Is angular momentum conserved in a non-inertial frame?

In summary: It seems that you are asking about angular momentum conservation in an accelerating reference frame.In an accelerating reference 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
  • #1
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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!
 
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  • #2
Leo Liu said:
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?
 
  • #3
A.T. said:
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.
 
  • #4
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.
 
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  • #5
Leo Liu said:
Around the latter.
The angular momentum around some accelerating point is definitely not conserved. Consider some trivial cases with a single point mass.
 
  • #6
Vanadium 50 said:
Angular momentum is the same way.
Could you explain this point using equations?
A.T. said:
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?

1607872737004.png
 
  • #7
Leo Liu said:
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?
 
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  • #8
A.T. said:
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.
 
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  • #10
A.T. said:
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?
 
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  • #11
Leo Liu said:
ot conserved.

Not invariant.
 
  • #12
Vanadium 50 said:
Not invariant.
Could you tell me what you meant by "noninvariant"? Does it have something to do with symmetry?
 
  • #13
Leo Liu said:
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.
 
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  • #14
Leo Liu said:
Could you tell me what you meant by "noninvariant"?

See message #4.
 
  • #15
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.
 

1. What is angular momentum?

Angular momentum is a physical quantity that describes the rotational motion of an object. It is a vector quantity defined as the product of an object's moment of inertia and its angular velocity.

2. What is a non-inertial frame?

A non-inertial frame is a reference frame that is accelerating or rotating. In this type of frame, objects appear to be subject to fictitious forces that are not present in an inertial frame.

3. Is angular momentum conserved in a non-inertial frame?

No, angular momentum is not conserved in a non-inertial frame because fictitious forces, such as the Coriolis force, can cause changes in an object's angular momentum.

4. Why is angular momentum conserved in an inertial frame but not in a non-inertial frame?

In an inertial frame, there are no fictitious forces acting on the object, so the total angular momentum remains constant. However, in a non-inertial frame, fictitious forces can cause changes in angular momentum, making it non-conserved.

5. What is an example of a non-inertial frame affecting angular momentum?

An example of a non-inertial frame affecting angular momentum is when a person is spinning on a rotating platform. The person's angular momentum will appear to change in the non-inertial frame due to the Coriolis force, even though their actual angular momentum remains constant in an inertial frame.

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