Answer to Question #193719 in Mechanics | Relativity for Rocky Valmores

Question

  A potter’s wheel—a thick stone disk of radius 0.500 m and mass 100 kg—is freely rotating at 50.0 revs/min. The potter can stop the wheel in 6.00 s by pressing a wet rag against the rim and exerting a radially inward force of 70.0 N. Find the effective coefficient of kinetic friction between wheel and rag.

Answer 1

$m = 100kg$

$R = 0.5m$

$N = 70$

$\omega_i = 50(\dfrac{rev}{min})(\dfrac{2\pi rad}{1 rev})(\dfrac{1min}{60s}) = 5.24rad/s$

$\omega_f = 0$

$t = 6s$

The moment of inertia of a solid disk,

$I = \dfrac{1}{2}mR^2$

After substituting the given values:

$I = \dfrac{1}{2}(100)(0.5)^2 = 12.5kgm^2$

Using the equation:

$\omega_f = \omega_i + \alpha t$

$\alpha = \dfrac{0-5.24}{6} = -0.873rad/s^2$

Now, using $\tau = I \alpha$

Putting values of $\alpha$ and $I$

$\tau = (12.5)(-0.873) = -10.9$

The magnitude of torque due to kinetic frictional force is:

$\tau = fR$

$f = \dfrac{\tau}{R}$

$f = \dfrac{-10.9}{0.5} = -21.8$

The kinetic friction force $f = \mu_kN$

Hence,

$\mu_k = \dfrac{f}{N}$

$\mu_k= \dfrac{21.8}{70}= 0.311$