# How to find Origin for Angular Momentum?

I am studying angular momentum and I get the concept that it has a meaning or it is defined only with respect to a certain point (often called as origin). I was looking at the realtionship between torque and angular momentum around an origin i.e. net torque = Rate of change of angular momentum (equation 11.29 in the picture). I think i get this relationship.

Where I got confused when author adds a caution (full quote below) to the equation saying that:

If the center of mass of the system is not accelerating relative to an inertial frame, that origin can be any point. However, if it is accelerating, then it must be the origin.

I am not sure if I get that statement and caution. Not sure what the author means by the “origin” here. Can't the origin be any point that we choose?

Let $$\vec{\tau}_{\mathrm{net}}$$ represent the net external torque, the vector sum of all external torques on all particles in the system. Then we can write Eq. 11-28 as $$\begin{equation} \vec{\tau}_{\mathrm{net}} = \frac{d \vec{L}} {dt} \qquad \text{(system of particles)} \tag{11-29} \end{equation}$$ which is Newton's second law in angular form. It says

The net external torque $$\vec{\tau}_{\mathrm{net}}$$ acting on a system of particles is equal to the time rate of change of the system's total angular momentum $$\vec{L}$$.

Eq. 11.29 is analogous to $$\vec{F}_{\mathrm{net}} = d\vec{P}/dt$$ (Eq. 9-27) but requires extra caution: Torques and the system's angular momentum must be measured relative to the same origin. If the center of mass of the system is not accelerating relative to an inertial frame, that origin can be any point. However, if it is accelerating, then it must be the origin. For example, consider a wheel as the system of particles. If it is rotating about an axis that is fixed relative to the ground, then the origin for applying Eq. 11-29 can be any point that is stationary relative to the ground. However, if it is rotating about an axis that is accelerating (such as when the wheel rolls down a ramp), then the origin can be only at its center of mass.

## 3 Answers

The rate of change of angular momentum of a body about a point P is equal to the torque about P only if one of the following is true:

1. the point P is at rest

2. the point P is the centre of mass of the body

3. the point P is moving parallel to the centre of mass.

Proof: Let P have position vector $${\bf R}$$, then $${\bf L} = \sum_i ({\bf r}_i -{\bf R})\times m_i \dot {\bf r_i}$$ so $$\dot {\bf L} = \sum_i (\dot {\bf r}_i -\dot {\bf R})\times m_i \dot {\bf r_i}+ \sum_i ({\bf r}_i -{\bf R})\times m_i \ddot {\bf r_i}\\ = -\dot {\bf R}\times {\bf P}_{\rm cofm} + \sum_i ({\bf r}_i -{\bf R})\times {\bf F}_i, \quad (\dot {\bf r}_i\times \dot {\bf r}_i=0)\\ =- \dot {\bf R}\times {\bf P}_{\rm cofm}+ {\boldsymbol \tau} .$$ So $$\dot {\bf L}= {\boldsymbol \tau}$$ needs $$\dot {\bf R}\times {\bf P}_{\rm cofm}=0$$. This requires one of the three conditions above.

• Isn't the first term in your second equation identical to the definition of ${\bf L}$ above it? Or is there something I'm not seeing. – garyp Jun 13 '20 at 16:36
• @garyp There are dots, denoting time derivatives, on the $r_i$ and $R$... – mike stone Jun 13 '20 at 16:58
• Thanks. They are nearly invisible on my laptop. – garyp Jun 14 '20 at 2:31
• @mike stone so if an object is rotating around an axis which is not moving but not passing through center of mass, this equation wont be valid.? How do we handle those cases.? – user31058 Jun 15 '20 at 14:05
• Why won't it be valid? Rember if there is a fixed axis the axle will exert a torque about the cof m. – mike stone Jun 15 '20 at 16:49

The author is just telling you how to choose the point from the angular momentum will be calculated, for it is not unique. The caution just tells you that there are some situations in which you have to choose the center of mass as the origin for the angular momentum, This is the case of an accelerating frame of reference, in which Newton's laws have to be modified to be satisfied. On an accelerating frame of reference, you have to take into account other effects such as Coriolis effect or centrifugal acceleration. In that case, the equation $$\vec{\tau}_{net}=\frac{d\vec{L}}{dt}$$ is only satisfied when the origin from where the angular momentum is calculated is equal to the center of mass of the system, because I doesn't take into account these other effects. You can learn more about non intertial frames of references here (the rotating frame of reference is a good example of an accelerating one).

