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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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