Rotational Equilibrium and Rotational Dynamics problem

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SUMMARY

The discussion centers on calculating the effective coefficient of kinetic friction between a potter's wheel and a wet rag. The wheel, with a radius of 0.50 m and a moment of inertia of 12 kg m², is initially rotating at 50 revolutions per minute. The potter applies a radial force of 70 N to stop the wheel within 6.0 seconds. By applying the angular version of Newton's second law, the relationship between torque, moment of inertia, and angular acceleration is established to derive the coefficient of kinetic friction.

PREREQUISITES
  • Understanding of angular motion and Newton's second law
  • Knowledge of moment of inertia calculations
  • Familiarity with torque and its relation to rotational dynamics
  • Basic concepts of friction, particularly kinetic friction
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  • Study the application of Newton's second law in rotational dynamics
  • Learn how to calculate torque in rotational systems
  • Research the relationship between normal force and kinetic friction
  • Explore the effects of angular acceleration on rotational motion
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The Question:

A potter's wheel having a radius of 0.50 m and a moment of inertia of 12 kg m^2 is rotating freely at 50 rev/min. The potter can stop the wheel in 6.0 sec by pressing a wet rag against the rim and exerting a radially inward force of 70 N. Find the effective coefficient of kinetic friction between the wheel and the rag.

What I know:

I know with the info given, I must use the angular version of Newton's 2nd law.

Sum (Torque) = I(Moment of Inertia) Alpha (Angular Acceleration).

But I am unsure how this even relates to the coeff. of kinetic friction.

Any help?
Thanks In Advance
 
Physics news on Phys.org
The friction force is the tangential force producing the torque that stops the wheel. How does kinetic friction relate to the normal force, which is the force with which the potter pushes against the rim?
 

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