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

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Jun 11th, 2015

Thank you for the opportunity to help you with your question!

Torque may be expressed in two useful ways for this problem: 

(1) τ = Iα 
(2) τ = rF 

Substituting (2) into (1): 

rF = Iα 

Solved for I: 

I = rF / α 
= (0.330m)(250N) / 1.090rad/s² 
= 8.175 kg∙m² 

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

I = 0.5 mr² 
m = 2I / r² 
= 2(8.175kg∙m²) / (0.330m)² 
= 49.55 kg 

Its angular velocity after 5.1s is: 

ω = ω₀ + αt 
= 0 + (1.090rad/s²)(5.1s) 
= 5.6 rad/s 

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 11th, 2015

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