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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)
= 5.6 rad/s
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