F=ma 2011 #24: Effects of Increasing Bearing Width & Radius

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The discussion revolves around the effects of increasing the bearing width (δ) and radius (R) on the power required to rotate a turntable supported by a Teflon ring. Participants express confusion about the setup and the implications of the changes, particularly regarding how friction interacts with the dimensions of the ring. It is noted that understanding the friction force's location is crucial for analyzing the problem. The consensus suggests that the diagram should clarify the situation, and prior knowledge of friction principles is assumed. Ultimately, the focus is on how these dimensional changes impact the power requirements without altering other factors.
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Homework Statement



A turntable is supported on a Teflon ring of inner radius R and outer radius R+δ (δ  R), as shown in the diagram.
To rotate the turntable at a constant rate, power must be supplied to overcome friction. The manufacturer of the
turntable wishes to reduce the power required without changing the rotation rate, the weight of the turntable, or
the coefficient of friction of the Teflon surface. Engineers propose two solutions: increasing the width of the bearing
(increasing δ), or increasing the radius (increasing R). What are the effects of these proposed changes?


(A) Increasing δ has no significant effect on the required power; increasing R increases the required power.
(B) Increasing δ has no significant effect on the required power; increasing R decreases the required power.
(C) Increasing δ increases the required power; increasing R has no significant effect on the required power.
(D) Increasing δ decreases the required power; increasing R has no significant effect on the required power.
(E) Neither change has a significant effect on the required power.

http://www.aapt.org/physicsteam/2012/upload/exam1-2011-1-3-answers_1.pdf (number 24)

Homework Equations


P=Fv=Tw


The Attempt at a Solution


I have no idea how to start this... Also I don't understand the situation. Is the turntable like a box on top of the ring? Or do they mean they have a vinyl disc on top of the ring? I have no idea what they're talking about.
 
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What difference does it make if the turntable is a vinyl disk or some sort of box?
What matters is how the friction with the teflon ring depends on the dimensions of the ring.

Anyway - the diagram should give you a hint.
 
I have no idea where the friction force would be located... Would it be concentrated at the middle of the ring width?
 
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