AP Physics 1 · Lesson 2

Angular momentum and its conservation

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A rigid object spinning about a fixed axis has angular momentum L=IωL = I\omega. A small object moving in a straight line also has angular momentum about a point: L=mvr⊥L = mvr_\perp, where r⊥r_\perp is the perpendicular distance from the point to the line of motion. That value stays the same all along a straight path.

A net torque changes angular momentum: the angular impulse τΔt\tau\Delta t equals ΔL\Delta L, just as impulse changes linear momentum. If the net external torque on a system is zero, its angular momentum is conserved: Iiωi=IfωfI_i\omega_i = I_f\omega_f.

Worked example Easy

A gyroscope rotor with rotational inertia 0.0090 kg·m² spins at 400 rad/s. What is the size of its angular momentum?

  1. 1.8 kg·m²/s
  2. 3.6 kg·m²/sAnswer
  3. 720 kg·m²/s
  4. 1,440 kg·m²/s

How to solve it

  1. For a rigid object about a fixed axis, L=IωL = I\omega.
  2. L=(0.0090)(400)=3.6L = (0.0090)(400) = 3.6 kg·m²/s.

Why each choice is right or wrong

  • A. Incorrect. 1.8 kg·m²/s puts in a factor of one-half, which belongs in the kinetic energy, not in L=IωL = I\omega.
  • B. Correct. L=Iω=(0.0090)(400)=3.6L = I\omega = (0.0090)(400) = 3.6 kg·m²/s.
  • C. Incorrect. 720 kg·m²/s is the number for the rotational kinetic energy, 12Iω2\frac{1}{2}I\omega^2, which is in joules.
  • D. Incorrect. 1,440 kg·m²/s squares the angular speed; angular momentum uses ω\omega once.

Trap Angular momentum conserved does not mean rotational energy conserved

When a spinning object pulls mass inward, LL stays the same, but K=L2/(2I)K = L^2/(2I) rises because I falls. The extra energy comes from the work done pulling the mass in. Do not set 12Iiωi2\frac{1}{2}I_i\omega_i^2 equal to 12Ifωf2\frac{1}{2}I_f\omega_f^2.

Explain Explain it

Task verbs: calculate, explain

To use conservation, choose a system on which no outside torque acts about the axis, write its total angular momentum before and after, and set them equal. When something lands on or grabs a spinning object, add its angular momentum before (mvr⊥mvr_\perp, or zero if it starts at rest) and its share of rotational inertia after (mr2mr^2).

A spinning body that changes shape keeps its angular momentum, so its angular speed changes in inverse proportion to its rotational inertia: halve I and ω\omega doubles.

Worked example Medium

A satellite spins freely at 1.6 rad/s with rotational inertia 450 kg·m². It then folds in two long antenna booms, lowering its rotational inertia to 180 kg·m². No outside torque acts. What is its new angular speed?

  1. 0.64 rad/s
  2. 1.6 rad/s
  3. 2.5 rad/s
  4. 4.0 rad/sAnswer

How to solve it

  1. No outside torque, so angular momentum is conserved.
  2. L=(450)(1.6)=720L = (450)(1.6) = 720 kg·m²/s before and after.
  3. ωf=720/180=4.0\omega_f = 720/180 = 4.0 rad/s.

Why each choice is right or wrong

  • A. Incorrect. 0.64 rad/s flips the ratio of the rotational inertias, which would make the satellite slow down as its mass moves in.
  • B. Incorrect. 1.6 rad/s assumes the spin rate is unchanged, but L stays fixed while I drops, so ω\omega must rise.
  • C. Incorrect. 2.5 rad/s keeps the rotational kinetic energy constant instead of the angular momentum.
  • D. Correct. Iiωi=IfωfI_i\omega_i = I_f\omega_f: (450)(1.6)=(180)ωf(450)(1.6) = (180)\omega_f, so ωf=4.0\omega_f = 4.0 rad/s.

Calculate Your turn

A figure skater spins on smooth ice at 5.0 rad/s with her arms out, giving a rotational inertia of 1.2 kg·m². She pulls her arms in, lowering her rotational inertia to 0.80 kg·m². Treat the ice as exerting no torque about her spin axis. Calculate her new angular speed and how much her rotational kinetic energy increases.

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Compare with a model answer

A full answer does this

  • Gives the new angular speed as 7.5 rad/s
  • Gives the increase in rotational kinetic energy as 7.5 J

Model answer

With no torque about the axis, angular momentum is conserved: (1.2)(5.0)=(0.80)ωf(1.2)(5.0) = (0.80)\omega_f, so ωf=6.0/0.80=7.5\omega_f = 6.0/0.80 = 7.5 rad/s. Her rotational kinetic energy goes from 12(1.2)(5.0)2=15\frac{1}{2}(1.2)(5.0)^2 = 15 J to 12(0.80)(7.5)2=22.5\frac{1}{2}(0.80)(7.5)^2 = 22.5 J, an increase of 7.5 J, supplied by the work her muscles do pulling her arms in.

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