Two Conceptual questions about rotational mechanics

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SUMMARY

The discussion centers on the implications of a non-slipping cord on a pulley in rotational mechanics, specifically in the context of a sphere rotating around a vertical axis. It is established that the assumption of a non-slipping cord allows for the simplification of calculations by ignoring friction. Additionally, the final speeds of the cord and the mass are confirmed to be the same due to the tension created by the mass and the moment of inertia of the sphere, ensuring that the system behaves consistently.

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babayevdavid
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Hi,

I see in many of these physics problems it is explicitly stated that "the cord on the pulley does not slip." [1] What is the implication of this? Why is it important?

Also, there is a problem here with a diagram that I will try to explain in words and then have a question about. (Assuming frictionless axes) There is a sphere rotating around a vertical axis through its center. Around its equator is tied some massless cord that continues horizontally from the sphere onto a pulley (I and radius of the sphere and pulley are given). The cord hangs over the pulley and at the end of the cord is a mass. If the mass were to be dropped a certain distance vertically from rest, and I wanted to find its final speed, and I used the sum of all kinetic energy present = the initial potential energy of the mass at rest, [2] is it okay to substitute the omegas in that setup for V/r ? Why is that okay? It is a valid assumption that the cord will have the same final speed throughout the system?

Thanks in advance!

-David
 
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The bit about the rope not slipping is usually so that you can ignore friction.

It is a valid assumption that the cord will have the same final speed throughout the system

Yes. You have two things keeping it in tension, the mass pulling in one direction and the moment of inertial of the sphere "pulling" in the other. If it doesn't stretch the rope will all be going at the same speed.
 

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