alleryStart learning →A helicopter rotor of moment of inertia 1067 kg m^2 turns at 300 rpm. What is its rotational kinetic energy?
A5.26 x 10^5 J
B4.80 x 10^7 J
C1.60 x 10^5 J
D2.00 x 10^5 J
Answer & Solution
Correct answer: A. 5.26 x 10^5 J
1. Convert the spin rate first, since the energy formula needs rad/s.
2. omega = (300 rev/min)(2 pi rad/rev)(1 min / 60 s) = 31.4 rad/s.
3. Rotational kinetic energy is one half I omega squared.
4. Substituting, KE = (0.5)(1067 kg m^2)(31.4 rad/s) squared = (0.5)(1067)(986).
5. That gives KE = 5.26 x 10^5 J.
6. Using 300 rpm without converting gives about 4.80 x 10^7 J, roughly a hundred times too large.
_Source: OpenStax College Physics (CC BY 4.0), Ch 10 "Rotational Motion and Angular Momentum", section 10.4 Rotational Kinetic Energy: Work and Energy Revisited_
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