Is angular momentum conserved about a moving axis?

Question

Consider a circular body (disc, ring, sphere, etc.) rolling on a flat horizontal surface. I've been told that the angular momentum of a rigid body executing combined rotation and translational motion is given as L(cm) + Mvr, where L(cm) is the angular momentum about the center of mass, and the term Mvr is linear momentum, about a "fixed point" from where the position vector 'r' is being measured. Also, 'v' is the velocity of the center of mass.

I do grasp that the net torque of the rolling body about the "non-fixed" point of instantaneous contact at every instance is zero, and thus its angular momentum is conserved about that "moving" axis of rotation. What I don't get is how we can conserve angular momentum in a moving frame of reference. Angular momentum is measured about a fixed point in the ground frame, and if we just measure its angular momentum at every instance about that moving axis, the term 'position vector of the center of mass,' 'r,' won't hold much significance . Can angular momentum be conserved about any free axis, provided that net torque is zero or is it necessary for it to be a fixed one?. Please share your insights and clarify this.

Answer 1

from the FBD

torques about point A

$$I_c\,\ddot \phi+m\,\dot v\,r=0$$ with the rolling condition $~\dot v=r\,\ddot\phi~$ $$I_c\,\ddot \phi+m\,r\,\ddot\phi\,r=0\quad\Rightarrow\\ \frac{d}{dt}\underbrace{[(I_c+m\,r^2\,)\dot\phi]}_{L}=0$$

thus the angular momentum $~L=~$ constant (conserved)

torques about point C $$I_c\,\ddot\phi-F_c\,r=0$$

with $~F_c=-m\,\dot v=-m\,r\,\ddot\phi$ you obtain that

$$\frac{d}{dt}\underbrace{[(I_c+m\,r^2\,)\dot\phi]}_{L}=0$$

thus the angular momentum about this axis is also conserved .

you can take any point on the disc , you obtain that the angular momentum is conserved

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take arbitrary point on the disk e.g. $~B~$ that moving with $~\phi~$

the sum of the torques about point B is:

$$I_c\,\ddot\phi+m\,\dot v\,r\cos(\phi)-F_c\,(r-r\cos(\phi))=0$$

with $~F_c=-m\,\dot v~,\dot v=r\,\ddot\phi~$ you obtain

$$\frac{d}{dt}\underbrace{[(I_c+m\,r^2\,)\dot\phi]}_{L}=0$$

the angular momentum $~L~$ is conserved

Answer 2

Unlike translational momentum which is descriptive of the whole body, rotational momentum (aka angular momentum) is descriptive of a specific point in space.

When summed about an arbitrary point A that might be fixed to the ground, fixed to the body, or moving in an arbitrary fashion the following relationships describe the rate of change of momentum

$$ \begin{array}{r|l} \text{translational momentum} & \tfrac{{\rm d}}{{\rm d}t} \boldsymbol{p} = \boldsymbol{F} \\ \text{rotational momentum} & \tfrac{{\rm d}}{{\rm d}t} \boldsymbol{L}_A = \boldsymbol{\tau}_A - \boldsymbol{v}_A \times \boldsymbol{p} \\ \end{array} $$

where $\boldsymbol{F}$ is the net force on a body, $\boldsymbol{p}$ is the linear momentum vector, $\boldsymbol{\tau}_A$ is the net torque on the body summed at the arbitrary point, and $\boldsymbol{L}_A$ is the angular momentum vector of the body summed also at the arbitrary point. Finally, $\boldsymbol{v}_A$ is the velocity vector of the arbitrary point.

_{See equation 3.90 in these Dynamics Lecture Notes which uses point B as the arbitrary point.}

What is important is the conditions listed above when $\boldsymbol{\tau}_A = \tfrac{\rm d}{{\rm d}t} \boldsymbol{L}_A$

• The arbitrary point is fixed to the observer
• The arbitrary point is fixed on the body center of mass
• The arbitrary point is moving in a direction parallel to the center of mass motion.

So to answer the question. To conserve angular momentum $\boldsymbol{L}_A$ at some arbitrary point, you must have $\boldsymbol{\tau}_A = \boldsymbol{v}_A \times \boldsymbol{p}$.

Answer 3

your formula is only right, if r is the radius to the touching point otherwise it should be $M*\vec{v}\times\vec{r}$