# About $\frac{d L}{d t}=\tau$ in a rotating center of mass frame

Statement: A Man of mass M stands on a railroad car that is rounding an unbanked turn of radius R at speed v. His center of mass is height L above the car, and his feet are distance d apart. The man is facing the direction of motion.

Torque about the center of mass is: $$-N_{1} \frac{d}{2}+N_{2} \frac{d}{2}+\left(f_{1}+f_{2}\right) L=0$$

Question:

$$-N_{1} \frac{d}{2}+N_{2} \frac{d}{2}+\left(f_{1}+f_{2}\right)L=0$$

The above equation uses the centre of mass of the man as the point where the equation $$\frac{d L}{d t}=\tau$$ has been used.

This equation is true in an inertial frame, but the centre of mass is rotating and hence non inertial, then how is the equation correct?

This equation is true in an inertial frame, but the centre of mass is rotating and hence non inertial, then how is the equation correct?

The centre of mass in itself is just a point, there is no unique frame attached to the centre of mass. Usually when we say COM frame, there is no rotation and we take a frame which has centre of mass at origin in all times.

Basically, when the net force is zero, the torque calculation about any point in space must agree. So, take an inertial frame and calculate torques about the COM. Proof of Claim By Selene Routley

• If the COM is accelerating (like in this problem) then the sum of forces is not zero. Jun 14 at 17:45
• @JAlex sure, the com can accelerate but I am not attaching my frame to the com. I am taking some inertial frame outside and taking com as the point about which I take the torques. Jun 14 at 17:46
• Yes, this is the proper approach. The COM is the summation point in the sense of $$\vec{L} = \sum_i \vec{r}_i \times (m_i \vec{v}_i)$$ adding all the individual moment of momenta with $\vec{r}_i$ the location of each particle with respect to the COM. Jun 14 at 17:49
• The equation in my question suggest that the equation has been written taking the origin as the CoM,this is accelerating! Do you mean this equation is written in an instantaneous inertial frame which has origin at the com? Also the net sum of forces isn't zero in this case. Jun 15 at 4:07
• Whatever point you take as origin of your coordinate system (and in turn, inertial frame), you can always calculate the torques about another point from the frame. So, take an inertial frame outside and calculate torque, you'll find it zero. This is equivalent to the instantenous intertial frame idea. On the net force not being zero... hmmm are you saying that because of the centripetal acceleratioN? @Kashmiri next time onwards please @ me Jun 27 at 10:17

You can calculate the torque with respect to any given point in any given coordinate system. In particular, you can calculate the torque relative to the center of mass with all vectors represented in the inertial coordinate system. Only because you choose some point to calculate the torque with respect to, doesn't mean you have to change the coordinate system and move it to that point. You just have to take cross products of vectors pointing from the chosen point to the points at which various forces are applied and add them together. The differential equation, you mentioned however, requires an inertial coordinate system. In an non-inertial one, you just have to correct it with the corresponding fictitious torques so something like this: $$\text{Inertial: } \frac{d \vec{L}}{dt} \, = \, \vec{\tau}$$ $$\text{Non-inertial: } \frac{d \vec{L}}{dt} \, = \, \vec{\tau} \, + \, \vec{\tau}_{\text{fict}}$$

For example, in rigid body dynamics, in an inertial frame, you have $$\frac{d \vec{L} }{dt} \, = \, \vec{\tau}$$ But in the frame attached firmly to the body and centered at the center of mass $$\frac{d \vec{L}}{ dt} \, = \, \vec{\tau} \, + \, \vec{L} \times \vec{\omega}$$ where $$\vec{\omega}$$ is the angular velocity expressed in the body-fixed frame. $$\vec{L}$$ and $$\vec{\tau}$$ are also expressed in the body-fixed frame.

Your case: you have a cart with a stationary man on top of it, moving uniformly along a circle of radius $$R$$ with constant velocity of magnitude $$v$$.

