6.4 Conservation of Angular Momentum

1. Example 1Calculator allowed

   A spinning skater

   A skater spins at 2.0 rev/s with arms out, where I = 3.0 kg·m². She pulls her arms in, reducing I to 1.2 kg·m². Find her new spin rate and compare her kinetic energy before and after.

   Show the solutionHide the solution

   1. 1Step 1: No outside torque (ignoring friction), so $I_i\omega_i = I_f\omega_f$: $\omega_f = \dfrac{(3.0)(2.0)}{1.2} = 5.0$ rev/s. You can stay in rev/s here, because the units cancel.
   2. 2Step 2: For energy you need rad/s: $\omega_i = 4\pi$ rad/s and $\omega_f = 10\pi$ rad/s.
   3. 3Step 3: $K_i = \tfrac{1}{2}(3.0)(4\pi)^2 \approx 237$ J and $K_f = \tfrac{1}{2}(1.2)(10\pi)^2 \approx 592$ J.
   4. 4Step 4: Her kinetic energy rises by a factor of 2.5. Her muscles did work pulling her arms in.

   Answer: 5.0 rev/s; kinetic energy rises from about 237 J to 592 J

2. Example 2Calculator allowed

   Clay hitting a hanging rod (exam-level)

   A uniform rod of mass 1.2 kg and length 0.90 m hangs at rest from a frictionless pivot at its top end. A 0.10 kg lump of clay moving horizontally at 6.0 m/s hits the bottom end and sticks. Find the angular velocity just after the collision and the kinetic energy lost.

   Show the solutionHide the solution

   1. 1Step 1: Conserve angular momentum about the pivot. Before: the clay's $L = mvr_\perp = (0.10)(6.0)(0.90) = 0.54$ kg·m²/s; the rod has none.
   2. 2Step 2: After: $I = \tfrac{1}{3}ML^2 + mL^2 = \tfrac{1}{3}(1.2)(0.81) + (0.10)(0.81) = 0.324 + 0.081 = 0.405$ kg·m².
   3. 3Step 3: $\omega = \dfrac{0.54}{0.405} \approx 1.33$ rad/s.
   4. 4Step 4: Kinetic energy before: $\tfrac{1}{2}(0.10)(6.0)^2 = 1.8$ J. After: $\tfrac{1}{2}(0.405)(1.33)^2 \approx 0.36$ J. About 1.44 J is lost.
   5. 5Step 5: Linear momentum isn't conserved: it's 0.60 kg·m/s before and about 0.84 kg·m/s after, because the pivot pushes on the rod during the hit.

   Answer: ω ≈ 1.33 rad/s; about 1.44 J of kinetic energy is lost

3. Example 3Calculator allowed

   Walking to the edge of a platform

   A uniform disk-shaped platform (mass 100 kg, radius 2.0 m) turns freely at 1.0 rad/s with a 50 kg person standing at its center. The person walks out to the edge. What is the new angular velocity? Treat the person as a point mass.

   Show the solutionHide the solution

   1. 1Step 1: Platform: $I = \tfrac{1}{2}(100)(2.0)^2 = 200$ kg·m². At the center, the person adds nothing to I.
   2. 2Step 2: At the edge, the person adds $mR^2 = (50)(2.0)^2 = 200$ kg·m², so the total is 400 kg·m².
   3. 3Step 3: $\omega_f = \dfrac{(200)(1.0)}{400} = 0.50$ rad/s.

   Answer: 0.50 rad/s

Common mistakes

• !Conserving linear momentum in a collision with a pivoted object. The pivot exerts an outside force, so use angular momentum about the pivot.
• !Assuming kinetic energy is conserved when I changes. It isn't: it rises when I decreases.
• !Forgetting to add the particle's mr² to the rod's I after it sticks.
• !Taking angular momentum about the center of mass instead of the pivot, where the outside force acts.

On the exam

• ✓Justify conservation by naming the system and explaining why the net external torque is zero, for example "the pivot force acts at the axis, so it exerts no torque about it".
• ✓Expect questions that ask whether L, K and linear momentum each increase, decrease or stay the same. Answer each one separately.