My current understanding of angular momentum and torques are that they are both defined for an arbitrary point, $o$, (pick the point and calculate, different points can have different values) and the angular momentum about that point of a rigid body, $L_o$, is given by:

$ L_o = \sum_{i}r_{oi}\times p_i $

where the sum is over all points of the body. Similarly the torque about that point, $G_o$, is given by:

$ G_o = \sum_{j}r_{oj}\times F_j $

where the sum is over all the external forces acting on the body. The distances are from point $o$ to the point that the force is acting on the rigid body.

There are also formulae in terms of momenta of inertia: $L=I\omega$ and $G=I\dot{\omega}$, where $I$ will be calculated for a particular point. My understanding says that these are valid only for a specific point, $a$, which is usually called the "axis of rotation" or something similar. You cannot choose this point arbitrarily (even if you think using parallel axis theorem can tell you the momenta of inertia about an arbitrary point). My question is about clearly defining the characteristics of this point where these formulae are valid.

I understand that for $L=I\omega$ is valid at the point where the total velocity of any point in the body is given by $v_i = \omega\times r_{ai}$ - from this it follows that the axis of rotation should be instantaneously stationary ($v_a = 0$).

Similarly to use $G=I\dot{\omega}$ I have impression that the rotation axis should be instantaneously not accelerating - the linear acceleration determined by $\sum_j F_j$ should be exactly opposite to the tangential acceleration of the torque calculated about the centre of mass.

In problems involving rolling without slipping, the contact point is both instantaneously stationary and not accelerating, so it's clear that the rotation axis is the contact point. But I think it's possible to have situations where there is no point (within or outside the body) that is both instantaneously at rest and not accelerating. Take a sphere moving at constant speed and rotating with a speed equal to that velocity (imagine "rolling in space"). The axis of rotation is at the bottom of the sphere as that point is instantaneously stationary. There's no torque so all points have zero tangential acceleration. Now apply a torque about the centre of mass using two equal but opposite forces. There's no linear acceleration, so the point that is not accelerating is the centre of mass. But the centre of mass is not instantaneously at rest. So in this situation is the axis that $G=I\dot{\omega}$ is valid for a different point that $L=I\omega$ is valid for? If these are different points, is there a separate name for these two points?

In summary, am I correct to conclude that the rotation axis isn't strictly a point that is both simultaneously at rest and not accelerating?



2 Answers 2


In short:

  • The angular momentum (about the origin of the coordinate system) of a particle that moves along a trajectory $r(t)$ is $L=r\times\dot r\,.$ For simplicity lets assume the particle has mass one. When we have several particles (not necessarily part of a rigid body) their total angular momentum is $L=\sum_ir_i\times\dot r_i\,.$ This is your first formula. It is very general as it does not require a rigid body and holds for the points performing any type of motion.

  • The angular velocity $\omega_i$ of particle $i$ is not so easy to define in general. It requires to distinguish between translational and rotational motion. In other words: it requires to extract from $\dot r_i$ the component that is not translational. My favourite book on that subject [1] achieves this by defining the socalled transferred velocity (p. 125). To do this a moving coordinate system is introduced. In the case of no translational motion the angular velocity $\omega_i$ is defined by $\dot r_i=\omega_i\times r_i\,.$ A priori it depends on $i\,.$

For simplicity I assume not translational motion from now on.

  • When the particles all are part of a rigid body one can find an angular velocity $\omega$ such that they all satisfy $\dot r_i=\omega\times r_i\,.$ This angular velocity does not depend on $i\,.$

The total angular momentum of the rigid body is then: $$ L=\sum_ir_i\times(\omega\times r_i)=-\sum_ir_i\times(r_i\times\omega)\,. $$ Another important step is now that this equation should be formulated in a coordinate system that rotates with the body: $$ L^{\text{body}}=\sum_iR_i\times(\Omega\times R_i)=-\sum_iR_i\times(R_i\times\Omega)\,. $$ See [1] for details. Using the trick $$ a\times b=\left(\matrix{0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0}\right)\left(\matrix{b_1\\b_2\\b_3}\right) $$ it is fairly easy to see that the linear map $\Omega\to L^{\text{body}}$ is given by a symmetric matrix $I$ which we call inertia tensor of the rigid body. In other words, the formula $$ L^{\text{body}}=I\Omega $$ holds under all assumptions made so far.

