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Lesson 29 · Rotational energy and momentum
The same two accounts, spinning
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The same two accounts, spinning

The translation table finishes here. Mass mm becomes moment of inertia II; speed vv becomes angular velocity ω\omega; and the two great bookkeeping quantities of mechanics come across unchanged in shape. Kinetic energy 12mv2\tfrac{1}{2}mv^{2} becomes KErot=12Iω2KE_{rot} = \tfrac{1}{2} I \omega^{2} — KE equals one-half I omega squared — where II is in kg⋅m2\mathrm{kg \cdot m^2}, ω\omega is in rad/s\mathrm{rad/s}, and the answer is in joules, the same joules as everywhere else.

Momentum p=mvp = mv becomes angular momentum L=IωL = I\omega — L equals I omega — in kg⋅m2/s\mathrm{kg \cdot m^2/s}. No ½, no square: momentum never had either, and its rotational twin does not acquire them. Telling the two apart on an exam paper is a units question and nothing more. Joules means energy; kg⋅m2/s\mathrm{kg \cdot m^2/s} means momentum.

The square in the energy is where flywheels earn their keep: double the speed and a wheel banks four times the energy, while its angular momentum merely doubles. That is why a modest steel disk spun fast can smooth a punch press through a stroke that would otherwise stall the motor — and why the same disk is a serious object to be standing beside when a bearing lets go.

IωKE

KErot=12Iω2KE_{rot} = \tfrac{1}{2} I \omega^{2}

  • KErotKE_{rot}= Rotational kinetic energy (energy)
  • II= Moment of inertia (moment of inertia)
  • ω\omega= Angular velocity (angular speed)
Rotational Kinetic Energy solver →
IωL

L=IωL = I \omega

  • LL= Angular momentum (angular momentum)
  • II= Moment of inertia (moment of inertia)
  • ω\omega= Angular velocity (angular speed)
Angular Momentum (L = Iω) solver →