# Validity of rotational Newton's second law in a changing instantaneously inertial frame

A standard textbook question is to ask about some rigid body (say, a 2D disk) rolling down an incline without slipping (cf. John Taylor's Classical Mechanics, Problem 3.35).

The standard approach is either to analyze the motion from the centre of mass (CM) frame, or from the point of contact $$P$$ at which there is no slipping. I'd like to focus on the validity of the latter approach. In particular, one uses "Newton's rotational second law" (effectively, just a statement about the derivative of the total angular momentum $$\bf{L}$$ of the system which uses Newton's second law in the derivation to connect acceleration and force). However, it is invariably noted during the derivation of $$\dot{\bf{L}}=\bf{\Gamma}$$ that the result holds only in an inertial frame precisely because Newton's second law has been invoked (with the standard "coincidence"/exception of the result holding in the CM frame being noted).

At any rate, and if this is the case, why can we use Newton's rotational second law in $$P$$'s frame? The point $$P$$ of contact moves (indeed, accelerates, with respect to time). I suppose that, at an instant of time, we can consider $$P$$ as the point on the fixed ground which is in contact with the disk. Is this how we get around the difficulty?

• Can you please provide the exact statement of problem 3.35 in the Taylor text? Is this for no slip of a symmetric object (please see my answer)? Oct 2, 2021 at 21:17
• @JohnDarby See my message in the discussion room. TL;DR there is no slip and there is symmetry.
– gmz
Oct 2, 2021 at 21:56
• Yes, that is what I think. Also this is 2D rotation in a plane, not general 3D rotation. Oct 2, 2021 at 22:12
– gmz
Oct 5, 2021 at 6:18
• There are two approaches to solve for the motion of the object, without using the CM, in the answers. You can either use the (i) balance law of angular momentum or (ii) derive angular momentum/Newton's second law in another frame (whose orientation is fixed). Among these, John Darby (ii), Claudio Saspinski (ii), and Adam Herst (ii), who derives Newton's second law in another frame, analyze the frame attached to a point on the circumference of the disk, and come to the conclusion that Newton's second law is only valid in the case of non-slip. (cont.)
– gmz
Oct 8, 2021 at 3:26

The validity in $$p$$'s frame is yet another interesting coincidence. We just have to derive Newton's 2nd law for rotation relative to an inertial frame, but keeping track of the intermediate non-inertial frame. First, let $$s$$ be any particle in the object, and $$g$$ be any inertial frame. I'll use subscript pairs to indicate properties of one frame/point relative to another, and I'll use $$\vec\tau$$ for torque. Then we have

$$\vec F_{s} = m_{s} \vec a_{sg} = m_{s}(\vec a_{sp} + \vec a_{pg}) \\ \vec r_{sp} \times \vec F_{s} = m_{s}(\vec r_{sp} \times \vec a_{sp} + \vec r_{sp} \times \vec a_{pg}) \\ \vec\tau_{sp} = m_{s}(\vec r_{sp} \times \frac{d}{dt}\vec v_{sp}) + m_{s}\vec r_{sp} \times \vec a_{pg} \\ \vec\tau_{sp} = m_{s}(\frac{d}{dt}(\vec r_{sp} \times \vec v_{sp}) - (\frac{d}{dt}\vec r_{sp}) \times \vec v_{sp}) + m_{s}\vec r_{sp} \times \vec a_{pg} \\ \vec\tau_{sp} = \frac{d}{dt}(\vec r_{sp} \times m_{s}\vec v_{sp}) + m_{s}\vec r_{sp} \times \vec a_{pg} \\ \vec\tau_{sp} = \frac{d}{dt}\vec L_{sp} + m_{s}\vec r_{sp} \times \vec a_{pg}$$

Since $$p$$ is in general non-inertial, we now have the corrective term on the right. Now, sum over all particles:

$$\Sigma_{s} \vec\tau_{sp} = \Sigma_{s}\frac{d}{dt}\vec L_{sp} + \Sigma_{s} (m_{s}\vec r_{sp} \times \vec a_{pg}) \\ \vec\tau_{net,p} = \frac{d}{dt} \vec L_{net,p} + (\Sigma_{s} m_{s}\vec r_{sp}) \times \vec a_{pg} \\ \vec\tau_{net,p} = \frac{d}{dt} \vec L_{net,p} + M_{tot}\vec r_{com,p} \times \vec a_{pg}$$

