A large industrial flywheel, used to store energy, has a 250 kg*m^2 moment of in
ID: 1911705 • Letter: A
Question
A large industrial flywheel, used to store energy, has a 250 kg*m^2 moment of inertia about its axle. The flywheel is initially spinning at a speed of 52 rad/s (500 rpm) when the power goes out, leaving it subject to a 3.0 N*m net frictional torque.
(a) Will it spin for at least an hour before finally stopping?
(b) Calculate the amount of energy that was initially stored in the flywheel. Show your work.
Explanation / Answer
t = Ia a = t / I The moment of inertia for a solid disk is mr²/2, so: a = t / mr²/2 = 2t / mr² = 2(50N•m) / (250kg)(0.75m)² = 0.71rad/s² Then, the time may be found from the rotational kinematics: ? = ?0 + at-------------------->?0 = 0 t = ?/a The units between angular velocity and angular acceleration need to be the same, so change 1200rpm to rad/s: 1200rpm = 1200rev/min(2prad/1.0rev)(1.0min/60s) = 130rad/s Therefore: t = 130rad/s / 0.71rad/s² = 180s (3.0min)
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