Whats the equation for rotational momentum?

In summary, when calculating angular momentum, you must consider the point of origin. For an object rotating on its own axis, the angular momentum is given by L=Iw, where I is the moment of inertia. For an object rotating around another point, the angular momentum is given by L=rxp, where r is the distance from the point of origin to the center of mass and p is the linear momentum. If an object is rotating on both its own axis and around another point, the total angular momentum is the sum of these two terms. The method of calculating the moment of inertia depends on the chosen point of origin, but the two methods are equivalent.
  • #1
toesockshoe
265
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.
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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?
 

1. What is rotational momentum?

Rotational momentum is a measure of the amount of rotational motion an object possesses. It is determined by the mass and distribution of mass of an object and its rotational velocity.

2. What is the equation for rotational momentum?

The equation for rotational momentum is L = Iω, where L represents rotational momentum, I represents moment of inertia, and ω represents angular velocity.

3. How is rotational momentum different from linear momentum?

Rotational momentum is a measure of rotational motion, while linear momentum is a measure of translational motion. Rotational momentum is dependent on an object's distribution of mass and rotational velocity, while linear momentum is dependent on an object's mass and linear velocity.

4. What are the units for rotational momentum?

The units for rotational momentum are kg*m^2/s. This can also be written as kg*m^2/s^2 * rad, where rad represents radians.

5. How is rotational momentum conserved?

Rotational momentum is conserved when there is no external torque acting on an object. This means that the total amount of rotational momentum remains constant, even if the object's shape or orientation changes.

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