A potter's wheel having a radius 0.52 m and a moment of inertia 15 kg·m2 is rotating freely at 53 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 73 N.
Find the effective coefficient of kinetic friction between the wheel and the wet rag.
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Verified answer
Torque = iα
F(total) * r=iα
F(friction)=Fn*μ
Normal force = 73 N
Friction *radius = iα
Converting rev/min to rad/sec.
53*2*π/60= 5.55 rad/sec
α=((5.55)/(6)) = .0925 rad/sec^2
friction = 15*.0925/.52 = 2.67 N
F(friction)=Fn*μ
μ=2.67/73 =.0366
I hope this helps.