Introductory rotational dynamics/energy conservation question

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
kinslow31
Messages
3
Reaction score
0

Homework Statement


Hey guy's I am 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.

Homework Equations



kinematics
Wnc = ΔRotational Energy

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
 
Physics news on Phys.org
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).
 
What force is exerting the torque on the wheel?
 
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
 
kinslow31 said:
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.