Magnitude: $L = rmv\sin\theta = r_{\perp}mv = mrv_{\perp}$

For rotation about fixed axis: $L = I\omega$

Relationship to Torque

$\vec{\tau} = \frac{d\vec{L}}{dt}$

(Rotational analog of $\vec{F} = \frac{d\vec{p}}{dt}$)

For fixed axis with constant $I$: $\tau = I\frac{d\omega}{dt} = I\alpha$

Conservation of Angular Momentum

When $\vec{\tau}_{ext} = 0$:

$\vec{L} = \text{constant}$

$I_i\omega_i = I_f\omega_f$

Example: Figure Skater

Skater pulls arms in, reducing moment of inertia:

$I_i\omega_i = I_f\omega_f$

If $I_f < I_i$, then $\omega_f > \omega_i$ (spins faster)

Energy changes: $KE_i = \frac{1}{2}I_i\omega_i^2, \quad KE_f = \frac{1}{2}I_f\omega_f^2$

$KE_f = \frac{1}{2}I_f\left(\frac{I_i}{I_f}\omega_i\right)^2 = \frac{I_i}{I_f}\cdot\frac{1}{2}I_i\omega_i^2 = \frac{I_i}{I_f}KE_i$

Energy increases! (Work done by internal forces)

Angular Momentum of System

$\vec{L}_{total} = \sum_i \vec{L}_i = \sum_i \vec{r}_i \times \vec{p}_i$

About center of mass: $\vec{L} = \vec{L}_{cm} + \vec{L}_{orbital}$

where:

• $\vec{L}_{cm} = I_{cm}\vec{\omega}$ (spin)
• $\vec{L}_{orbital} = \vec{r}_{cm} \times M\vec{v}_{cm}$ (orbital)

Central Force Motion

For central force (directed toward/away from fixed point):

$\vec{\tau} = \vec{r} \times \vec{F} = 0$

(because $\vec{F}$ parallel to $\vec{r}$)

Therefore: $\vec{L}$ = constant

Consequences:

1. Motion confined to a plane
2. Areal velocity constant (Kepler's second law)
3. $r^2\dot{\theta} = \frac{L}{m}$ = constant

Areal Velocity

Area swept out per unit time:

$\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt} = \frac{1}{2}r^2\omega = \frac{L}{2m}$

(Constant for central forces)

Collisions and Angular Momentum

For collision, if $\vec{\tau}_{ext} = 0$ about some point, then $\vec{L}$ conserved about that point.

Example: Putty Ball Hitting Rod

Ball of mass $m$, speed $v$ hits rod of length $L$, mass $M$ at distance $d$ from pivot.

Before: $L_i = mvd$ (ball's angular momentum)

After: $L_f = I_{total}\omega$

where $I_{total} = \frac{1}{3}ML^2 + md^2$ (rod + stuck ball)

$mvd = \left(\frac{1}{3}ML^2 + md^2\right)\omega$

$\omega = \frac{mvd}{\frac{1}{3}ML^2 + md^2}$

Precession

Spinning top with angular momentum $\vec{L}$ tilted at angle $\theta$:

Gravitational torque: $\tau = mgr\sin\theta$

This causes precession (axis rotates) with angular velocity:

$\Omega = \frac{\tau}{L\sin\theta} = \frac{mgr}{L} = \frac{mgr}{I\omega}$

Gyroscopic Motion

Gyroscope resists changes in orientation due to angular momentum conservation.

Applied torque $\vec{\tau}$ causes change: $\Delta\vec{L} = \vec{\tau}\Delta t$

Direction of $\Delta\vec{L}$ perpendicular to both $\vec{L}$ and $\vec{\tau}$, causing precession.

Angular Impulse

$\int_{t_1}^{t_2} \vec{\tau} \, dt = \Delta\vec{L}$

Analog of linear impulse $\int \vec{F} \, dt = \Delta\vec{p}$

Student at edge: $I_{student,i} = mR^2 = (60)(2.0)^2 = 240$ kg·m²

Total initial: $I_i = 200 + 240 = 440 \text{ kg·m}^2$

$L = I_i\omega_0 = (440)(0.5)$

$\boxed{L = 220 \text{ kg·m}^2\text{/s}}$

(b) Final angular velocity:

Student at center: $I_{student,f} = 0$

$I_f = I_{platform} = 200 \text{ kg·m}^2$

Conservation of angular momentum: $L_i = L_f$ $440(0.5) = 200\omega_f$

$\omega_f = \frac{220}{200}$

$\boxed{\omega_f = 1.1 \text{ rad/s}}$

(c) Change in KE:

$KE_i = \frac{1}{2}I_i\omega_0^2 = \frac{1}{2}(440)(0.5)^2 = 55 \text{ J}$

$KE_f = \frac{1}{2}I_f\omega_f^2 = \frac{1}{2}(200)(1.1)^2 = 121 \text{ J}$

$\Delta KE = 121 - 55$

$\boxed{\Delta KE = +66 \text{ J}}$

Energy increased! Student did work walking inward against fictitious centrifugal force.