A clay cylinder of radius 20 cm on a potter's wheel spins at a constant rate of 10 rev/s. The potter applies a force of 10 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 that the potter has to deliver to the wheel to keep it rotating at this constant rate?

Given:

• Radius of the clay cylinder, \( r = 20 \) cm = \( 0.20 \) m
• Angular velocity, \( \omega = 10 \) rev/s
• Coefficient of friction, \( \mu = 0.1 \)
• Force applied by the potter, \( F = 10 \) N

Calculations:

First, let's calculate the tangential velocity:

\[ v = r \times \omega \\ v = 0.20 \times 10 = 2 \, \text{m/s} \]

Now, let's calculate the normal force:

\[ \text{Acceleration} = \frac{{v^2}}{{r}} = \frac{{(2 \, \text{m/s})^2}}{{0.20 \, \text{m}}} = 20 \, \text{m/s}^2 \]

Now, let's find the normal force:

\[ \text{Mass} = \frac{{\text{Force}}}{{\text{Acceleration}}} \\ \text{Mass} = \frac{{10 \, \text{N}}}{{20 \, \text{m/s}^2}} = 0.5 \, \text{kg} \] \[ \text{Normal Force} = \text{Mass} \times \text{Acceleration} = 0.5 \, \text{kg} \times 20 \, \text{m/s}^2 = 10 \, \text{N} \]

Finally, let's calculate the frictional force:

\[ \text{Frictional Force} = \text{Coefficient of Friction} \times \text{Normal Force} = 0.1 \times 10 \, \text{N} = 1 \, \text{N} \]

Now, we have the force and velocity, so let's calculate the power:

\[ \text{Power} = \text{Force} \times \text{Velocity} = 10 \, \text{N} \times 2 \, \text{m/s} = 20 \, \text{W} \]