Orbital Angular Momenta - Conversion to a Different Origin

In summary, we have two objects with angular momenta \mathbf{L}_1 and \mathbf{L}^\prime_2 about rotation centers O_1 and O_2 respectively. O_2 has a relative velocity \mathbf{v}_2 to O_1. We want to find the total angular momentum of the system with respect to O_1. In general, we can express this as the sum of the individual angular momenta, but for the specific case of a rotating sphere with its axis parallel to the vector \mathbf{r} \times \mathbf{v}_2, we can simplify the expression to include the total mass and the center of mass.
  • #1
MisterX
764
71
Suppose we have some object with angular momenta [itex]\mathbf{L}_1[/itex] about rotation center [itex]O_1[/itex] and we have another object with angular momentum [itex]\mathbf{L}^\prime_2[/itex] about rotation center [itex]O_2[/itex]. [itex]O_2[/itex] has some velocity [itex]\mathbf{v}_2[/itex] relative to [itex]O_1[/itex]. Then we wonder what is the angular momentum of the second object with respect to [itex]O_1[/itex] so that we may express the total angular momentum of the system. Is there anything to say in general about this?

What about the case where the second object is a sphere, and it is rotating on its axis parallel to [itex]\mathbf{r} \times \mathbf{v}_2[/itex], where [itex]\mathbf{r}[/itex] is a vector from [itex]O_1[/itex] to [itex]O_2[/itex] ? For example consider a rotating sphere in a circular orbit, where the sphere's axis of rotation is parallel to the plane of the orbit. What is the total angular momentum about the center of the circular orbit?
 
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  • #2
I don't know why this didn't occur to me until I took a shower.

[itex]\mathbf{x} = \mathbf{x}^\prime + \mathbf{O}_2[/itex]

[itex][/itex]

[itex]\mathbf{L}_2 = \sum m_i \mathbf{x}_i \times \dot{\mathbf{x}}_i = \sum m_i (\mathbf{x}_i^\prime + \mathbf{O}_2)\times \dot{\mathbf{x}}_i [/itex]

[itex]= \sum m_i (\mathbf{x}_i^\prime + \mathbf{O}_2)\times \left(\dot{\mathbf{x}}^\prime_i + \dot{\mathbf{O}}_2\right)[/itex]

[itex]=\mathbf{L}_2^\prime + \sum m_i \mathbf{O}_2 \times \dot{\mathbf{O}}_2 + \sum m_i \mathbf{x}_i^\prime \times \dot{\mathbf{O}}_2 + \sum m_i \mathbf{O}_2 \times \dot{\mathbf{x}}^\prime_i [/itex]

So at least the first two terms have a nice form.

[itex]=\mathbf{L}_2^\prime + M_{total} \mathbf{O}_2 \times \dot{\mathbf{O}}_2 + \sum m_i \mathbf{x}_i^\prime \times \dot{\mathbf{O}}_2 + \sum m_i \mathbf{O}_2 \times \dot{\mathbf{x}}^\prime_i [/itex]
 

1. What is orbital angular momentum?

Orbital angular momentum (L) is a quantum mechanical property of a particle or system of particles that describes the rotational motion of the particles around a fixed point, known as the origin. It is a vector quantity and is related to the distribution of mass and velocity of the particles.

2. How is orbital angular momentum calculated?

The magnitude of orbital angular momentum is calculated using the formula L = r x p, where r is the position vector and p is the linear momentum of the particle. The direction of the vector is given by the right-hand rule, with the thumb pointing in the direction of the angular velocity and the fingers pointing in the direction of r x p.

3. Can orbital angular momentum be converted to a different origin?

Yes, orbital angular momentum can be converted to a different origin. This is known as the origin shift or translation of the origin. The conversion is done by subtracting the new origin position vector from the original position vector, and then using the new position vector in the calculation of orbital angular momentum.

4. Why is it important to convert orbital angular momentum to a different origin?

Converting orbital angular momentum to a different origin is important because it allows for a more accurate description of the rotational motion of a particle or system. It also allows for easier comparison and analysis of different systems that may have different origins.

5. Are there any limitations to converting orbital angular momentum to a different origin?

Yes, there are limitations to converting orbital angular momentum to a different origin. This conversion is only valid for stationary systems or systems with a constant velocity. It is not applicable for systems with changing velocities or accelerating systems, as it would require additional calculations to account for the changing motion.

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