Finally, rearranging so that the corrective term appears as the "fictitious torque" as mentioned in Claudio's answer:

$$\frac{d}{dt} \vec L_{net,p} = \vec\tau_{net,p} + (\vec r_{p,com} \times M_{tot} \vec a_{pg})$$

where $$\vec r_{com,p}$$ is the position of the center of mass relative to $$p$$. That's why it coincidentally disappears when you choose $$p$$ as the center of mass. But what if $$p$$ is the contact point on the object? Then, $$\vec r_{com,p}$$ is normal to the incline. The trickier part is $$\vec a_{pg}$$, but we do know that $$\vec a_{pg} = \vec a_{p,com} + \vec a_{com,g}$$. The motion relative to the center of mass is circular, so $$\vec a_{p,com}$$ has a tangential component up the incline, and a centripetal part normal to it. But the tangential part has magnitude $$\frac{dv_{p,com}}{dt}$$ where $$v_{p,com}$$ is the speed of $$p$$ relative to the COM. Meanwhile, $$\vec a_{com,g}$$ has magnitude $$\frac{dv_{com,g}}{dt}$$ down the slope. But since we are rolling without slipping, the speeds $$v_{p,com}$$ and $$v_{com,g}$$ are always equal (not just at this instant). Hence the tangential part of $$\vec a_{p,com}$$ cancels $$\vec a_{com,g}$$, leaving $$\vec a_{pg}$$ with only the centripetal part normal to the incline. This is parallel to $$\vec r_{com,p}$$, zeroing out their cross product. So the corrective term disappears in this frame, as you were hoping for.

• Nice, clear discussion. I reached the same conclusion in the inertial frame; what about the solution in the non-inertial frame where the point of contact is not accelerating? Please see my response , and the earlier response by @Claudio Saspinski and comment as appropriate. Oct 6, 2021 at 22:29
• @JohnDarby Thanks, yeah, I upvoted yours earlier but just now had a chance to comment on it. Yours and Claudio's are obviously equivalent and were posted first of course, but I didn't realize that until I worked through the derivation and had already posted it, thinking I had an original answer! Oct 7, 2021 at 3:11

The balance of angular momentum is postulated first and foremost about some fixed point $$O$$: $$\dot{\bf L}_O = {\bf \Gamma}_O$$, where $${\bf L}_O$$ is the angular momentum of the rigid body about $$O$$ and $${\bf \Gamma}_O$$ is the net torque acting on the body about $$O$$.

The angular momentum about point $$O$$ is defined as

$${\bf L}_O = \int_{\mathcal{B}} {\bf r} \times {\bf v} dm$$,

where $${\bf r}$$ is the position vector relative to $$O$$ of the mass element $$dm$$ belonging to the body $$\mathcal{B}$$ which has velocity vector $${\bf v}$$. Since the body is rigid, we have the relationship $${\bf v} = {\bf v}_G + \omega \times ({\bf r} - {\bf r}_G)$$, where $$G$$ indicates the center of mass. Inserting this relationship into the definition of $${\bf L}_O$$, we find

$${\bf L}_O = {\bf r}_G \times {\bf G} + {\bf L}_G$$,

where $${\bf G} = m {\bf v}_G$$ is the linear momentum of the rigid body.

Noting that $${\bf v}_G \times {\bf G} = {\bf 0}$$, and that $${\bf \Gamma}_O = {\bf \Gamma}_G + {\bf r}_G \times {\bf F}$$, where $${\bf F}$$ is the resultant force acting on the body, we also then have $$\dot{\bf L}_G = {\bf \Gamma}_G$$ as a consequence of the balance of angular momentum about $$O$$ as well as the balance of linear momentum $$\dot{\bf G} = {\bf F}$$.

Let $$P$$ be your point of contact, an accelerating material point. The relationship $${\bf v}_P = {\bf v}_G + \omega \times ({\bf r}_P - {\bf r}_G)$$ holds due to rigidity of the disk. We deduce again a similar relationship as before:

$${\bf L}_P = {\bf L}_G + ({\bf r}_G - {\bf r}_P) \times {\bf G}$$.

This time though a time derivative yields

$$\dot{\bf L}_P = \dot{\bf L}_G + ({\bf v}_G - {\bf v}_P) \times {\bf G} + ({\bf r}_G - {\bf r}_P) \times {\bf F}$$.

In any case, since $${\bf G}$$ points in $${\bf v}_G$$, we have $${\bf v}_G \times {\bf G} = {\bf 0}$$. So in any case,

$$\dot{\bf L}_P = \dot{\bf L}_G - {\bf v}_P \times {\bf G} + ({\bf r}_G - {\bf r}_P) \times {\bf F}$$.

