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

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urfunzzntobby

Science

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## Explanation & Answer

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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)

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

Sngva Zraunm (1755)
Cornell University
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