Does this help to rethink the subject? I have the impression that to answer all your questions it is inevitable to go into many details that [1] has laid out much better.

[1] V. Arnold, Mathematical Methods of Classical Mechanics.

  • $\begingroup$ Thanks, I'll see if I can find a copy of the text you recommend, but your argument relies on a point existing such that the instantaneous velocity of every point in a rigid body is given by $\omega \times r_i$ - this point you arranged to be at origin and i think assumed the c.o.m. is there too, but for a rigid body with translational motion it wont be. But there is a point that will have zero velocity and at that point I believe $L=I\omega$ can be used. When force is present there is a specific point where the acceleration is zero, and at this point (and only this point) $G=I\dot{\omega}$. $\endgroup$ Jan 18, 2022 at 22:23
  • $\begingroup$ Another popular text that handles this is a mechanics book by Goldstein. I recommend to dive deep into these matters before looking at translational motion. In particular get familiar with the difference of space and body coordinates (see last edit). I get the impression that you want too many answers too quickly. $\endgroup$
    – Kurt G.
    Jan 19, 2022 at 10:30

Your question has two parts that need to be addressed.

Rotational Momentum

For a rigid body consisting of particles that move together, angular momentum summed at a point o is defined as follows

$$ \boldsymbol{L}_o = \sum_i \boldsymbol{r}_{io} \times (m_i \boldsymbol{v}_i) \tag{1}$$

At this point, the motion of each particle needs to be decomposed as a translation of the center of mass $\boldsymbol{v}_c$ and a rotation about the center of mass by $\boldsymbol{\omega}$. The relative position of each particle being $\boldsymbol{r}_{ic}$ and $\sum_i m_i \boldsymbol{r}_{ic}=0$.

$$ \boldsymbol{v}_i = \boldsymbol{v}_c + \boldsymbol{\omega} \times \boldsymbol{r}_{ic} \tag{2}$$

Translational momentum is defined as $$\boldsymbol{p} = \sum_i m_i \boldsymbol{v}_i = m\, \boldsymbol{v}_c \tag{3}$$

Rotational momentum is defined as above and evaluated as follows

$$\begin{gathered}\boldsymbol{L}_{o}=\sum_{i}\left(\boldsymbol{r}_{io}\times m_{i}\left(\boldsymbol{v}_{c}+\boldsymbol{\omega}\times\boldsymbol{r}_{ic}\right)\right)\\ =\sum_{i}\left(\left(\boldsymbol{r}_{ic}+\boldsymbol{r}_{co}\right)\times m_{i}\left(\boldsymbol{v}_{c}+\boldsymbol{\omega}\times\boldsymbol{r}_{ic}\right)\right)\\ =\boldsymbol{r}_{co}\times\sum_{i}\left(m_{i}\left(\boldsymbol{v}_{c}+\boldsymbol{\omega}\times\boldsymbol{r}_{ic}\right)\right)+\sum_{i}\left(\boldsymbol{r}_{ic}\times m_{i}\left(\boldsymbol{v}_{c}+\boldsymbol{\omega}\times\boldsymbol{r}_{ic}\right)\right)\\ =\boldsymbol{r}_{co}\times\sum_{i}\left(m_{i}\right)\boldsymbol{v}_{c}+\boldsymbol{r}_{c}\times\sum_{i}\left(m_{i}\left(\boldsymbol{\omega}\times\boldsymbol{r}_{ic}\right)\right)+\sum_{i}\left(m_{i}\boldsymbol{r}_{ic}\right)\times\boldsymbol{v}_{c}+\sum_{i}\left(\boldsymbol{r}_{ic}\times m_{i}\left(\boldsymbol{\omega}\times\boldsymbol{r}_{ic}\right)\right)\\ =\boldsymbol{r}_{co}\times\boldsymbol{p}+\sum_{i}\left(\boldsymbol{r}_{ic}\times m_{i}\left(\boldsymbol{\omega}\times\boldsymbol{r}_{ic}\right)\right) \end{gathered}$$

$$\boldsymbol{L}_{o} = \boldsymbol{r}_{co}\times\boldsymbol{p}+\mathrm{I}_{c}\boldsymbol{\omega} \tag{4}$$

