Is it possible to use L= r x p to find angular momentum here

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Homework Help Overview

The discussion revolves around the application of angular momentum formulas, specifically L = r x p and L = I ω, in the context of a rotating wheel in contact with the ground. Participants are exploring the conditions under which these formulas can be applied.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the applicability of L = r x p for a rotating wheel, considering the effects of friction and the wheel's moment of inertia. There is an exploration of whether L = MRV can be used when the object is not rotating.

Discussion Status

The discussion is ongoing, with participants providing insights into the conditions for using different angular momentum formulas. Some guidance has been offered regarding the relationship between rotation and linear motion, but no consensus has been reached on the specific application in this scenario.

Contextual Notes

There is a mention of the wheel's rotation about the instantaneous contact point and the relevance of the parallel axis theorem, indicating that assumptions about the motion and forces acting on the wheel are being examined.

Warlic
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Homework Statement



upload_2015-12-3_13-6-0.png
[/B]

Homework Equations



L = I ω
L= r x p

The Attempt at a Solution



For b)

My proffesor found the moment of inertia at the point at which the wheel touches the ground, and used the formula L = I ω
upload_2015-12-3_13-7-47.png
What I don't understand is why can't one use L= r x p to solve the problem.
Wouldn't L then simply be L=RMV? [/B]
 

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Warlic said:
What I don't understand is why can't one use L= r x p to solve the problem.
Wouldn't L then simply be L=RMV?
If the wheel were sliding without friction along the road then that would be true. Then the wheel could be modeled as a point mass M moving past the point of interest at a perpendicular distance R.

But the wheel is rotating (about the instantaneous contact point on the road). So the wheel's moment of inertia and the parallel axis theorem is called for.
 
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gneill said:
If the wheel were sliding without friction along the road then that would be true. Then the wheel could be modeled as a point mass M moving past the point of interest at a perpendicular distance R.

But the wheel is rotating (about the instantaneous contact point on the road). So the wheel's moment of inertia and the parallel axis theorem is called for.
I see, thank you. So if the object is not rotating, I can use L=MRV, where R is the distance to the center of mass of the object?
 
Warlic said:
I see, thank you. So if the object is not rotating, I can use L=MRV, where R is the distance to the center of mass of the object?
Yes. Then it's just a non-rotating rigid body undergoing linear motion.
 
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