Let us fix an inertial coordinate system, centered at the center of the circular arc formed by the train-tracks, $$x$$ and $$y$$ coordinate axes aligned with the ground and the $$z$$ axis perpendicular to it. Let $$\vec{r} \, = \, \vec{r}(t)$$ be the vector pointing from the origin of the system to the projection of the man's center of mass onto the cart (the cart follows the circular train-tracks in the $$x, y$$ horizontal plane). Then, because the cart movies uniformly along a circle, $$\frac{d^2\vec{r}}{dt^2} \, = \, - \frac{v^2}{R^2} \vec{r}$$

The angular velocity of the cart in the inertial system, is $$\vec{\omega} \, = \, \frac{v}{R} \, \vec{e}_z$$ Since the man is motionless relative to the cart, i.e. he moves the same way the cart does, his angular velocity, relative to the inertial system, is equal to the cart's angular velocity $$\omega$$.

Consequently, the angular momentum of the man in the inertial coordinate system looks like $$\vec{L} \, = \, I \,\vec{\omega}\,= \, I_{z} \, \frac{v}{R} \, \vec{e}_z$$ where $$I_z$$ is the inertia moment of the man calculated along the vertical axis passing through his center of mass, which is always parallel to the $$z$$ axis of the inertial coordinate system, so $$I_z$$ is a constant. This particular configuration is special. Had the man been tilted somehow, so that his inertia axes were not as in the picture, then the equation would have needed an extra term for a time-changing angular momentum. But because of how the man stands on the cart, the angular momentum and the angular velocity are aligned (point in the same direction). This is because one axis of inertia of the man is aligned with the $$z-$$coordinate axis of the inertial coordinate system, while the other two inertia axes always rotate parallel to the $$x,y-$$coordinate plane of the inertial coordinate system.

The forces, in the inertial coordinate system, acting on the man on the cart are $$N_1 \vec{e}_z, \,\,\, N_2 \vec{e}_z, \,\,\, - \, \frac{f_1}{R}\vec{r}, \,\,\, - \, \frac{f_2}{R}\vec{r}, \,\,\, -\,\vec{mg}\,\vec{e}_z$$

The vector $$\vec{r}_G$$ from the origin of the coordinate system to the man's mass center is $$\vec{r}_G \, = \, \vec{r} \, + \, L \, \vec{e}_z$$

The vectors from the origin of the coordinate system to the feet of the man are

$$\vec{r}_1 \, =\, \left(1 + \frac{d}{2R}\right)\vec{r}$$ $$\vec{r}_2 \, =\, \left(1 - \frac{d}{2R}\right)\vec{r}$$

because $$|\vec{r}| = R$$ and so $$\frac{\vec{r}}{R}$$ is a unit vector, aligned with $$\vec{r}$$ at all times $$t$$. Hence $$\vec{r}_i \, = \, \vec{r} \,\pm\, \frac{d}{2} \, \frac{\vec{r}}{R}$$

So the vectors from the center of mass of the man to his feet are

$$\vec{r}_1 - \vec{r}_G \, = \,\,\, \left(1 + \frac{d}{2R}\right)\vec{r} \, - \, \vec{r} \, - \, L \, \vec{e}_z \, = \, \frac{d}{2R}\,\vec{r} \, - \, L \, \vec{e}_z$$ $$\vec{r}_2 - \vec{r}_G \, = \, \left(1 - \frac{d}{2R}\right)\vec{r} \, - \, \vec{r} \, - \, L \, \vec{e}_z \, = \, -\, \frac{d}{2R}\,\vec{r} \, - \, L \, \vec{e}_z$$

Now, let's calculate the torques: $$(\vec{r}_1 - \vec{r}_G) \times ( \,N_1\,\vec{e}_z \,) \, = \, \frac{N_1d}{2R} \,\vec{r} \times \vec{e}_z$$ $$(\vec{r}_2 - \vec{r}_G) \times ( \,N_2\,\vec{e}_z \,) \, = \, - \, \frac{N_2d}{2R} \,\vec{r} \times \vec{e}_z$$ $$(\vec{r}_1 - \vec{r}_G) \times \left( \, - \, \frac{f_1}{R}\vec{r} \,\right) \, = \, \frac{f_1L}{R} \,\vec{e}_z \times \vec{r}$$ $$(\vec{r}_2 - \vec{r}_G) \times \left( \, - \, \frac{f_2}{R}\vec{r} \,\right) \, = \, \frac{f_2L}{R} \,\vec{e}_z \times \vec{r}$$ The torque of gravity with respect to center of mass is zero.

