A flywheel rotates at an angular velocity (omega) = 31.4 rad / s. The moment of inertia is 63.6 kg / m². Inhibitory forces act on it. Find the braking moment M, under the action of which the flywheel stops after 20 S.
τ=Iϵ→τ=IΔωΔt=63.6⋅0−31.420=−99.85 (N⋅m)\tau=I\epsilon\to\tau=I\frac{\Delta\omega}{\Delta t}=63.6\cdot\frac{0-31.4}{20}=-99.85\ (N\cdot m)τ=Iϵ→τ=IΔtΔω=63.6⋅200−31.4=−99.85 (N⋅m) . Answer
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