Is Angular Momentum Dependent on Choice of Origin?

In summary, the formula for angular momentum of a particle with mass m at time t and displacement vector r(t) relative to origin O is given by ##\vec l=\vec r\times m\vec v##. The angular momentum does depend on the choice of origin, as shown by the difference in angular momentum when changing the origin. However, if the translation is independent of time, the magnitude and direction of the velocity vector remain unchanged and the angular momentum does not change.
  • #1
rtsswmdktbmhw
38
2

Homework Statement


At time t, particle with mass m has displacement ##\vec r (t)## relative to origin O. Write a formula for its angular momentum about O and discuss whether this depends on choice of origin.

The second part is what I'm more unsure of.

Homework Equations

The Attempt at a Solution


##\vec l=\vec r\times\vec p## so at time t, ##\vec l=\vec r(t)\times m\vec v##

For the second part, I wasn't sure what to do. I tried and got as far as this:
##\vec l_2=\vec r_2\times m\vec v_2##
Then ##\vec l-\vec l_2=(\vec r\times m\vec v)-(\vec r_2\times m\vec v_2)##
What now?
 
Physics news on Phys.org
  • #2
The point you're missing is that when you change the origin, the velocity doesn't change(of course if the translation is independent of time):
[itex]
\vec v '=\frac{d}{dt}\vec r'=\frac{d}{dt}(\vec r+\vec a)=\frac{d}{dt}\vec r=\vec v
[/itex]
 
  • #3
What happens if you try to describe the displacement of the particle r(t) from other point as a combination of motions (using the relative movement).

Something like this:
From origin (point O) you know the displacemente is r(t)
From other origin (point A) you know the displacement is rA(t)
Point A does not moves relative to Oringin O

So r(t) can be described also as OA + rA(t)
What does this say to you?
 
  • #4
I don't understand what you're asking!
 
  • #5
Shyan said:
The point you're missing is that when you change the origin, the velocity doesn't change(of course if the translation is independent of time):
[itex]
\vec v '=\frac{d}{dt}\vec r'=\frac{d}{dt}(\vec r+\vec a)=\frac{d}{dt}\vec r=\vec v
[/itex]
Hm, well the magnitude doesn't change but wouldn't the direction with respect to the second origin be different? But, I gave it a shot (below).

tonyxon22 said:
What happens if you try to describe the displacement of the particle r(t) from other point as a combination of motions (using the relative movement).

Something like this:
From origin (point O) you know the displacemente is r(t)
From other origin (point A) you know the displacement is rA(t)
Point A does not moves relative to Oringin O

So r(t) can be described also as OA + rA(t)
What does this say to you?
Okay, so let ##\vec r_2=\vec r+\vec a## and ##\vec v_2=\vec v## then ##\vec l_2=(\vec r+\vec a)\times m\vec v## and so ##\vec l_2-\vec l=(\vec r+\vec a)\times m\vec v- \vec r\times m\vec v=\vec a\times m\vec v## so depends on where the origin is with respect to the first origin?
 
  • Like
Likes tonyxon22
  • #6
rtsswmdktbmhw said:
Hm, well the magnitude doesn't change but wouldn't the direction with respect to the second origin be different? But, I gave it a shot (below).
No, both the direction and the magnitude of the velocity vector remain unchanged. My proof was vectorial!
rtsswmdktbmhw said:
Okay, so let r⃗ 2=r⃗ +a⃗ \vec r_2=\vec r+\vec a and v⃗ 2=v⃗ \vec v_2=\vec v then l⃗ 2=(r⃗ +a⃗ )×mv⃗ \vec l_2=(\vec r+\vec a)\times m\vec v and so l⃗ 2−l⃗ =(r⃗ +a⃗ )×mv⃗ −r⃗ ×mv⃗ =a⃗ ×mv⃗ \vec l_2-\vec l=(\vec r+\vec a)\times m\vec v- \vec r\times m\vec v=\vec a\times m\vec v so depends on where the origin is with respect to the first origin?
That's true but because in general, [itex] \vec v [/itex] changes from point to point in both direction and magnitude, you can say what happens so in general Angular momentum does change in a time independent translation of origin.
 
  • #7
Ok, thank you very much for the help!
 

1. What is angular momentum?

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

2. What is the angular momentum problem?

The angular momentum problem is a theoretical issue in physics that arises when trying to explain the origin of the angular momentum of celestial bodies, such as planets and stars. It is also known as the "angular momentum catastrophe" or the "missing angular momentum problem."

3. How is the angular momentum problem related to the formation of the solar system?

The angular momentum problem arises when trying to explain the angular momentum of the solar system, which is much higher than what can be accounted for by the current theories of planetary formation. This has led to various proposed solutions and theories, such as the "giant impact hypothesis" and the "dust cloud collapse model."

4. What are some proposed solutions to the angular momentum problem?

Some proposed solutions to the angular momentum problem include the "giant impact hypothesis," which suggests that the early solar system experienced a series of collisions that redistributed angular momentum, and the "dust cloud collapse model," which proposes that the solar system formed from a spinning cloud of gas and dust.

5. Why is solving the angular momentum problem important?

Solving the angular momentum problem is important because it can provide insight into the formation and evolution of the solar system and other celestial bodies. It can also help refine our understanding of the laws of physics and how they apply to large-scale systems.

Similar threads

Replies
3
Views
803
  • Introductory Physics Homework Help
Replies
9
Views
636
  • Introductory Physics Homework Help
Replies
12
Views
832
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
817
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
Back
Top