Question

A potter's wheel having a radius of 0.500 m and mass 45.0 kg is rotating freely at 50.0 rev/min in a clockwise direction. The potter can stop the wheel by pressing a wet rag against the outside rim of the wheel and exerting a radially inward force of 70.0 N. Since this force is in the radial direction, it alone cannot cause the wheel to slow its spinning. What can happen, however, is that a normal force can result from this inward force which results in a constant kinetic friction force. The friction force produces a constant torque that, in turn, causes the wheel to slow down. a) Assume that the wheel is a disk. Compute its rotational inertia. b) Suppose that after the wet rag is applied to the rim, it takes 6.00 seconds for the wheel to come to a stop. What is the angular acceleration of the wheel? c) What net torque was exerted on the wheel, d) What normal force was exerted? e) What was the coefficient of kinetic friction between the wheel and the rag?

Answer 1

Given:-

Radius,r=0.5m

Mass,m=45 kg

Initial angular speed,w_i=50rev/mt= 5.23 rad/s                    [1rev/mt= 2*pie/60 rad/s]

Force,F=70N

a]

Moment of Inertia of disk= mr²= 45*0.5²=11.25 kg m²

b]

Time,t=6s

W_f= 0 rad/s

Use the kinematic equation, $w_{f}=w_{i}+\alpha t$

$\alpha =\frac{w_{f}-w_{i}}{t}$

?=0-5.23/6= -0.872 rad/s²

c]

Net Torque= Moment of inertia* angular acceleration

$\tau =I*\alpha$

Torque,T= 11.25 kg m² *-0.872 rad/s² = -9.81Nm

d]

Normal force, F_n= mv²/r= mrw² = 45*0.5* 5.23 =117.7N

e] Coeffieint of kinetic friction between the wheel and the rag is, $\mu =\frac{v^{2}}{rg}=\frac{rw^{2}}{g}$

$\mu =\frac{0.5*5.23^{2}}{9.8}=\mathbf{1.4}$