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Physics / Chapter 10: Rotation / Section 10-5

Kinetic Energy of Rotation — Question 5

Section 10-5 · Physics pages 1460–1461

Question 5 Opposite spins do not cancel kinetic energy Two independent coaxial rotors have inertias 0.20 and 0.30 kg m2 and angular velocities +10 and −10 rad/s. I1 = .20, ω1 = +10 I2 = .30, ω2 = −10 Two independent coaxial rotors have opposite signed rates; SI values are shown. (a) Find their total kinetic energy. (b) A student adds signed angular velocities and obtains zero energy. Explain the error and find the energy if only the second rotor stops.
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Question 5 - Solution Opposite spins do not cancel kinetic energy I1 = .20, ω1 = +10 I2 = .30, ω2 = −10 (a) Derivation and calculation Add the energies of the independent degrees of freedom: K = K1 + K2 = 1 2I1ω2 1 + 1 2I2ω2 2. K = 1 2(0.20)(10)2 + 1 2(0.30)(−10)2 = 10 + 15 = 25 J . (b) Result and interpretation The rates do not cancel in a scalar energy sum. There is no common rigid-body rate for these independent rotors. If rotor 2 alone stops, Kf = 1 2(0.20)(10)2 + 1 2(0.30)(0)2 = 10 J . The energy removed is 25 −10 = 15 J. Physical check Reversing either rotor alone leaves its squared speed and kinetic energy unchanged.

Question 5

Opposite spins do not cancel kinetic energy

Two independent coaxial rotors have inertias 0.200.20 and 0.30kgm20.30\,\mathrm{kg\,m^2} and angular velocities +10+10 and −10rad/s-10\,\mathrm{rad/s}.

Diagram shown in the original SVG worksheet.

  1. Find their total kinetic energy.

  2. A student adds signed angular velocities and obtains zero energy. Explain the error and find the energy if only the second rotor stops.

Question 5 - Solution

Opposite spins do not cancel kinetic energy

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Add the energies of the independent degrees of freedom: K=K1+K2=12I1ω12+12I2ω22.K=K_1+K_2=\tfrac12I_1\omega_1^2+\tfrac12I_2\omega_2^2. K=12(0.20)(10)2+12(0.30)(−10)2=10+15=25J.\boxed{K=\tfrac12(0.20)(10)^2+\tfrac12(0.30)(-10)^2=10+15=25\ \mathrm J}.

(b) Result and interpretation

The rates do not cancel in a scalar energy sum. There is no common rigid-body rate for these independent rotors. If rotor 2 alone stops, Kf=12(0.20)(10)2+12(0.30)(0)2=10J.\boxed{K_f=\tfrac12(0.20)(10)^2+\tfrac12(0.30)(0)^2=10\ \mathrm J}. The energy removed is 25−10=1525-10=15 J.

Physical check

Reversing either rotor alone leaves its squared speed and kinetic energy unchanged.

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