  1. Rotational momentum is evaluated with (4) in general. Only in the special case where point o does not translate and $\boldsymbol{v}_i = \boldsymbol{\omega} \times \boldsymbol{r}_{io}$ you can state that $$\boldsymbol{L}_o = \mathrm{I}_o \boldsymbol{\omega}$$

    On the other hand, the rotational momentum about the center of mass can always be evaluated as $$\boldsymbol{L}_c = \mathrm{I}_c \boldsymbol{\omega} \tag{5}$$

Equations of motion

The equations of motion (at the center of mass) are derived from $\sum \boldsymbol{F} = \frac{\rm d}{{\rm d}t} \boldsymbol{p}$ and $\sum \boldsymbol{\tau}_c = \frac{\rm d}{{\rm d}t} \boldsymbol{L}_c$ where $\boldsymbol{\tau}_c$ is net torque summed at the center of mass.

$$\begin{aligned}\sum\boldsymbol{F} & =\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{p} & \sum\boldsymbol{\tau}_{c} & =\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{L}_{c}\\ & =m\,\boldsymbol{a}_{c} & & ={\rm I}_{c}\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{ \omega}+\boldsymbol{\omega}\times{\rm I}_{c}\boldsymbol{\omega} \end{aligned} \tag{6} $$

Equipollent Torque

Without loss of generality, consider a single force $\boldsymbol{F}$ applied at the summation point o that moves with the body and a torque $\boldsymbol{\tau}_o$. From (6) the net torque at o must be

$$\begin{aligned}\boldsymbol{\tau}_{o} & =\boldsymbol{\tau}_{c}+\boldsymbol{r}_{co}\times\boldsymbol{F}\\ \boldsymbol{\tau}_{o} & =\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{L}_{c}+\boldsymbol{r}_{co}\times\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{p} \end{aligned} $$

But we want to relate the above to $\frac{\rm d}{{\rm d}t} \boldsymbol{L}_o$ in order to express the equations of motion summed at o

$$\begin{aligned}\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{L}_{o} & =\tfrac{{\rm d}}{{\rm d}t}\left(\boldsymbol{L}_{c}+\boldsymbol{r}_{co}\times\boldsymbol{p}\right)\\ & =\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{L}_{c}+\left(\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{r}_{co}\right)\times\boldsymbol{p}+\boldsymbol{r}_{co}\times\left(\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{p}\right) \end{aligned}$$

  • if point o is fixed and $\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{r}_{co} = \boldsymbol{v}_c$ then the equations of motion are

$$ \boldsymbol{\tau}_{o}=\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{L}_{o} \tag{7}$$

  • but for the general case where point o moves over time then $\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{r}_{co} = \boldsymbol{v}_c - \boldsymbol{v}_o$ and the equations of motion are

$$ \boxed{ \boldsymbol{\tau}_{o}=\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{L}_{o}+\boldsymbol{v}_{o}\times\boldsymbol{p} } \tag{8} $$

  • $\begingroup$ Thanks for this detail. I'm still trying to digest it - can I ask why (6) implies $\tau_0 = \tau_c + r_{c0}\times F$ for a Force acting through point 0? I can't quite see how that follows from (6)? $\endgroup$ Jan 25, 2022 at 14:49
  • 1
    $\begingroup$ Look at equation (6) for torque; first term should have $I_c {d\omega \over dt}$ instead of $I_c \omega$. $\endgroup$
    – John Darby
    Jun 23, 2022 at 3:27
  • $\begingroup$ @JohnDarby - yes. I fixed it. $\endgroup$ Jan 8 at 23:18

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