Angular Momentum: Conditions & Answers

AI Thread Summary
The discussion centers on the conditions for applying the angular momentum formula L = Iω, which can be used around any axis, provided the correct moment of inertia (I) for that axis is used. It highlights that I is not a simple scalar but a tensor, varying with the axis of rotation due to mass distribution. The conversation draws parallels between angular momentum and linear momentum, emphasizing that while mass is constant, moment of inertia varies based on the object's shape and how mass is distributed. Specific examples, such as the rotation of a wooden pole and ballet dancers, illustrate how mass proximity to the rotation axis affects ease of rotation. Understanding these principles is crucial for accurately applying angular momentum concepts in classical mechanics.
enippeas
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First of all, sorry for my poor english
I am studying classical mechanics and have a problem with angular momentum. I am looking many books but i can't find an exactly answer. Under what conditions can i apply L = I ω; So far i understand that i can apply this formula around either the cm or a fixed point. Is that correct; Is there any other point; Thanks
 
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That is the definition of angular momentum for an object.

You only need to choose which is the REFERENCE AXIS for the rotations considered, because the inertia momentum I is not simply a scalar quantity, but is more complicated than that. Technically it's a tensor. It means that the value of I is different depending on which directions you are looking at it.

As a help, notice the great similarity between L=Iw (angular momentum) and p=mv (momentum). They are indeed the same thing, except that the first is for rotations and the second for traslations.

However, while m (mass) is just one simple value that works whatever the direction of the movement, I is not as simple.

Hint: think of an object with a long shape, such a wooden pole. Trying to make it spin requires an easier effort if you'r spinning it around its own axis rather than trasversally. That is represented by the fact that I' (referring to the pole's axis) is smaller than I'' (referring to an axis perpendicular to the pole and crossing the pole's middle point). If you try to rotate the pole horizontally and by holding one extremity rather than the middle point, it's even more difficult: I''' (ref to an axis perpendicular to the pole but crossing one of the extremity) is even greater.

To get the same rotational speed, you need a bigger effort L when I is greater.

Conversely, with equal effort L you speed up more an object that has a small I.

I depends on how the object's mass is distributed, relative to the rotation axis. The closer the mass is located, or "lumped" around the axis, the smaller I and hence the "easier" the rotation. Typical example of this is the ballet dancers, who first spin themselves with arms extended, then they raise their arms above their head (this way, they move part of their body mass closer to the rotational axis) and get a faster rotation.
 
Welcome to PF!

enippeas said:
Under what conditions can i apply L = I ω; So far i understand that i can apply this formula around either the cm or a fixed point. Is that correct; Is there any other point; Thanks

Hi enippeas ! Welcome to PF! :smile:

L = I ω is valid about any axis.

(but not about a point)

But of course we must use the I for that axis, which will always be larger than the I for the parallel axis through the c.o.m. :smile:

From the PF Library:
The parallel axis theorem:

The Moment of Inertia of a body about an axis is

I = (I_C\,+\,md^2)

where m is the mass, d is the distance from that axis to the centre of mass, and I_C is the Moment of Inertia about the parallel axis through the centre of mass.

(oh, and it's "I have looked in many books, but I couldn't find an exact answer." :smile: )
 
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