Question #114143

The wheel on the grinder is a uniform 0.9kg disk of 8cm radius it costs uniformly to rest from 1400rpm in a time of 35 sec. How large a friction torque slows its motion?

Expert's answer

The Newton's second law says

"\\tau=I\\alpha"The moment of inertia of a disk

The angular acceleration

"\\alpha=\\frac{\\Delta\\omega}{\\Delta t}=\\frac{1400\/60\\times 2\\pi}{35}=4.19\\:\\rm rad\/s^2"Hence, the torque

"\\tau=2.88\\times 10^{-3}\\times 4.19=0.012\\:\\rm N\\cdot m"
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