## introductory rotational dynamics/energy conservation question

1. The problem statement, all variables and given/known data
Hey guy's im kind of struggling with this i would appreciate any help

A potter's wheel having a radius 0.49 m and a moment of inertia of 12.1 kg · m2 is rotating freely at 52 rev/min. The potter can stop the wheel in 6.0 s by pressing a wet rag against the rim and exerting a radially inward force of 74 N. Find the effective coefficient of kinetic friction between the wheel and the wet rag.

2. Relevant equations

kinematics
Wnc = ΔRotational Energy

3. The attempt at a solution

Ok here so i started this one trying to do an energy conservation 74Δx(or θ) + uknΔx = (1/2)iω^2

To represent the non conservative forces on one side and the change in rotational kinetic energy on the other, Δx being the stopping distance that can be obtained from kinematics

then i figured there might be a radial component instead of a torque

ƩFr = n + 74 = v^2/r*m

so the radial component gives u an n i can sub back in the conservation equation and it points in the same direction, is this anywhere near right or should i be representing the forces over a torque
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 Mentor You can figure out what the stopping torque was using the info given. You can figure out what kind of *tangential* force would have to act to provide this torque. The coeff of friction relates this tangential force to the radial force (which is normal to the surface).
 Mentor Blog Entries: 1 What force is exerting the torque on the wheel?

## introductory rotational dynamics/energy conservation question

so i guess i would be assuming acceleration is constant get it from α then
(.49)74-.49(Ffric)=Iα
are the radial force and the friction both part of the x dimension? they just get separate equations so
-ukn = ma and n = (v^2)/(r)*m-74

Mentor
Blog Entries: 1
 Quote by kinslow31 so i guess i would be assuming acceleration is constant get it from α then (.49)74-.49(Ffric)=Iα
Only the tangential force exerts a torque.