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Question

A potter’s wheel having a radius of 0.50 m and a moment of inertia of 12 kgm2k 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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Answered 2 years ago
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We are given following data for potter’s wheel:

r=0.5 mr=0.5\text{ m}

I=12 kg m2I=12\text{ kg m}^2

t=6 st=6\text{ s}

F=70 NF=70\text{ N}

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

Angular acceleration is equal to:

I=2πωt=2π0.8336=0.87 rad s2I=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:

τ=μFr=Iα\tau=\mu\cdot F\cdot r=I\cdot \alpha

Solving it for effective coefficient of kinetic friction:

μ=IαFr=120.87700.5=0.3\mu=\dfrac{I\cdot \alpha}{F\cdot r}=\dfrac{12\cdot 0.87}{70\cdot 0.5}=\boxed{ 0.3}

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