Question

A clay cylinder of radius 17 cm on a potter's wheel spins at a constant rate of 12 rev/s. The potter applies a normal force of 11.3 N to the clay with his hands where the coefficient of friction is 0.1 between his hands and the clay. What is the power (in W) that the potter has to deliver to the wheel to keep it rotating at this constant rate?

Answer 1

Answer:

0.37 W

Explanation:

e potter provide torque which then causes power developed in the wheel to rotate.

The power delivered, P is found by

$P=\tau\omega$ where $\omega$ is angular speed on the wheel and \tau is torque on the wheel

Frictional force $F_{k}$ developed between the hands of the potter and the wheel is given by

$F_{k}=\mu F_{N}$ where $\mu$ is coefficient of friction and $F_{N}$ is perpendicular force on the wheel caused by the potter’s hands

Torque on the wheel is given by

$\tau=RF_{K}$ where R is radius of the wheel

$\tau=R\mu F_{N}$

To convert the wheel spin rate from rev/s to rads/s

$\omega=\frac {12rev/s}{2\pi rev/rad}$= 1.909859 rad/s

Substituting the above \omega into the equation of power

P=$\omega\tau=\omega R\mu F_{N}$

P=1.909859 rad/s *0.17m*0.1*11.3N=0.366884 W

Power delivered by the potter to the wheel is  0.37W