# Rolling without slipping taking the contact point as pivot

I'm confused about this "rolling without slipping" kind of situation. Or better in this case the object is rolling and slipping, just use the label "rolling without slipping" to identify the kind of problem.

Suppose to have a disk with initial velocity $v$ and angular velocity $\omega$. The motion is to the right but the angular velocity is counterclockwise. There are no forces acting on the disk besides the kinetic friction $\mathbf{f}$.

Things are ok if I take as pivot point the center of mass. $$\{\begin{matrix} - \mathbf{f} = m\mathbf{a_{CM}}\\ - \mathbf{r} \times \mathbf{f} =I_{cm} \mathbf{\alpha} \end{matrix}\tag{1}$$

But if I take the point $O$ on the ground, then the kinetic friction has zero torque.

$$\{\begin{matrix} - \mathbf{f} = m\mathbf{a_{CM}}\\ 0 =I_{O} \mathbf{\alpha} \end{matrix}\tag{2}$$

I assumed that the angular velocity (and so $\alpha$) is the same it I take as pivot the center of mass or the point $O$.

If this is the case than parallel axis theorem can be used and $$I_O=I_{cm}+m \mathbf{r}^2$$

But there is a contradiction since I get $\alpha=0$ from $(2)$ and $\alpha\neq0$ from$(1)$.

How can that be? Maybe $\alpha$ is not the same in the two cases?

• Think of the friction force as a function of slip velocity. When slipping is zero friction is zero, otherwise friction acts in a way to minimize slipping. – John Alexiou Apr 21 '16 at 14:45
• In $(1a)$ and $(2a)$ it's $\mathbf f$, not $-\mathbf f$. A vector already "contains" its orientation. In $(1b)$ you have to specify which $\mathbf r$ this is. If this is $\mathbf{CO}$ (with $C$ the c.m.), then again no minus. – L. Levrel Apr 21 '16 at 19:54

The equation of motion

$$\text{torque about stationary geometrical point O} = \text{moment of inertia w.r.t. O} \times \text{angular acceleration w.r.t. O}$$

is valid only if the motion of the body is planar rotation around an axis that passes through O. This is the case if the point O is taken to be point of contact of the body when rolling without slipping, but not when rolling with slipping. Generally valid version of torque-angular momentum theorem is

$$\text{torque about stationary geometrical point O} = \ = \frac{d}{dt}\left(\text{angular momentum w.r.t. stationary geometrical point O}\right).$$

If the body is rolling with slipping, there is no stationary geometrical point O on the ground for which the angular momentum could be written as $I_O\omega_O$ with $I_O$ constant in time and the latter equation doesn't reduce to the former one.

As an unbalanced force, $\mathbf{f}$ acts to accelerate the disk. Since it is located at the bottom of the disk, O must accelerate as well and is therefore in a non-inertial frame of reference.

That non-inertial frame will have a fictitious forces appear that oppose acceleration. We can draw a force $\mathbf{f'}$ that acts through the center of mass in the opposite direction of $\mathbf{f}$.

Because it acts through the center of mass, it provides a torque relative to O, and is able to reduce the angular velocity.

• Nice way of curing the ill-definedness of the problem! – L. Levrel Apr 21 '16 at 19:57

When doing this sort of problem you can add two forces acting at the centre of mass whose resultant is zero.
This system of three forces can now be viewed in the following way. The frictional force $f$ is exactly equivalent to a force of the same magnitude whose line of action passes through the centre of mass of the disc (shown in blue) and a pair forces $f$ shown in red which constitute a couple.
The blue forces produces the linear acceleration of the centre of mass of the disc and the couple produce the torque on the disc and hence the angular acceleration of the disc.

• Thanks for the answer, but I can do what you desctived only after the choice of the pivot point. Now if I choose $O$ the torque of $f$ must be 0, hence I don't think I can use the procedure you stated in this case. Otherwise what would be the influence of the pivot choice? – Sørën Apr 21 '16 at 13:35
• The fantastic thing about the torque due to a couple is that it is totally independent of the point about you wish to find the torque. In the diagram above if the radius of the disc is $r$ the the torque about the centre of mass, about the point of contact between the disc and the ground, the very top of the disc is always the same $Fr$. Draw a diagram of a couple and then arbitrarily choose any point on your piece of paper and work out the torque due to the couple. – Farcher Apr 21 '16 at 14:57

I assumed that the angular velocity (and so $α$) is the same it I take as pivot the center of mass or the point $O$.

If this is the case than parallel axis theorem can be used

You are mixing rotation and circular translation. Angular velocity is defined with respect to the (instantaneous) axis of rotation, which you cannot choose at will. It is imposed by the kinematics of the body in the chosen frame of reference; it has the property of having no (instant) velocity. Here, $O$ (as a point of the rim) is moving in both the CM frame and the plane frame.

You may apply König's theorems about any fixed axis of your choice. Let's choose any point $A$ of the plane. I call $C$ the c.m., $\hat x$ is rightwards, $\hat y$ is into the figure, $\hat z$ is upwards. König's 1st theorem: $$\vec L_A=I_C\vec ω+\vec{AC}×m\vec v=(I_Cω+rmv_x)\hat y.$$

Torque of $\vec f$ with respect to $A$ is zero, hence $\vec L_A$ is constant, so does $I_Cω+rmv_x$. Then $$I_C\dot ω+rma_x=0,$$ which is consistent with your set of equations $(1)$ obtained in the c.m. frame.