So the vector equation of motion for angular momentum/angular velocity is

$$\frac{d}{dt} \vec{L} \, = \,\frac{d}{dt}\left( I_{z} \, \frac{v}{R} \, \vec{e}_z\right) \,= \, \frac{N_1d}{2R} \,\vec{r} \times \vec{e}_z \, - \, \frac{N_2d}{2R} \,\vec{r} \times \vec{e}_z \,+\,\frac{f_1L}{R} \,\vec{e}_z \times \vec{r} \, +\, \frac{f_2L}{R} \,\vec{e}_z \times \vec{r}$$

However, since $$\frac{v}{R}$$ is a constant,

$$\frac{d}{dt}\left( I_{z} \, \frac{v^2}{R} \, \vec{e}_z\right) \,= \, 0 \, = \, \left(\,- \, \frac{N_1d}{2R} \, + \, \frac{N_2d}{2R} \,+\,\frac{f_1L}{R} \, +\, \frac{f_2L}{R} \,\right) \, \vec{e}_z \times \vec{r}$$

which yields

$$- \, \frac{N_1d}{2R} \, + \, \frac{N_2d}{2R} \,+\,\frac{f_1L}{R} \, +\, \frac{f_2L}{R} \, = \, 0$$ and if you multiply both sides of the latter identity by $$R$$, you get exactly

$$- \, \frac{N_1d}{2} \, + \, \frac{N_2d}{2} \,+\, ({f_1} \, +\, {f_2})\,L \, = \, 0$$

So I guess you have some reason to be unsatisfied with the textbook's explanation, because they were sloppy. The total torque would not have been zero, had the motion been accelerating along the circle or some more general motion of the cart. The only reason the latter equation holds is because the angular velocity and the angular momentum are constant vectors for this specific uniform motion along a circle!

Maybe If you would like, you can read this answer of mine, where I try to derive rigorously the mathematical modelling behind the motion of a rigid body and how that motion can be decomposed into translational motion of the center of mass plus rotation around the center of mass.

• Hello, you calculated L about the centre of track and equated it with the torque about the center of mass. How's that true? Jul 5 at 5:28
• Also you wrote $\vec{r}_1 \, =\, \left(1 + \frac{d}{R}\right)\vec{r}$ I can't see how. Can you please explain. Jul 5 at 5:29
• @Kashmiri read my edits. The angular momentum is $\vec{L} = I \, \vec{\omega} = I \, \frac{v}{R} \vec{e}_z$. The question is what is $I$. In this problem, $I$ is block-diagonal matrix, because of the way the inertia axes of the man are aligned with the inertial coordinate system. and therefore $I \vec{e}_z = I_z \vec{e}_z$, where $I_z$ is a number and is the man's moment of inertia calculated relative to the axis parallel to the $z-$coordinate axis through his center of mass. Jul 5 at 13:41
• Is the $L$ about the center of the circular arc formed by the train-tracks ? I think so ,because as man rotates about the centre of the track as a rigid body it's angular momentum is $I\omega$ Jul 6 at 4:40
• @Kashmiri No, it's not the center of the train-tracks. The angular momentum and the torques have to be calculated relative to the same point and in this case the point is the man's center of mass. The angular momentum is $\vec{L} = I \vec{\omega}$. The angular velocity $\vec{\omega}$ is the same relative to any point. The relationship of the angular momentum to a point comes from the inertia matrix $I$ (distribution of mass in space). I took the inertia $I$ to be relative to the man's center of mass. And the torques are calculated relative to it Jul 6 at 11:20

$$f_1$$ and $$f_2$$ are the centrifugal forces in the non-inertial frame. Once you also include the non-inertial forces then that equation is correct.

