Conservation of Angular Momentum — Worked Solutions

These solutions belong to Conservation of Angular Momentum. Each step states why it is useful. Try the questions first, then use this page to check your reasoning.

Question 1

A rotating system pulls its mass inward with negligible external torque. Which statement is correct?

1. Negligible external angular impulse gives I_iω_i = I_fω_f. Why: Internal shape-changing forces do not change the system total L.

2. Reducing I allows ω to increase while L stays constant. Why: Internal work can change kinetic energy, so energy need not stay constant.

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Question 2

Two coaxial disks exchange angular momentum through contact friction. Which system can have constant total L if bearing torque is negligible?

1. Select both disks and identify contact friction as internal. Why: Each frictional angular impulse is accompanied by an equal opposite transfer to the other disk.

2. With negligible external bearing torque, total L of both disks is constant. Why: Selecting only one disk leaves the other disk’s friction as an external transfer.

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Question 3

At fixed nonzero angular momentum, inertia is halved. How does rotational kinetic energy change?

1. At fixed L, replace ω by L/I in K = ½Iω². Why: This removes angular speed in favor of the conserved quantity.

2. K = L²/(2I), so halving I doubles K. Why: The extra energy must come from internal work or another energy source during the shape change.

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Question 4

A rotating system initially has I_i = 2.0 kg·m² and ω_i = +3.0 rad/s. Its inertia falls to 1.0 kg·m² while an external braking angular impulse of −1.0 kg·m²/s acts. Find ω_f.

1. L_i = I_iω_i = 2.0(3.0) = +6.0 kg·m²/s. Why: The initial configuration supplies the initial angular momentum.

2. Add external angular impulse: L_f = 6.0 − 1.0 = +5.0 kg·m²/s. Why: The external brake means total L of the selected system is not constant.

3. ω_f = L_f/I_f = 5.0/1.0 = +5.0 rad/s. Why: The reduced inertia increases speed even though the system loses angular momentum.

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Question 5

A rotating stool system has effective inertia 2.4 kg·m² with arms out and 0.80 kg·m² with arms in. Video gives angular speeds 2.0 ± 0.10 rad/s before the change and 5.5 ± 0.10 rad/s afterward; treat each quoted interval as possible measurement values and the inertias as exact for this task. Are these data consistent with zero external angular impulse? Determine the sign of any supported transfer and propose a control measurement that could test whether bearing drag explains it.

1. Initial L has possible values 2.4(1.90) to 2.4(2.10), or 4.56–5.04 kg·m²/s. Final L lies between 0.80(5.40) and 0.80(5.60), or 4.32–4.48 kg·m²/s. Why: Use the supplied speed intervals rather than assuming the central values are exact; positive constant inertia scales each interval.

2. The intervals do not overlap. Final L is below initial L even at the closest endpoints, so the data support a negative external angular impulse rather than zero. Why: The largest possible final value, 4.48, is below the smallest possible initial value, 4.56. Increasing angular speed does not by itself establish constant L.

3. As a control, record the same stool coasting for the same duration with a fixed mass configuration and similar rotation speeds. Compare the measured L decrease with the shape-change trial. Why: Keeping configuration fixed separates a background drag trend from the inertia change. Additional uncertainty in real inertia estimates must also be assessed before a real experiment makes the same claim.

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