In the case of a disk rolling without slip, where $$P$$ is the point of contact, we have additionally that $${\bf v}_P = {\bf 0}$$, which contains a pair of non-holonomic constraints and one holonomic constraint. Hence, for this particular case,

$$\dot{\bf L}_P = \dot{\bf L}_G + ({\bf r}_G - {\bf r}_P) \times {\bf F}$$.

Inserting this into the balance of angular momentum about the center of mass, we find

$$\dot{\bf L}_P = {\bf \Gamma}_P$$

for this very very special particular case.

Notice that this answer avoids any discussion of "frames." You are free to express the vectorial quantities in any frame that you want. Indeed, there is no notion of a "frame" belonging to a point. Frames can be embedded in rigid bodies, however. What is required in this discussion is book-keeping on angular momenta and torques. These quantities are intrinsically about some point. In fact, some authors use the terms moment of linear momentum and moment of force, which may prove elucidating. The moment of linear momentum and force about what point?

• Your point of contact is indeed stationary at the time of analysis, removing the need to include any complicating terms. However, since you do take a time derivative, $\dot{\textbf{L}}_P$ is incorrect as stated.
– gmz
Oct 5, 2021 at 4:46
• Would you care to elaborate on where it is incorrect? I was pretty explicit in my derivation.
– Evan
Oct 5, 2021 at 4:48
• I would love to! I've been thinking about this question a lot, but can we do this in a separate room?
– gmz
Oct 5, 2021 at 4:52
• You can put your thoughts down in the room and I will have time to respond tomorrow. But it would be good for the record and for the people to see where you claim the derivation to be wrong.
– Evan
Oct 5, 2021 at 4:56
• It's quite difficult to quote, would it be alright if I edit and put numbering beside your equations (as in $\textbf{L}$ blah blah ... (1), etc.)?
– gmz
Oct 5, 2021 at 4:57

### Summary

To use the rotational law with a moving summation point A you need to adjust it to

$$\boldsymbol{\Gamma}_A = \dot{ \boldsymbol{L} }_A + \boldsymbol{v}_A \times \boldsymbol{p}$$

Look at equation 3.83 in these lecture notes for this exact equation, as well as development of the equations.

To use the simpler form $$\boldsymbol{\Gamma}_A = \dot{ \boldsymbol{L} }_A$$ the summation point must meet of the following criteria

• Point A is not moving, $$\boldsymbol{v}_A = 0$$
• Point A is the center of mass, $$\boldsymbol{v}_A \times \boldsymbol{v}_C = 0$$
• Point A is co-moving with the center of mass, $$\boldsymbol{v}_A \times \boldsymbol{v}_C = 0$$
• Point A is moving parallel to the center of mass, $$\boldsymbol{v}_A \times \boldsymbol{v}_C = 0$$

### Proof

Given $$\boldsymbol{\Gamma}_C = \dot{ \boldsymbol{L} }_C$$ where C designates the center of mass, transfer this relation to another arbitrary point A.

Angular momentum summed at the arbitrary point is defined by the transformation law from the value at the center of mass

$$\boldsymbol{L}_A = \boldsymbol{L}_C + (\boldsymbol{r}_C - \boldsymbol{r}_A) \times \boldsymbol{p} \tag{1}$$

Similarly for torque summed at the arbitrary point

$$\boldsymbol{\Gamma}_A = \boldsymbol{\Gamma}_C + (\boldsymbol{r}_C - \boldsymbol{r}_A) \times \boldsymbol{F} \tag{2}$$

Now take the time derivative of (1)

$$\dot{\boldsymbol{L}}_A = \dot{ \boldsymbol{L}}_C + (\boldsymbol{r}_C - \boldsymbol{r}_A) \times \dot{\boldsymbol{p}} + (\boldsymbol{v}_C - \boldsymbol{v}_A) \times \boldsymbol{p} \tag{3}$$

and use Newton's laws $$\boldsymbol{F} = \dot{\boldsymbol{p}}$$ and $$\boldsymbol{\Gamma}_C = \dot{ \boldsymbol{L}}_C$$ into (3) to get

$$\dot{\boldsymbol{L}}_A = \boldsymbol{\Gamma}_C + (\boldsymbol{r}_C - \boldsymbol{r}_A) \times \boldsymbol{F} - \boldsymbol{v}_A \times \boldsymbol{p} \tag{4}$$ $$\dot{\boldsymbol{L}}_A = \boldsymbol{\Gamma}_A - \boldsymbol{v}_A \times \boldsymbol{p} \tag{5}$$

