System of Particles and Rotational Motion · Lesson 14 of 15
Fixed-Axis Angular Momentum and Its Conservation
“Distinguish axial angular momentum from the full vector and explain variable-inertia spinning systems.”
• Derive L_z = Iω from particle contributions. • State when L is parallel to ω. • Derive the variable-I torque law. • Explain spin-rate changes and their energy source.
Start with one particle, then sum
Take the fixed axis as z and an origin O on it. For one particle P, split its position as OP = OC + CP, where C lies on the axis at the centre of its circular path. The CP × mv contribution is along z and has magnitude mr⊥v = mr⊥²ω. The OC × mv contribution is perpendicular to z. Hence even a particle moving in a circle need not have its full angular momentum along the axis when the reference origin is away from its circle’s centre.
For a body whose mass distribution has the appropriate symmetry about the chosen axis, equal opposite particle contributions cancel the transverse components. Then L is along the axis and the vector relation L = Iω is valid. More generally a principal axis also has this property. A body driven around an arbitrary nonprincipal axis can have a nonparallel L. This is a significant limit to the simple analogy with p = mv, where p and v are always parallel for an ordinary particle of positive mass.
Torque changes the axial component
For an asymmetric body held to a fixed axis, support constraints may exert perpendicular torques to keep that axis from tilting. Projecting onto the allowed axis avoids assuming those transverse torques vanish. The complete vector equation remains dL/dt = τ_ext; the one-axis equation concerns only its z component. A zero axial torque guarantees constant L_z, while conservation of the full vector requires the full external torque to vanish.
Changing the mass distribution changes spin rate
On a freely turning swivel chair, extending the arms puts more mass farther from the axis and increases I. The angular speed falls. Bringing the arms inward reduces I and increases angular speed. Skaters, dancers, divers and acrobats use the same principle. For a diver in flight, changing body shape changes the distribution of mass about the spin axis; negligible external torque allows angular momentum to remain approximately constant.
Problem
A system turns at 40 rpm. Folding the arms reduces its total moment of inertia to 2/5 of its original value. Find final speed and energy ratio.
- 1.Angular-momentum conservation gives ω_f = (I_i/I_f)ω_i = (5/2)(40) = 100 rpm.
- 2.K_f/K_i = (I_f/I_i)(ω_f/ω_i)² = (2/5)(5/2)² = 5/2.
- 3.The rotational kinetic energy rises by 150% of its initial value. The child does internal work, converting chemical energy into mechanical energy while moving the arms inward.
- 4.If 2/5 refers only to the child’s I rather than the total including the turntable, add the unchanged turntable inertia before calculating the ratio. The simple 100 rpm answer assumes the stated factor applies to total rotating inertia or that platform inertia is negligible.
Angular-momentum conservation does not mean energy must be constant. It states the absence of an external angular impulse about the relevant axis. Internal work can change rotational energy, and friction can dissipate energy while transferring angular momentum to the surroundings. Always specify system boundaries before deciding which quantity is conserved.
Quiz
For an arbitrary rotation axis, L is necessarily parallel to ω?
A general body can have transverse angular momentum; the simple vector relation requires an appropriate axis.
The always applicable fixed-axis scalar relation is what?
Summing the axial particle contributions gives I_zω.
If total I halves with negligible axial torque, ω changes how?
Iω is conserved, so ω changes inversely with I.
At conserved L, reducing I changes K how?
K = L²/(2I) for the simple spin model.
A child pulls arms inward and gains rotational energy. The source is what?
Internal muscular work supplies the energy. The correct choice is the internal-work option.
Key Takeaways
• L_z = I_zω concerns the axis component; vector parallelism requires an appropriate axis. • Use d(Iω)/dt when the mass distribution changes. • Conserve angular momentum only for the stated system and torque condition. • A spin-rate increase can draw energy from internal work.