Is angular momentum conserved in a non-inertial frame?

[Leo Liu]

Question: If we place the frame of reference on an accelerating point, does the total rotational momentum still remain the same?

I attempted to solve this question by manipulating the equations as shown below.
$$\text{Define that }\vec r_i=\vec R+\vec r_i'\text{, where r is the position vector starting from a noninertial frame, R is the position of CM in that frame, r' is the distance between the particle to the CM}$$
$$\text{According to Konig theorem: }\vec\tau=\vec R\times M\ddot{\vec R}+\sum_i \vec r_i'\times m_i\ddot{\vec r}_i'$$
$$\text{Since the position vector does not depend on the accelerating frame, it is a constant, and we obtain:}$$
$$\vec\tau=\vec R\times M\ddot{\vec R}+\vec{Constant}$$
It seems that as long as the equation above equals to 0, the rotational momentum of this system is conserved.
Does this imply that the frame doesn't have to be noninertial to make the rotational momentum conserved?

Thank you and merry early Christmas!

 

[A.T.]

Question: If we place the frame of reference on an accelerating point, does the total rotational momentum still remain the same?

Are you asking about the total angular momentum around the total center of mass, or around some fixed point in the accelerating reference frame?

 

[Leo Liu]

Are you asking about the total angular momentum around the total center of mass, or around some fixed point in the accelerating reference frame?

Around the latter.

 

[Vanadium 50]

You don't seem to be asking if it is conserved. You seem to be asking if it is invariant: the same for all observers.

Energy, for example, is conserved, but is not invariant, even in inertial frames. Angular momentum is the same way.

 

[A.T.]

Around the latter.

The angular momentum around some accelerating point is definitely not conserved. Consider some trivial cases with a single point mass.

 

[Leo Liu]

Angular momentum is the same way.

Could you explain this point using equations?

The angular momentum around some accelerating point is definitely not conserved.

I am sorry. I think I failed to express my idea well. What I meant was that if we had an accelerating frame, whether the angular momentum of the orange body below would still be conserved (measured using the accelerating point). If not, what's wrong with my derivation?

 

[A.T.]

I am sorry. I think I failed to express my idea well. What I meant was that if we had an accelerating frame, whether the angular momentum of the orange body below would still be conserved (measured using the accelerating point).

View attachment 274218

Just compute the AM for this case in the accelerating frame. Is it constant?

 

[Leo Liu]

Just compute the AM for this case in the accelerating frame. Is it constant?

Well I don't know. If it satisfies $\vec R\times M\ddot{\vec R}=\vec C'$ then yes, otherwise no.

 

[A.T.]

Well I don't know.

Why don't you just apply the definition of AM?

https://en.wikipedia.org/wiki/Angular_momentum#Definition_in_classical_mechanics

 

[Leo Liu]

Why don't you just apply the definition of AM?

https://en.wikipedia.org/wiki/Angular_momentum#Definition_in_classical_mechanics

Okay. I will give it a shot on a single particle as opposed to on an extended body.
$$\vec L=\vec r\times m\dot{\vec r}$$
$$\frac{d\vec L}{dt}=\vec r\times m\ddot{\vec r}+\dot{\vec r}\times m\dot{\vec r}=0$$
$$\vec r\times m\ddot{\vec r}=0$$
$$\text{So I suppose there are two ways that can make the equation above valid, and they are:}$$
$$\begin{cases}
\text{The acceleration of the frame is along the same line as the position vector between the frame and the paricle.}\\
\\
\text{Neither does the frame nor the particle accelerates in another inertial frame.}\\
\end{cases}$$
So I have proved that unless their motions are in 1D, the system's rotational momentum measured about the accelerating frame is not conserved.

Thank you for your hint. Is my conclusion correct?

 

[Vanadium 50]

ot conserved.

Not invariant.

 

[Leo Liu]

Not invariant.

Could you tell me what you meant by "noninvariant"? Does it have something to do with symmetry?

 

[A.T.]

So I suppose there are two ways that can make the equation above valid, and they are:

The acceleration of the frame is along the same line as the position vector between the frame and the paricle.

You seem to conflate frame of reference and point of reference (also in your picture). A frame of reference accelerates in a direction. A point accelerates along a line. But I think you have the right idea here.

 

[Vanadium 50]

Could you tell me what you meant by "noninvariant"?

See message #4.

 

[jbriggs444]

In a accelerating frame one has fictitious forces. If the acceleration is linear, so is the fictitious force. The fictitious force applied to an object is given by the mass of the object times the acceleration of the frame.

Corresponding to the fictitious force is a fictitious torque. The moment arm of the fictitious torque on an object is given by the offset of the position of the center of mass of the object from the reference axis in a direction perpendicular to the acceleration of the frame.

If the acceleration is non-zero, and the mass of the object is non-zero and the offset of the object's center of mass from the reference axis is non-zero then angular momentum will not be constant. It will not be conserved.

If, instead, one is thinking in terms of acceleration shifting perspective from one inertial reference frame to another inertial frame that is infinitesimally different then one might prefer to say that the angular momentum is not the same in the two frames. It is not invariant.

In this case it is two words and two different ways of looking at the same thing.