And now solve for $$\boldsymbol{\Gamma}_A$$

• Could you provide validity of the transformation laws? Are $\textbf{p}$ and $\textbf{F}$ taken wrt. to the CM, $\textbf{A}$, or the laboratory frame?
– gmz
Oct 4, 2021 at 22:09
• Momentum $\boldsymbol{p}$ is always equal to mass time the velocity of the CM and thus does not have a specific point of summation. The same for the force as its value does not change with the point of summation as torque does. All the quantities need to be on the same basis vectors (same orientation) in order to be able to add/subtract etc. Oct 4, 2021 at 22:51
• (1) and (2) are extensions of the angular momentum for a particle $\boldsymbol{r} \times \boldsymbol{p}$ and the torque of a particle $\boldsymbol{r} \times \boldsymbol{F}$ onto rigid bodies, and can be proved with the summation of all the particles that are contained in a body. Oct 4, 2021 at 22:53
• Does $\textbf{L}_C$ refer to the angular momentum of the CM or the angular momentum of the body relative to the CM?
– gmz
Oct 4, 2021 at 23:01
• Yes $\boldsymbol{L}_C$ is angular momentum summed at the CM. Oct 4, 2021 at 23:22

... my question is why can we use Newton's rotational second law in $$P$$'s frame? The point $$P$$ of contact moves (indeed, accelerates, with time). I suppose that, at an instant of time, we can consider $$P$$ as the point on the fixed ground which is in contact with the disk.

When in doubt, be careful; assumptions here do not appear as they seem.

Let $$\textbf{r}_{\text{s}}$$ be the vector connecting our lab frame to some other stationary frame of reference, and let all positions in the lab frame $$\textbf{r}_i$$ be represented as $$\textbf{r}_\text{s} + \textbf{δ}_i$$, where $$\textbf{δ}_i = \textbf{r}_i - \textbf{r}_\text{s}$$.

Angular momentum of a body, in that other frame of reference, is defined by $$\textbf{L} = \int_{\,V}\textbf{δ}_{dm} \times (\dot{\textbf{δ}}_{dm}\, dm) = \int_{\,V}(\textbf{r}_{dm} - \textbf{r}_\text{s})\times\dot{\textbf{r}}_{dm}\,dm$$

With that in mind, let's examine the validity of switching stationary frames. For constant different $$\textbf{r}_\text{s}$$, it might be of value to find the geometric shape such that $$\textbf{L}$$ remains invariant.

Then the angular momentum, for different $$\textbf{r}_\text{s}$$, becomes \begin{align} \textbf{L} &= \int_{\,V}(\textbf{r}_{dm}\times\dot{\textbf{r}}_{dm})\,dm - \int_{\,V}(\textbf{r}_\text{s}\times\dot{\textbf{r}}_{dm})\,dm \\ &= \int_{\,V}(\textbf{r}_{dm}\times\dot{\textbf{r}}_{dm})\,dm - \textbf{r}_\text{s}\times\left(\int_{\,V}dm\right)\left(\frac{\int_{\,V}\dot{\textbf{r}}_{dm}\,dm}{\int_{\,V}dm}\right) \\ &= \int_{\,V}(\textbf{r}_{dm}\times\dot{\textbf{r}}_{dm})\,dm - \textbf{r}_\text{s}\times M\dot{\textbf{r}}_\text{CM} \\ \end{align}

Due to the cross product, the endpoints of $$\textbf{r}_\text{s}$$ that correspond to the same $$\textbf{L}\,$$ effectively draws a line parallel to whichever direction $$\dot{\textbf{r}}_\text{CM}$$ points in.

Hence, if we were to consider a disk rolling down a slope, we can set the endpoints of $$\textbf{r}_\text{s}$$ to sit on the contact points the disk rolls through, since this is a straight line. As the disk rolls by each endpoint, we can peer in each stationary frame and safely analyze the torques and forces as expected, in a snapshot of time; the contributions $$\dot{\textbf{L}}=\textbf{Γ}$$ considered correspond to a commensurate change in the same angular momentum across our stationary $$\textbf{r}_\text{s}$$ vectors.

$$\rule[0]{300pt}{0.4pt}$$

It's important to note that if the endpoint of $$\textbf{r}_s$$ is legitimately translating along with the contact point of the disk, then the angular momentum obtained is different from the angular momentum we analyzed through jumping across stationary frames—we must take into account $$\dot{\textbf{r}}_\text{s} \neq 0$$, and the equivalence between $$\dot{\textbf{L}}$$ and $$\textbf{Γ}$$ is not quite true.