• Hi, the author solves the problem in an inertial frame Jul 6 at 4:41

Torque and angular momentum is summed about the COM (point) but using common basis-vectors (orientation) as a non-rotating frame.

These problems assume there is an inertial coordinate system coincident with the COM at each time frame. Sometimes this is called a co-moving reference frame and it is valid so long at it is not accelerating and not rotating.

• We can do this problem in an non inertial frame attached with the CM but with its axis parallel to an inertial frame. Then the equation holds true. Jun 21 at 14:18

David Morin from his Classical mechanics book :

Let the position of the origin be $$\mathbf{r}_{0}$$ (see Fig. 8.16), and let the positions of the particles be $$\mathbf{r}_{i}$$. The vectors $$\mathbf{r}_{0}$$ and $$\mathbf{r}_{i}$$ are measured with respect to a given fixed coordinate system. The total angular momentum of the system, relative to the(possibly accelerating) origin $$\mathbf{r}_{0}$$, is $$\mathbf{L}=\sum_{i}\left(\mathbf{r}_{i}-\mathbf{r}_{0}\right) \times m_{i}\left(\dot{\mathbf{r}}_{i}-\dot{\mathbf{r}}_{0}\right)$$ Therefore, \begin{aligned} \frac{d \mathbf{L}}{d t} &=\frac{d}{d t}\left(\sum_{i}\left(\mathbf{r}_{i}-\mathbf{r}_{0}\right) \times m_{i}\left(\dot{\mathbf{r}_{i}}-\dot{\mathbf{r}}_{0}\right)\right) \\ &=\sum_{i}\left(\dot{\mathbf{r}}_{i}-\dot{\mathbf{r}}_{0}\right) \times m_{i}\left(\dot{\mathbf{r}_{i}}-\dot{\mathbf{r}}_{0}\right)+\sum_{i}\left(\mathbf{r}_{i}-\mathbf{r}_{0}\right) \times m_{i}\left(\ddot{\mathbf{r}_{i}}-\ddot{\mathbf{r}}_{0}\right) \\ &=0+\sum_{i}\left(\mathbf{r}_{i}-\mathbf{r}_{0}\right) \times\left(\mathbf{F}_{i}^{\mathrm{ext}}+\mathbf{F}_{i}^{\mathrm{int}}-m_{i} \ddot{\mathbf{r}}_{0}\right) \end{aligned}because $$m_{i} \ddot{\mathbf{r}}_{i}$$ is the net force (namely $$\mathbf{F}_{i}^{\text {ext }}+\mathbf{F}_{i}^{\text {int }}$$ ) acting on the $$i$$ th particle. But a quick corollary to Problem $$8.9$$ is that the term involving $$\mathbf{F}_{i}^{\mathrm{int}}$$ vanishes (as you should check). And since $$\sum m_{i} \mathbf{r}_{i}=M \mathbf{R}$$ (where $$M=\sum m_{i}$$ is the total mass, and $$\mathbf{R}$$ is the position of the center of mass), we have $$\frac{d \mathbf{L}}{d t}=\left(\sum_{i}\left(\mathbf{r}_{i}-\mathbf{r}_{0}\right) \times \mathbf{F}_{i}^{\mathrm{ext}}\right)-M\left(\mathbf{R}-\mathbf{r}_{0}\right) \times \ddot{\mathbf{r}}_{0}$$ The first term here is the external torque, measured relative to the origin $$\mathbf{r}_{0}$$. The second term is something we wish would go away. And indeed, it usually does. It vanishes if any of the following three conditions is satisfied.

