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

Question

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?

Answer 1

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.

Answer 2

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 $\boldsymbol{\Gamma}_A$

Answer 3

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?

Answer 4

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.

Answer 5

... 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.

Answer 6

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.