Whats the equation for rotational momentum?

Click For Summary

Homework Help Overview

The discussion revolves around the concept of angular momentum in physics, specifically the equations L=Iw and L=rxP. Participants are exploring when to apply each equation and the implications of different reference points in calculating angular momentum.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to differentiate between the use of L=Iw for rotational motion about an object's own axis and L=rxP for motion relative to another object's axis. Questions arise about the definitions of moment of inertia and the reference points used in calculations.

Discussion Status

The discussion is ongoing, with some participants providing insights into the relationship between the two equations and the conditions under which they apply. There is an exploration of the parallel axis theorem and its relevance to the problem, indicating a productive exchange of ideas.

Contextual Notes

Participants are considering the implications of different reference points for calculating angular momentum, particularly in relation to the center of mass and fixed origins. There is an acknowledgment of the complexity involved in these calculations.

toesockshoe
Messages
265
Reaction score
2

Homework Statement


I'm confused with when to use L=Iw (inertia times angular speed) for momentum and when to use L=rxP (r cross p) for inertia. Can someone please explain to me what each one is?

Homework Equations


Just a conceptual question.

The Attempt at a Solution


Just a conceptual question.
 
Physics news on Phys.org
Angular momentum is always calculated relative to a point, called the origin. The angular momentum is different depending on what point you choose to consider the origin. So let's take the example of the Earth. The Earth spins on its own axis. That produces an angular momentum relative to the center of the Earth of

[itex]\vec{L}_{spin} = I \vec{\omega}[/itex]

But the Earth also is orbiting around the sun. So if we are computing the origin to be the sun, there is an additional term, the orbital angular momentum, given by:

[itex]\vec{L}_{orbital} = \vec{r} \times \vec{p}[/itex]

The total angular momentum about the sun is the sum of these two terms:

[itex]\vec{L}_{total} = \vec{L}_{spin} + \vec{L}_{orbital}[/itex]

By picking the center to be the center of mass of the object, you can make the orbital angular momentum zero.
 
stevendaryl said:
Angular momentum is always calculated relative to a point, called the origin. The angular momentum is different depending on what point you choose to consider the origin. So let's take the example of the Earth. The Earth spins on its own axis. That produces an angular momentum relative to the center of the Earth of

[itex]\vec{L}_{spin} = I \vec{\omega}[/itex]

But the Earth also is orbiting around the sun. So if we are computing the origin to be the sun, there is an additional term, the orbital angular momentum, given by:

[itex]\vec{L}_{orbital} = \vec{r} \times \vec{p}[/itex]

The total angular momentum about the sun is the sum of these two terms:

[itex]\vec{L}_{total} = \vec{L}_{spin} + \vec{L}_{orbital}[/itex]

By picking the center to be the center of mass of the object, you can make the orbital angular momentum zero.
oh ok, so basically you use L=Iw if an object is rotating on resect to its own axis and use L=rxp if an object is rotating respect to another objects axis (and if an object does both,,,, rotate on its own axis as well as another object) then you add the two correct?
 
toesockshoe said:
oh ok, so basically you use L=Iw if an object is rotating on resect to its own axis and use L=rxp if an object is rotating respect to another objects axis (and if an object does both,,,, rotate on its own axis as well as another object) then you add the two correct?
Sort of. It depends what you mean by I here. If you are defining I as the MoI about the object's mass centre then yes, you need to add the contribution from the linear motion. Alternatively, you can use the parallel axis theorem to find the MoI about the reference axis, then you don't need to add in the linear contribution. The two methods are equivalent.
 
haruspex said:
Sort of. It depends what you mean by I here. If you are defining I as the MoI about the object's mass centre then yes, you need to add the contribution from the linear motion. Alternatively, you can use the parallel axis theorem to find the MoI about the reference axis, then you don't need to add in the linear contribution. The two methods are equivalent.
hey haruspex, you also replied to my other problem where I asked someone to check my solution to a rotations problem. So in that case, for the r component of the orbital angular momentum be the distance from the point to the center of the stick or would it be from the point to the fixed origin (which was where the center of the stick INITIALLY was)... it would be to where the fixed origin (where the center of the stick INITIALLY was) right?
 

Similar threads

Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 71 ·
3
Replies
71
Views
5K
Replies
335
Views
17K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K