1. $$\mathbf{R}=\mathbf{r}_{0}$$, that is, the origin is the $$\mathrm{CM}$$.
2. $$\vec{r}_{0}=0$$, that is, the origin is not accelerating.
3. $$\left(\mathbf{R}-\mathbf{r}_{0}\right.$$ ) is parallel to $$\ddot{\mathbf{r}}_{0}$$. This condition is rarely invoked.

If any of these conditions is satisfied, then we are free to write $$\frac{d \mathbf{L}}{d t}=\sum\left(\mathbf{r}_{i}-\mathbf{r}_{0}\right) \times \mathbf{F}_{i}^{\mathrm{ext}} \equiv \sum_{i} \tau_{i}^{\mathrm{ext}}$$

In this problem we've the reference frame at the center of mass, hence case 1 applies, and we can write $$\frac{d L}{d t}=\tau$$ about the centre of mass. Since $$\frac{d L}{d t}$$ is zero in the COM frame we arrive at the required equation $$\tau$$=0

Consider the frame with the origin resting relative to the ground, and the $$OX$$, $$OY$$ axes pointing, say, east and south. This is inertial frame, and we can write the torque around arbitrary point.

We want to do it for the current position of the guy's center of mass. So we put the origin of the coordinate system to this immediate position. This allows to simplify equations a bit, by getting rid of the gravity force. The expression for the torque in this case you already have. So the question boils down whether the angular momentum of the guy in this frame is constant.

On the other hand you can say that the guy just rotates as whole along the axis going through the center of the rails circle, thus it is possible to describe the motion by angular momentum only. The expression for it in the inertial frame with the origin located on the aforementioned axis at the same altitude as the com will be

$$L = \sum_i m_i \vec{r}_i \times \dot{\vec r}_i ,$$

and it is constant. If we write the angular momentum around the point $$\vec{r}_0$$ for which we calculated the torque before, it modifies as

$$L' = \sum_i m_i (\vec{r}_i - \vec{r}_0) \times (\dot{\vec r}_i - \dot{\vec r}_0) .$$

Note that $$\dot{\vec r}_0 = 0$$ because $$\vec{r}_0$$ is the immediate position of the com in the inertial frame. Finally one need to show that

$$L - L' = \vec{r}_0 \times \sum_i m_i \dot{\vec r}_i$$

is constant. The last expression is by definition velocity of the center of mass so we get $$\vec{r}_0 \times M \dot{\vec{r}}_c$$. After taking the derivative we obtain that

$$\dot{L}' = - \vec{r}_0 \times M \ddot{\vec{r}}_c$$

Using that $$\vec{r}_0$$ is radial and $$\ddot{\vec r}_c$$ is also radial we find that $$L'$$ is constant. Hence, you get that the total torque shall be zero.

Finally, I want to give more explanation regarding the location of the origin of the frame of reference in which we defined $$L$$. It was sloppy in my previous version of the answer.

So if the body rotates around the axis with constant angular velocity, its angular momentum with respect to this axis is constant. The angular momentum around the axis is defined as $$\sum_i r_{i,\perp} v_i$$, where the $$r_{i,\perp}$$ is the distance from the point to the axis, and $$v_i$$ is it's velocity (which is tangential).

If we talk about the angular momentum in vectorial form as in the first equation, in principle its direction will depend on the selected origin point (as you can also see from the derivation above). If we choose the origin not at same height (z coordinate) as the com then the angular momentum will precess around the vertical axis. To be more precise, the above statement is only correct when the rotation axis is parallel to the one of the principle axes of the moment of inertia tensor. In general case you can't rely on the fact that the angular momentum with respect to com remains constant when the body rotates with constant angular velocity, as suggested in this answer. I think more detailed explanation shall be done in this respect.

• $r_0$ is not radially inwards in line with $r_{cm}$. Actually $r_0$ has a z component pointing upwards while as the acceleration of com is centripetal and hence directed inwards Jul 6 at 3:56
• @Kashmiri good point, there might be mistake somewhere in my derivation, because the answer can not depend on the height of the origin of the coordinate system. Jul 6 at 9:03
• You might want to see my reply to this question, actually an excerpt from David Morins text above. Jul 6 at 10:11
• @Kashmiri yes, this another option to go in the frame which is moving with the com, but not rotating (i.e. with OX, OY always directing to east and south). In this frame the guy is again rotating with constant angular velocity. Jul 6 at 10:40
• Yes and hence dL/dt =0=torque. Jul 6 at 11:54

Sorry for my poor English. It's not my native language !

