Question

A potter's wheel having a radius 0.47 m and a moment of inertia of 14.2 kg · m2 is rotating freely at 53 rev/min. The potter can stop the wheel in 8.0 s by pressing a wet rag against the rim and exerting a radially inward force of 68 N. Find the effective coefficient of kinetic friction between the wheel and the wet rag.

Answer 1

the effective coefficient of kinetic friction between the wheel and the wet rag is 0.31

Step-by-step explanation:

given information:

radius, r = 0.47 m

moment of inertia, I = 14.2 kg
`m^(2)`

angular velocity, ω0 = 53 rev/min = 53 x 2π/60 = 5.5 rad/s

time, t = 8.0 s

inward force, N = 68 N

τ = I α

F r = I α, F is friction force, F = μ N

μ N r = I α

μ = I α / N r

We have to find α

ωt = ω0 + αt. ωt = 0 because the wheel stop after 8 s

0 = 5.5 + α 8

α = -5.5/8 = 0.69
`rad/s^(2)`

Now we can calcultae the coeffcient of kinetik friction

μ = I α / N r

= (14.2) (0.69) / (68) (0.47)

= 0.31