... the result holds only in an inertial frame precisely because Newton's second law has been invoked (with the standard "coincidence"/exception of the result holding in the CM frame being noted).

Differentiating $$\textbf{L}$$ with respect to time gives: \begin{align*} \dot{\textbf{L}} &= \int_{\,V}(\textbf{r}_{dm}-\textbf{r}_s)\times\ddot{\textbf{r}}_{dm}\,dm - \left(\int_{\,V}dm\right)\left(\frac{\int_{\,V}\textbf{r}_{dm}\,dm}{\int_{\,V}dm} - \textbf{r}_\text{s}\right)\times\ddot{\textbf{r}}_{s}\\ \end{align*}

If we define torque in the moving frame of reference with respect to $$\textbf{r}_\text{s}$$, as it appears from the lab frame, we have

$$\dot{\textbf{L}} = \textbf{Γ} - M\left(\textbf{r}_\text{CM} - \textbf{r}_\text{s}\right)\times\ddot{\textbf{r}}_{s}$$

which accounts for your CM frame exception.

The difference between these two outwardly similar but intrinsically disparate analyses is the source of much confusion. To clarify once again: the conventional analysis considers multiple stationary frames which have the illusion of moving if you track which frame you consider at each moment. Furthermore, the method is only valid if the CM moves in a straight line. Note that whether or not the body slips is irrelevant here.

On the other hand, a true moving $$\textbf{r}_\text{s}$$ is generally not so simple and should be avoided, unless the extra term $$M(\textbf{r}_\text{s} - \textbf{r}_\text{CM})\times \ddot{\bf{r}}_\text{s}$$ is trivial.

Fourth Updated response follows:

I assume the object is rolling without slipping in a plane (2D motion) and the object is symmetric (contact point P has the same position relative to the center of mass as the object moves).

I believe the text Mechanics by Symon addresses your question. Consider a system of particles viewed in an inertial coordinate frame (origin O fixed) with point Q not necessarily fixed in space with respect to O. If Q is the center of mass (CM), then as you state $${{d\vec L_Q} \over {dt}} = \vec \tau_Q$$ where $$\vec L_Q$$ is the angular momentum of the system and $$\vec \tau_Q$$ is the total torque from all the external forces on the system with respect to point Q.

For any point Q, Symon shows that the angular momentum about Q is

$${{d\vec L_Q} \over {dt}} = \vec \tau_Q -M(\vec R - \vec r_Q) \times \ddot {\vec r_Q}$$ (eqn. 1)

where $$M$$ the total mass, $$\vec R$$ is the position of the CM with respect to O, and $${\vec r_Q}$$ is the position of point Q with respect to O. Note: $$\vec r_Q -\vec R$$ is the position of point Q relative to the position of the CM. The additional term $$-M(\vec R - \vec r_Q) \times \ddot {\vec r_Q}$$ is zero if: Q is not accelerating in the inertial frame, Q is chosen to be the CM, or the acceleration of Q is along the line joining Q with the CM.

For your specific question, Q is your point P. P has zero velocity but not zero acceleration. However, as @Claudio Saspinski says in his response, for no slipping the acceleration of the point of contact is zero from the inclined plane's frame in the direction parallel to the plane... [but] there is centripetal acceleration directed toward the center. Since, the centripetal acceleration of P is along the line joining P with the CM, the cross product in the additional term is zero, so $${{d\vec L_P} \over {dt}} = \vec \tau_P$$ in the inertial frame.

The discussion above considers P moving in an inertial frame with fixed origin O. If the origin O is accelerating, you are using a non-inertial frame and fictitious forces/torques must be considered; for example, if you consider P to be at rest at the origin of a frame, that is a non-inertial frame. The Symon text also discusses this.

If you wish to consider P from a non-inertial frame moving (accelerating) with P, in this case fictitious forces/torques must be considered. Consider the case where P is at rest in the non-inertial frame. In Eqn (1) $$\ddot {\vec r_P}$$ is zero in this frame, but $$\vec \tau_P$$ includes the torque due to the fictitious forces present in this frame, in addition to the torque from external forces. As @Claudio Saspinski says in his response, the acceleration of the point of contact is zero from the inclined plane's [inertial] frame in the direction parallel to the plane... [but] there is centripetal acceleration directed toward the center [in the inertial frame]. However as he states, the resulting fictitious force does not contribute to the torque in the non-inertial frame attached to P, so $${{d\vec L_P} \over {dt}} = \vec \tau_P$$ in this non-inertial frame as well.

