Introduction
Angular momentum of rigid bodies combines contributions from all mass elements. For fixed-axis rotation, it simplifies to L = Iω. The general case requires tensor treatment. Conservation of angular momentum explains gyroscopic effects, figure skater spins, and stability of rotation.
Fixed Axis Rotation
For rotation about fixed principal axis: L = Iω. Direction along axis by right-hand rule. Simple scalar relationship valid when rotation axis is fixed in body and space. Magnitude proportional to both angular velocity and moment of inertia.
General Rigid Body
L = Σ r_i × p_i = Σ m_i r_i × (ω × r_i). Generally, L and ω are not parallel unless ω is along principal axis. Inertia tensor relates L and ω: L = I·ω where I is 3×3 matrix. For arbitrary rotation, L precesses about ω.
\nIntroduction
Angular momentum of rigid bodies combines contributions from all mass elements. For fixed-axis rotation, it simplifies to L = Iω. The general case requires tensor treatment. Conservation of angular momentum explains gyroscopic effects, figure skater spins, and stability of rotation.
Fixed Axis Rotation
For rotation about fixed principal axis: L = Iω. Direction along axis by right-hand rule. Simple scalar relationship valid when rotation axis is fixed in body and space. Magnitude proportional to both angular velocity and moment of inertia.
General Rigid Body
L = Σ r_i × p_i = Σ m_i r_i × (ω × r_i). Generally, L and ω are not parallel unless ω is along principal axis. Inertia tensor relates L and ω: L = I·ω where I is 3×3 matrix. For arbitrary rotation, L precesses about ω.
\nPrincipal Axes
Principal axes are directions where L is parallel to ω. For symmetric bodies: axes of symmetry are principal axes. About principal axis: L = Iω with scalar I. Choose coordinates aligned with principal axes to simplify calculations. Every rigid body has at least three mutually perpendicular principal axes through CM.
Conservation of Angular Momentum
If τ_ext = 0 about a point, L about that point is conserved. For rigid body: Iω = constant. If I changes (skater pulling arms in), ω changes to conserve L. Effects: spinning skater spins faster when arms pulled in; diver tucks to spin faster; collapsing cloud spins up as it shrinks.
Gyroscopic Effects
When torque is perpendicular to L, it causes precession rather than change in magnitude. dL = τdt is perpendicular to L, so L rotates. Precession rate: Ω = τ/L = mgr/(Iω). Gyroscopes resist changes to orientation - basis for navigation systems. Spinning tops precess instead of falling immediately.
\nSolved Example: Spinning Skater
A figure skater spins at 2 rev/s with arms extended. Her body (excluding arms) has I_body = 2.5 kg·m2. Each arm (modeled as rod) has mass m = 4 kg, length L = 0.7 m. Initially arms are extended horizontally (I_arm = (1/3)mL2 about shoulder, which is at distance d = 0.25 m from rotation axis). She pulls arms in vertically alongside body (I ≈ 0 for arms about body axis). Find new spin rate and kinetic energy ratio. Solution: Extended arms: I_arm,CM = (1/12)mL2 = (1/12)×4×(0.7)2 = 0.163 kg·m2. About vertical axis through body center: parallel axis gives I_arm = I_arm,CM + m(d2 + (L/2)2) - need to calculate moment about vertical axis when arm is horizontal. Simpler: treat each arm as point mass at distance r = d + L/2 = 0.25 + 0.35 = 0.6 m from axis. I_one_arm = mr2 = 4×(0.6)2 = 1.44 kg·m2. Both arms: 2.88 kg·m2. Total I_initial = 2.5 + 2.88 = 5.38 kg·m2. Arms pulled in: I_arms ≈ 0 (aligned with axis), I_final = 2.5 kg·m2. Conservation: I1ω1 = I2ω2 → 5.38×2 = 2.5×ω2 → ω2 = 10.76/2.5 = 4.3 rev/s. Spin rate more than doubles. KE ratio: K2/K1 = (1/2I2ω22)/(1/2I1ω₲) = (I2ω22)/(I1ω₲) = (2.5×(4.3)2)/(5.38×4) = 46.2/21.5 = 2.15. Skater does work pulling arms in, increasing kinetic energy by factor 2.15.
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