alleryStart learning →If all of the 5.26 x 10^5 J of rotational energy in a helicopter rotor could be used to lift the 1000 kg machine, how high would it rise?
A26.8 m
B53.7 m
C107 m
D5.37 m
Answer & Solution
Correct answer: B. 53.7 m
1. At the top of the climb all the rotational energy has become gravitational potential energy.
2. Setting one half I omega squared equal to m g h gives h = KE / (m g).
3. The stored energy is KE = 5.26 x 10^5 J.
4. The denominator is m g = (1000 kg)(9.80 m/s^2) = 9800 N.
5. So h = (5.26 x 10^5 J) / (9800 N) = 53.7 m.
6. Halving the energy again gives 26.8 m, a double count of the one half already inside KE.
7. The answer 5.37 m drops a factor of ten in the arithmetic.
_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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