Again, these conclusions are specific to rolling without slipping of a symmetric object in a plane.

• The Symon text Mechanics provides a rather lengthy derivation supporting my answer. If there is enough interest by the community I can provide it. Let's see what others say. Oct 2, 2021 at 17:19
• The non-zero term is correct, but I find the distinction between "a fixed point at a snapshot in time as $Q$" and "a moving point as $Q$" a bit blurry.
– gmz
Oct 2, 2021 at 19:07
• I agree. I do not consider Q as "a fixed point at a snapshot in time". For an inertial frame with a fixed origin, Q is any point you chose in that frame about which to evaluate the angular momentum; if Q is accelerating in that frame the additional term (earlier answer) is non-zero unless Q is either the center of mass or the acceleration of Q is along the line joining Q with the center of mass. If Q is accelerating but you consider Q to be not moving, then you are in a non-inertial reference frame and must consider fictitious forces/torques; as noted in the Symon text. Oct 2, 2021 at 19:47
• Mhm. Another problem that could be better addressed would be identifying what exactly happens to $\textbf{L}$ as we switch frames from one stationary $Q$ to another, I think that was what the OP was going for.
– gmz
Oct 2, 2021 at 19:50
• What do you mean by "If Q is accelerating but you consider Q to be not moving"? $Q$ should always be stationary with respect to $Q$'s reference frame. If you may, let us discuss this further in a separate room.
– gmz
Oct 2, 2021 at 19:52

According to the figure above, any point at the periphery of the roll has a momentarily velocity of $$v_{cm} - \omega R \sin(\phi)$$ in the direction of the inclined plane.

Using the rule of the derivative of a product, the acceleration of this point (also in the direction of the inclined plane) is $$\frac{dv_{cm}}{dt} - (R \sin(\phi)\frac{d \omega}{dt} + \omega R \cos(\phi) \frac{d\phi}{dt})$$

As $$\frac{d\phi}{dt} = \omega$$

the acceleration is $$\frac{dv_{cm}}{dt} - (R \sin(\phi)\frac{d \omega}{dt} + \omega^2 R \cos(\phi) )$$

When this point comes in contact with the plane, $$ϕ = \frac{\pi}{2}$$ and the expression simplifies to $$\frac{dv_{cm}}{dt} - R \frac{d \omega}{dt}$$

From the frame of the center, the plane has a momentarily velocity of $$-v_{cm}$$, and if there is no slip, the point of contact has the same velocity. So $$v_{cm} = \omega R$$, and the expression above vanishes.

The conclusion is that not only the velocity but also the acceleration of the point of contact is zero in the direction parallel to the plane, from the inclined plane's frame.

There is also a centripetal acceleration, which avoids what we can call the point of contact a true inertial (non-accelerated) frame. But, as its only acceleration is directed towards the center, the fictitious force that arises from this fact doesn't contribute to the torque, and the equation $$\mathbf {\dot L} = \mathbf {\Gamma}$$ can be used.

• While the velocity of the point of contact is indeed zero, the acceleration is not zero, as you can verify by differentiating the parametric equation of a cycloid. I don't know where in your answer your conclusion about the acceleration came from.
– gmz
Oct 4, 2021 at 21:32
• Yes, I made the same mistake; although the velocity of P is zero, the acceleration of P is not zero, otherwise there is no rolling. I need to reconsider my conclusion. Interested to see a response from @Claudio Saspinski to your comment. Oct 5, 2021 at 20:58
• @JohnDarby the center of the roll is accelerated, and there is also an angular acceleration. What doesn't mean that the acceleration of the point of contact must be different of zero. When I made the derivative of the velocity of a point in the periphery from the ground's frame, the zero acceleration when it contacts the plane came as conclusion. There is no logical contradiction with the fact that the object keeps rolling. Oct 5, 2021 at 21:43
• A key point for answering the question is whether or not the point of contact (with zero velocity with no slip) is accelerating with respect to the inclined plane's inertial frame. I follow your derivation, but just do not see physically how that point is not accelerating. What say you gmz, @ Evan, and @John Alexiou? Oct 5, 2021 at 22:03
• I just realized the answer I just posted is equivalent to this one, but this answer is simpler. But @gmz, if you want a more rigorous proof, check out my answer. Oct 6, 2021 at 14:39