Which textbook is that from? I don't find such advise to be well worded.

Let $$\vec{r}$$ be the displacement of a particle of mass $$m$$ from its centre of rotationa and $$\vec p$$ be the momentum of the particle. By definition,

$$\vec{L} := \vec{r} \times \vec{p}.$$

Then, by definition, the torque acting on $$m$$ is:

$$\vec{\tau} := \frac{\textrm{d}\vec{L}}{\textrm{d}t} = \frac{\textrm{d}\vec{r}}{\textrm{d}t} \times \vec{p} + \vec{r} \times \frac{\textrm{d}\vec{p}}{\textrm{d}t} = \vec{v} \times m\vec{v} + \vec{r} \times \vec{F},$$ so $$\vec{\tau} = \vec{r} \times \vec{F}$$ since $$\vec{v} \times \vec{v} \equiv0$$. By these relations you can see how torque and angular momentum are analogues of force and momentum.

Note that the angular momentum $$\vec{L}$$ defined above was taken about its centre of rotation. Now, let's see what happens if we modify the definition of $$\vec{L}$$ by adding any arbitrary vector to $$\vec{r}$$ (we'll call this new definition of angular momentum $$\vec{\mathcal L}$$):

$$\vec{\mathcal L} := \vec{R} \times \vec{p} = (\vec{r} + \vec{a}) \times \vec{p},$$ where $$\vec{R} = \vec{r} + \vec{a}$$ and $$\vec{r}$$ is again the displacement from its centre of rotation and $$\vec{a}$$ is an arbitrary vector (you can interpret $$\vec{a}$$ to be the displacement of the centre of rotation from any arbitrary choice of the origin of your axes).

What happens if we calculate the time derivative of this quantity?

$$\vec{\mathcal T} := \frac{\textrm{d}\vec{\mathcal L}}{\textrm{d}t} = \frac{\textrm{d}}{\textrm{d}t}(\vec{r} \times \vec{p}) + \frac{\textrm{d}}{\textrm{d}t}(\vec{a} \times \vec{p}).$$ We know that the first term equates to $$\vec{r} \times \vec{F}$$ from above. Let's continue with the second term: $$\vec{\mathcal T} = \vec{r} \times \vec{F} + \left(\frac{\textrm{d}\vec{a}}{\textrm{d}t} \times \vec{p} + \vec{a} \times \frac{\textrm{d}\vec{p}}{\textrm{d}t}\right) = \frac{\textrm{d}\vec{a}}{\textrm{d}t} \times \vec{p} + (\vec{r} + \vec{a}) \times \vec{F}.$$ So, if $$\vec{\mathcal L} = \vec{R} \times \vec{p}$$, then $$\vec{\mathcal T} = \left( \frac{\textrm{d}\vec{a}}{\textrm{d}t} \times \vec{p} \right) + \vec{R} \times \vec{F}.$$ If $$\vec{a}$$ is constant in time, or $$\textrm{d}\vec{a}/\textrm{d}t$$ is parallel to $$\vec{p}$$, then $$\vec{\mathcal T} = \vec{R} \times \vec{F}$$, and $$\vec{\mathcal L}$$ and $$\vec{\mathcal T}$$ obey the same relation to each other as $$\vec L$$ and $$\vec \tau$$ do. I think this is the point that the author was making.

I personally would forget the advise of that textbook. The key thing to remember about torque and angular momentum is that it is always angular momentum/torque about some point. What I called $$\vec{\mathcal L}$$ above is an angular momentum of the particle, it is simply its angular momentum about some point other than its centre of rotation. Usually what you want is $$\vec L$$, but it is not wrong to ask about $$\vec{\mathcal L}$$ if you want to know what the particle's momentum about the point defined by $$\vec{a}$$ is.