolling without slipping phenomenon implies that the relative velocity of the contact point of the circular body with respect to (w.r.t.) that of the ground surface is vanishing. In both situations depicted in the figure below, this assumption implies that $$f=\mu_s W$$.

In both thought experiments shown below, the rotational velocity and acceleration of the body are measured positive in the direction of a right hand screw being screwed out of the screen. The coordinate system used is $$XYZ$$ with the $$X$$ axis pointing to the right of the screen parallel to the ground surface and the $$Z$$ axis pointing vertically downwards. In both cases below, the COM of the body will (using our physical intuition in the thought experiment) accelerate towards the right of the page, that is, in the direction $$+X$$.

• Force driven wheel (figure on the left): The Newton's laws of motion imply that $$F-f=\frac{W}{g}a=-\frac{W}{g}\alpha R$$ and $$-fR=I\alpha$$ which implies that $$0\leq a$$, $$\alpha\leq 0$$. We observe that if the direction of the static friction force is reversed, we would obtain a contradiction since the rolling without slipping condition would be violated (because the direction of acceleration obtained would be opposite to that required in the known relationship $$\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in the direction of the applied force $$F$$ which points in the $$+X$$ direction, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $$m:=\frac{W}{g}$$, then $$I=m\frac{R^2}{2}$$, so that the equations of motion yield $$F=f$$. The static friction condition $$f\leq \mu_s W$$ therefore implies that $$F=3f\leq \mu_s W=3\mu_s mg$$, which provides the upper bound on the driving force which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $$a=\frac{2f}{m}$$ allowable under the rolling without slipping regime. In fact, this is the underlying reason why circular wheels are more efficient than non-circular ones.
• Torque driven wheel (figure on the right): The Newton's laws of motion imply that $$-\tau+fR=I\alpha$$ and $$f=\frac{W}{g}a=-\frac{W}{g}\alpha R$$, which implies that $$0\leq a$$, $$\alpha\leq 0$$. Clearly, assuming that the direction of the friction of force is opposite to that shown in the figure will lead to a contradiction violating the rolling without slipping condition (because the direction of acceleration obtained would be opposite to that required in the known relationship $$\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in opposite to the direction of the applied force $$F$$ in the figure on the left, that is in the direction $$-X$$, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $$m:=\frac{W}{g}$$, then $$I=m\frac{R^2}{2}$$, so that the equations of motion yield $$\tau=-\frac{3}{2}{f}{R}$$. The static friction condition $$f\leq \mu_s W$$ therefore implies that $$\tau\leq \frac{3}{2}\mu_s WR=\frac{3}{2}\mu_s mgR$$, which provides the upper bound on the driving torque which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $$a=\frac{f}{m}$$ allowable under the rolling without slipping regime.

$$\underline{\textit{Conclusions}}$$

1. The contact force referred to as kinetic friction is applied by one surface on another in the direction opposite to that of the relative velocity of the latter surface w.r.t. the former surface.
2. The contact force referred to as static friction is applied by one surface on another in the direction opposite to that of the 'impending' relative velocity of the latter surface. The direction of the impending velocity, which is a fictitious quantity, is in the direction of the relative acceleration (w.r.t. the surface applying the force) of the point of contact resulting from the dynamics in which the friction force of interest is fictitiously assumed to be vanished w.r.t. the former surface.
• Thanks for this really in depth answer! I now understand the differences in the direction of the tendency to slide considering a force driven wheel and a torque driven wheel. However, there is 1 thing I am still confused about: In the case of the second diagram with a wheel rolling down a slope, in my eyes that is both force driven (the gravitational force causing the wheel to move down the ramp) and torque driven (with the existence of fs). If the direction of your “impending” velocity is opposite in these two conditions yet both apply, which one determines where your “velocity” is directed? Commented Nov 16, 2021 at 23:59
• @CarolynSun consider a cylindrical wheel with planar symmetry about the center of mass (COM) on an incline. The total torque on the wheel referred to the COM, due to the gravitational force, is vanishing, because of the symmetry so that such a wheel is a force driven wheel, using the terminology of the answer. However, if a motor drives the wheel down slope (say, an electric bicycle), the direction of impending motion may be calculated by obtaining the resultant relative acceleration of the point of contact, caused by the force and torque acted on the wheel., w.r.t. the surface of contact. Commented Nov 17, 2021 at 0:33
• When both a translational force and torque act on the wheel and would theoretically induce acceleration in opposing directions how would you then obtain the resultant relative acceleration? I'm also a bit confused on how the torque on the wheel vanishes. The COM of the wheel would not rotate, but I don't see how it doesn't mean that the wheel doesn't experience a torque. If we refer back to the second diagram, the rolling object experiences a torque from friction, yet behaves in a "force-driven" matter. Commented Nov 17, 2021 at 6:34
• @CarolynSun although forces and torques applied to the wheel may induce accelerations in opposite directions, the resultant acceleration is calculated on incorporating all dynamic causes, that is, forces and torques. It is the direction of this resultant acceleration which is indicates the direction and type of friction force. In the case of the second figure, the symbol $\alpha$ is the acceleration caused by the resultant dynamic effects of the force and torque. Commented Nov 17, 2021 at 21:35
• I see! That makes a lot of sense. One more thing. How would you go about reconciling the forces and torques to find your resulting acceleration? Commented Nov 18, 2021 at 6:14

the particle P has a tendency to slip to the left as a wheel rotates in a clockwise manner.

That is true only if you assume information that is not given in the diagram (like something producing a torque on the axle). It could be that the wheel is just rolling along. In which case, there is no tendency to slip.

On a slope, the direction of tendency to slip is down the slope

While generally true, I would prefer to say that wheel is accelerated downward (leftward) due to gravity. That acceleration to the left will tend to force the slip.

what indicates the direction of the tendency to slip?

Slippage happens when you have translational acceleration that does not match rotational acceleration.

If you have a ball on a ramp, gravity is causing only translational acceleration. If you have a car at a green light, the engine is causing only rotational acceleration.

In each case, if there is sufficient friction, then the rotational and translational accelerations are joined together. But if you do not have enough friction, then the accelerations are uncoupled and slip occurs.