As friction takes place, it implies that the center of mass is decelerating. To simply apply the Newton's law, one has to stay in the inertial frame. Alternatively, if one discusses the problem in a non-inertial frame of reference, some inertial force $\vec{f}_{int}$ should be properly added to the analysis to restore the law's validity. In the following, let us discuss several different approaches to the problem.

(1) Let us consider an analysis carried out in a reference relatively motionless with respect to the center of the mass of the disk. It is an approach in the non-inertial frame and is the easiest approach to the problem. By definition, the inertial force in question acts on the center of mass with the same magnitude of the friction but on the opposite direction. We note that the inertial force will not change the equations with respect to the center of mass; but it does for the second case when the analysis is carried out on the pivot point on edge of the disk (see 2 below).

To be specific, the equation of motion with respect to the center of mass reads $$\vec{\tau}=\vec{r} \times (\vec{f}+\vec{f}_{int})=\vec{r} \times \vec{f} =I_{cm} \dot{\vec{\omega}} \tag{1}$$

As a side note, Farcher's answer is equivalent to this solution and is most desirable: The motion of the center of the mass of a rigid body is determined by the total mass and the sum of external forces; the rotation of the rigid body is determined by the moment of inertial (or better, the inertia tensor with respect to the center of mass) and the sum of external torques. Once we take the center of mass as the origin of the coordinate system and evaluate moment of inertial, we are dealing with non-inertial reference frame. However, this view of point does not explicitly involve the concept of inertial force because (i) the inertial force does not affect the motion of the center of mass and (ii) in this case the inertial force passes through the center of mass, so it does not contribute to the torque.

(2) Now let us try to write down the equation of motion with respect to the contact point $O$. To be specific, the contact point sits on edge of the disk vertically below the center of mass which moves at the same velocity/acceleration as the latter. In other words, we will discuss the problem in the same non-inertial frame of reference as above, but take the origin of the coordinate system to be the contact point. In the case, the inertial force explicitly comes into play. The equation of motion reads $$\vec{\tau}'=\vec{r}' \times (\vec{f}+\vec{f}_{int})=\vec{r}' \times \vec{f}_{int} =\dot{\vec{L}}=I_{cm} \dot{\vec{\omega}'}+\dot{\vec{L}}_{cm} \tag{2}$$ where $I_{cm} \dot{\vec{\omega}'}$ is the change rate of angular moment with respect of the center of mass, and $\dot{\vec{L}}_{cm}$ is that of the center of mass itself. We note here one can not make use of the expression for the moment of inertial (neither the parallel axis theorem) to evaluate the angular moment of the disk about point $O$, since the latter is not a fixed point (in time) on the body (c.f. Eq.(3) in 3 and c.f. 4 below).

It is not difficult to show that Eq.(2) is equivalent to Eq.(1), by noticing $\vec{f}_{int}=-\vec{f}$ (by the definition of inertial force), $\vec{r}'=-\vec{r}$ (due to the locations of the origin and the point where the force in question acts on), $\vec{\omega}=\vec{\omega}'$ (the same angular velocity of the disk with respect to the center of mass) and $L_{cm}=0$ (since the center of mass is motionless in this reference frame).

(3) Now, following the above line of thought, one might want to tackle the problem with respect to the fixed pivot point on the edge of the disk (which rotates together with the disk and coincides with the contact point $O$ only at a given instant). At a first thought, this might be desirable since now we are able to make use of the definition of $I_0$. However, this actually turns out to be the most complicated approach. On the one hand, to calculate the derivative of angular moment, it involves the concept of time derivative in rotating reference frame. In fact, in this case the derivative of the angular moment reads $$\dot{\vec{L}}=I_{o} \dot{\vec{\omega}'}+\vec{\omega}'\times \vec{L} \tag{3}$$ On the other hand, to calculate the torque it involves extra terms of inertial force such as Coriolis force, due to the fact that the coordinate system is not only non-inertail but also possessing radial and angular velocity with respect to an inertial coordinate system. It is not very trivial to show the equivalence, but it is implied. Please refer to "Classical Dynamics" by Goldstein for more information.

(4) Last but not least, there is another scenario when one carries out the analysis with respect to the instant contact point on the floor (since it is a fixed point on the floor, therefore it is also the contact point $O$ only at a given instant). The difference is that in this case, we are handling the problem in the inertial frame. As a result, we can safely write down the Newton's equation of motion without involving any inertial force. It is a good exercise to show that the angular moment of the system (the disk) indeed is indeed not changing with time, of course, with respect to the above fixed point on the floor: $$\vec{\tau}''=\vec{r}'' \times \vec{f}=0 \tag{4}.$$

In fact, one can explicitly show that the increase of the angular moment with respect to the center of mass exactly cancels with the decrease of the angular moment of the center of the mass with respect to the fixed point on the floor. Again by making use of the formula that the angular moment of a system of massive points can be written as a sum of the angular moment of the center of the mass and the angular moment of individual massive points with respect to the center of mass, the angular moment is therefore shown to be conserved. It is also noted that the definition of inertial moment does not make much sense in this last case either (in the sense that it is not a constant in time), because it should be defined with respect to a fixed point on the rigid body to be meaningful.

• About your note 1: the disk is slipping, there should be no relation between $\mathbf ω$ and $\mathbf v$. – L. Levrel Apr 21 '16 at 20:06