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KE rot = 1 2 2 . size 12{"KE" rSub { size 8{"rot"} } = { {1} over {2} } Iω rSup { size 8{2} } } {}

Both I size 12{I} {} and ω size 12{ω} {} were found above. Thus,

KE rot = 0.5 4.352 kg m 2 5.42 rad/s 2 = 64.0 J . size 12{"KE" rSub { size 8{"rot"} } =0 "." 5 left (4 "." "352"" kg" cdot m rSup { size 8{2} } right ) left (5 "." "42"" rad/s" right ) rSup { size 8{2} } ="64" "." 0" J"} {}


The final rotational kinetic energy equals the work done by the torque, which confirms that the work done went into rotational kinetic energy. We could, in fact, have used an expression for energy instead of a kinematic relation to solve part (b). We will do this in later examples.

Helicopter pilots are quite familiar with rotational kinetic energy. They know, for example, that a point of no return will be reached if they allow their blades to slow below a critical angular velocity during flight. The blades lose lift, and it is impossible to immediately get the blades spinning fast enough to regain it. Rotational kinetic energy must be supplied to the blades to get them to rotate faster, and enough energy cannot be supplied in time to avoid a crash. Because of weight limitations, helicopter engines are too small to supply both the energy needed for lift and to replenish the rotational kinetic energy of the blades once they have slowed down. The rotational kinetic energy is put into them before takeoff and must not be allowed to drop below this crucial level. One possible way to avoid a crash is to use the gravitational potential energy of the helicopter to replenish the rotational kinetic energy of the blades by losing altitude and aligning the blades so that the helicopter is spun up in the descent. Of course, if the helicopter’s altitude is too low, then there is insufficient time for the blade to regain lift before reaching the ground.

Problem-solving strategy for rotational energy

  1. Determine that energy or work is involved in the rotation .
  2. Determine the system of interest . A sketch usually helps.
  3. Analyze the situation to determine the types of work and energy involved .
  4. For closed systems, mechanical energy is conserved . That is, KE i + PE i = KE f + PE f . size 12{"KE" rSub { size 8{i} } +"PE" rSub { size 8{i} } ="KE" rSub { size 8{f} } +"PE" rSub { size 8{f} } } {} Note that KE i size 12{"KE" rSub { size 8{i} } } {} and KE f may each include translational and rotational contributions.
  5. For open systems , mechanical energy may not be conserved, and other forms of energy (referred to previously as OE size 12{ ital "OE"} {} ), such as heat transfer, may enter or leave the system. Determine what they are, and calculate them as necessary.
  6. Eliminate terms wherever possible to simplify the algebra .
  7. Check the answer to see if it is reasonable .

Calculating helicopter energies

A typical small rescue helicopter, similar to the one in [link] , has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it?


Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy.

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Source:  OpenStax, Mechanics. OpenStax CNX. Apr 15, 2013 Download for free at http://legacy.cnx.org/content/col11506/1.2
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