Question

This is given as a practice problem for an upcoming exam, can you please explain it with as much detail as possible?

A potter's wheel - a thick stone disk of radius .5m and mass 100kg - is freely rotating at 50 rev/min. The potter can stop the wheel in 6 s by pressing a wet rag against the rim and exerting a radially inward force of 70N.

a. Find the effective coefficient of kinetic friction between the wheel and the rag

b.What is the change in angular momentum of the potter's wheel?

c. If the angular momentum is conserved what happens to the angular momentum that used to be in the potter's wheel?

Answer 1

Initial angular velocity, i = 50 rev/min
= 50 / 60 rev/s = 0.833 rev/s
= 0.833 x 2 rad/s
= 5.235 rad/s
Final angular velocity, f = 0
Time taken, t = 6 s
Using the formula, f = i + t
Where is the angular acceleration.
= (f - i)/t
(0 - 0.833)/6
= - 0.139 rev/s²
Using the formula,
= it + 1/2t²,
Where is the displacement,
= 0.833 x 6 + 0.5 x (-0.139) x 6²
= 2.5 rev
= 2.5 x 2R m
= 2.5 x 2 x 0.5 m
= 7.854 m

Initial kinetic energy = work done by friction.
1/2 Ii² = Ff x distance [Ff is the frictional force]
Where I is the moment of inertia of the disk, I = 1/2 MR²
= 0.5 x 100 x 0.5² = 12.5 kgm2
Ff = [1/2 x 12.5 x (5.235)²] / 7.854
= 21.816 N

Ff = Coefficient of friction x Normal force
Coefficient of friction = Ff/Normal force
= 21.816 / 70
= 0.31

b)
Change in angular momentum = If - Ii
= 12.5 [0 - 5.235]
= - 65.45 kg rad/s

c)
Since there is an external torque applied here, angular momentum is not conserved.