Angular Momentum and Conservation

This lesson covers angular momentum, its conservation, and applications to collision and rotation problems.

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1 Definition of Angular Momentum

The angular momentum $L$L of a particle about an axis is:

$$L = mvr$$L=mvr

where $r$r is the perpendicular distance from the axis to the line of motion of the particle.

For a rigid body rotating about a fixed axis:

$$\boxed{L = I\omega}$$L=Iω​

Property	Detail
SI unit	kg m$^2$2 s$^{-1}$−1 (or N m s)
Vector	Direction along the axis of rotation (right-hand rule)
Depends on	Moment of inertia and angular velocity

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2 Relationship to Torque

Angular momentum is related to torque by:

$$\boxed{\tau = \frac{\mathrm{d}L}{\mathrm{d}t}}$$τ=dtdL​​

This is the rotational analogue of $F = \mathrm{d}p/\mathrm{d}t$F=dp/dt.

If $I$I is constant: $\tau = I\alpha$τ=Iα.

If $\tau = 0$τ=0: $L$L is constant — this gives us conservation of angular momentum.

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3 Conservation of Angular Momentum

When no external torque acts on a system, the total angular momentum is conserved:

$$\boxed{I_1 \omega_1 = I_2 \omega_2}$$I1​ω1​=I2​ω2​​

This applies when the moment of inertia changes (e.g. a spinning figure skater pulling their arms in).

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Example 1 — Spinning Ice Skater

A skater has moment of inertia 4 kg m$^2$2 while spinning at 2 rev s$^{-1}$−1 with arms extended. She pulls her arms in, reducing her moment of inertia to 1.5 kg m$^2$2. Find her new angular speed.

Solution

$I_1\omega_1 = I_2\omega_2$I1​ω1​=I2​ω2​

$4 \times 2 = 1.5 \times \omega_2$4×2=1.5×ω2​

$\omega_2 = 8/1.5 = 5.33$ω2​=8/1.5=5.33 rev s$^{-1}$−1.

Note: KE is not conserved. $KE_1 = \frac{1}{2}(4)(4\pi)^2 = 32\pi^2$KE1​=21​(4)(4π)2=32π2 J. $KE_2 = \frac{1}{2}(1.5)(10.67\pi)^2 = 85.3\pi^2$KE2​=21​(1.5)(10.67π)2=85.3π2 J. The skater does internal work by pulling in her arms.

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Example 2 — Turntable Problem

A disc of moment of inertia 0.5 kg m$^2$2 rotates at 10 rad s$^{-1}$−1. A ring of moment of inertia 0.3 kg m$^2$2 is dropped onto it. They rotate together. Find the common angular speed.

Solution

No external torque during the brief contact:

$I_1\omega_1 = (I_1 + I_2)\omega$I1​ω1​=(I1​+I2​)ω

$0.5 \times 10 = 0.8\omega$0.5×10=0.8ω

$\omega = 6.25$ω=6.25 rad s$^{-1}$−1.

KE lost: $\frac{1}{2}(0.5)(100) - \frac{1}{2}(0.8)(39.0625) = 25 - 15.625 = 9.375$21​(0.5)(100)−21​(0.8)(39.0625)=25−15.625=9.375 J.

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4 Angular Impulse

The angular impulse is the change in angular momentum:

$$\int \tau\,\mathrm{d}t = \Delta L = I\omega_2 - I\omega_1$$∫τdt=ΔL=Iω2​−Iω1​

For a constant torque: angular impulse = $\tau \Delta t$τΔt.

Example 3