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Question

# A potter’s wheel having a radius of 0.50 m and a moment of inertia of 12 $k g \cdot m ^ { 2 }$ is rotating freely at 50 rev/min. The potter can stop the wheel in 6.0 s by pressing a wet rag against the rim and exerting a radially inward force of 70 N. Find the effective coefficient of kinetic friction between the wheel and the wet rag.

Solution

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We are given following data for potter’s wheel:

$r=0.5\text{ m}$

$I=12\text{ kg m}^2$

$t=6\text{ s}$

$F=70\text{ N}$

$\omega=50\frac{\text{ rev}}{\text{ min}}=0.833\frac{\text{ rev}}{\text{ s}}$

Angular acceleration is equal to:

$I=2\pi\cdot \dfrac{\omega}{t}=2\pi\cdot \dfrac{0.833}{6}=0.87\frac{\text{ rad}}{\text{ s}^2}$

Torque is given by:

$\tau=\mu\cdot F\cdot r=I\cdot \alpha$

Solving it for effective coefficient of kinetic friction:

$\mu=\dfrac{I\cdot \alpha}{F\cdot r}=\dfrac{12\cdot 0.87}{70\cdot 0.5}=\boxed{ 0.3}$

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