A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is fre

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Question Description

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 230 N applied to its edge causes the wheel to have an angular acceleration of 0.818 rad/s2

(a) What is the moment of inertia of the wheel? 

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Explanation & Answer

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(1) τ = Iα 
(2) τ = rF 

Substituting (2) into (1): 

rF = Iα 

Solved for I: 

I = rF / α 
= (0.330m)(230N) / 0.818rad/s² 
= 62.1 kg∙m² 

Then, the moment of inertia for a solid uniform cylinder is: 

I = 0.5mr² 
m = 2I / r² 
= 2(62.1kg∙m²) / (0.330m)² 
= 1141kg 

Please let me know if you need any clarification. I'm always happy to answer your questions.

Sngva Zraunm (1755)
Boston College

Anonymous
I was having a hard time with this subject, and this was a great help.

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