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# Answer to Question #114143 in Mechanics | Relativity for Feker

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?
1
Expert's answer
2020-05-12T09:33:01-0400

The Newton's second law says

"\\tau=I\\alpha"

The moment of inertia of a disk

"I=\\frac{1}{2}mR^2=\\frac{1}{2}\\times 0.9\\times 0.08^2=2.88\\times 10^{-3}\\:\\rm kg\\cdot m^2"

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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