To begin with, it does not seem prudent to me not to write the point of application of the different "moments": they are functions of points.

Except in special cases (see below), the angular momentum theorem must be written at a fixed point. If we choose the point O we will have $$\frac{d\vec{L_O}}{dt}=\vec{\tau_{Oext}}$$

If the frame of reference is not inertial, the moment of inertia forces must be taken into account.

We place ourselves in the fixed frame of reference linked to the ground. We denote by G the center of mass. The angular momentum in O is related to that in G using König's theorem: $${\vec{L}}_O={\vec{L}}_G+\vec{OG}\ \land m\ {\vec{v}}_G$$ or $$\frac{d\vec{L_O}}{dt}=\frac{d\vec{L_G}}{dt}+\vec{OG} \land m\ {\vec{a}}_G$$

We also have the usual relation: $$\vec{\tau_{Oext}}=\vec{\tau_{Gext}} +\vec{R_{ext}}\bigwedge\vec{GO}$$ with $$\vec{R_{ext}}$$ the global external forces.

Since $$m\vec{a_G}=\vec{R_{ext}}$$ we find: $$\frac{d\vec{L_G}}{dt}=\vec{\tau_{Gext}}$$

Whatever the motion of the center of mass G, we can apply the angular momentum theorem at the center of mass.

We can prove this theorem in a more physical way by placing ourselves in the (non-inertial) frame of reference which originates from the center of mass and is in translation with respect to the previous frame of reference. In this frame of reference, in translation, the moment of the inertial force must be taken into account. As the frame of reference is in translation, we are dealing with a uniform force field resulting in $$\vec{R_{ie}}=-m\vec{a_G}$$. As this field is uniformly distributed, the point of application is G and its moment at point G is zero. So there remains $$\frac{d\vec{L_G}}{dt}=\vec{\tau_{Gext}}$$ without having to take the inertial forces into account.

Finally, we just have to apply this theorem to the standing man. There is a small subtlety already indicated by another member below and that we can develop a little: even if the speed of rotation is constant $$\vec{\omega}=\omega\vec{e_z}$$, the angular momentum in G is $$\vec{L_G}=\left[I\right]\vec{\omega}$$ if $$\left[I\right]$$ is the inertia tensor in G. This tensor is independent of time if it is written in an axis system linked to the solid. If the Oz axis is a main axis of inertia, we will have $$\left[I\right]=\left[\begin{matrix}&&0\\&&0\\0&0&I_{zz}\\\end{matrix}\right]$$ and so $$\vec{L_G}=I_{zz}\vec{\omega}$$ Which gives $$\frac{d\vec{L_G}}{dt}=\vec{0}$$

But if this is not the case (an arm up, an arm down for example), we could have $$\left[I\right]=\left[\begin{matrix}&&0\\&&I_{xz}\\0&I_{xz}&I_{zz}\\\end{matrix}\right]$$ on a basis linked to the man. In this case,$$\vec{L_G}=I_{zz}\omega\vec{e_z}+I_{xz}\omega\vec{e_X}$$ with $$\vec{e_X}$$ a unit vector linked to the man. We will then have $$\frac{d\vec{e_X}}{dt}=\vec{\omega}\bigwedge\vec{e_X}=\omega\vec{e_Y}$$ or $$\frac{d\vec{L_G}}{dt}=I_{xz}\omega^2\vec{e_Y}$$

In the latter case, we will have $$\vec{\tau_{Gext}}=I_{xz}\omega^2\vec{e_Y}$$ different from zero.

To conclude, the moment about the center of mass is zero only if Oz is a principal axis of inertia.