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 $\int \vec{F} \, dt = \Delta\vec{p}$

$(0.5)(10) + 0 = (0.5 + 0.3)\omega_f$

$5.0 = 0.8\omega_f$

$\boxed{\omega_f = 6.25 \text{ rad/s}}$

(b) Angular momenta:

Initial: $L_i = (0.5)(10) = \boxed{5.0 \text{ kg·m}^2\text{/s}}$

Final: $L_f = (0.8)(6.25) = \boxed{5.0 \text{ kg·m}^2\text{/s}}$ ✓

(c) Energy lost:

Initial rotational KE: $KE_i = \frac{1}{2}I_1\omega_{1i}^2 = \frac{1}{2}(0.5)(10)^2 = 25 \text{ J}$

Final rotational KE: $KE_f = \frac{1}{2}(I_1 + I_2)\omega_f^2 = \frac{1}{2}(0.8)(6.25)^2$

$KE_f = 0.4(39.1) = 15.6 \text{ J}$

$\Delta KE = 25 - 15.6$

$\boxed{\Delta KE = 9.4 \text{ J lost}}$

Lost to friction/heat during collision.

(a) Angular momentum vector:

$\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$

$\vec{L} = (2\hat{i} + 1\hat{j}) \times [(0.5)(3\hat{i} + 4\hat{j})]$

$\vec{L} = (2\hat{i} + 1\hat{j}) \times (1.5\hat{i} + 2\hat{j})$

Using $\hat{i} \times \hat{i} = 0$, $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{i} = -\hat{k}$, $\hat{j} \times \hat{j} = 0$:

$\vec{L} = (2)(2)\hat{k} + (1)(1.5)(-\hat{k})$

$\vec{L} = 4\hat{k} - 1.5\hat{k}$

$\boxed{\vec{L} = 2.5\hat{k} \text{ kg·m}^2\text{/s}}$

(b) Magnitude:

$|\vec{L}| = \boxed{2.5 \text{ kg·m}^2\text{/s}}$

(c) Rate of change:

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

$\boxed{\frac{d\vec{L}}{dt} = 5\hat{k} \text{ kg·m}^2\text{/s}^2}$

This means angular momentum increases at 5 kg·m²/s² in the +z direction.

After time t: $\vec{L}(t) = 2.5\hat{k} + 5t\hat{k} = (2.5 + 5t)\hat{k}$