**Problem Statement:**A potter’s wheel is a uniform solid disk of radius \( R = 0.300 \, \text{m} \) and mass \( M = 120 \, \text{kg} \). It is initially rotating at \( 12.0 \, \text{rad/s} \). The potter can stop the wheel in \( 6.00 \, \text{s} \) by pressing a wet rag against the rim exerting a normal force (radially inward) of \( 68.0 \, \text{N} \). The effect of pressing with the rag produces a *frictional torque* that brings the wheel to a stop. What is the frictional torque that acts on the wheel while it is slowing down?**Options:**- \( -125 \, \text{N} \cdot \text{m} \)- \( -25.5 \, \text{N} \cdot \text{m} \)- \( -50.0 \, \text{N} \cdot \text{m} \)- \( -12.5 \, \text{N} \cdot \text{m} \)- \( -10.8 \, \text{N} \cdot \text{m} \)

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Answered: A potter's wheel is a uniform solid…

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