Angular Momentum

Conservation of angular momentum, cross products, and applications

Angular Momentum

Definition

For a particle: $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$

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:

Motion confined to a plane
Areal velocity constant (Kepler's second law)
$r^2\dot{\theta} = \frac{L}{m}$ = 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 $\int \vec{F} \, dt = \Delta\vec{p}$

📚 Practice Problems

1Problem 1easy

❓ Question:

A disk (I = 0.5 kg·m²) rotates at ω₀ = 10 rad/s. A second disk (I = 0.3 kg·m²) initially at rest drops onto it, and they rotate together. Find: (a) the final angular velocity, (b) the initial and final angular momenta, and (c) the energy lost.

💡 Show Solution

Given:

I₁ = 0.5 kg·m², ω₁ᵢ = 10 rad/s
I₂ = 0.3 kg·m², ω₂ᵢ = 0

(a) Final angular velocity:

Conservation of angular momentum: $L_i = L_f$ $I_1\omega_{1i} + I_2\omega_{2i} = (I_1 + I_2)\omega_f$

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

$5.0 = 0.8\omega_f$

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

(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.

2Problem 2medium

❓ Question:

A student (mass 60 kg) stands at the edge of a rotating platform (mass 100 kg, radius 2.0 m, I = MR²/2). The platform rotates at 0.5 rad/s. The student walks to the center. Find: (a) the initial angular momentum, (b) the final angular velocity, and (c) the change in rotational kinetic energy.

💡 Show Solution

Given:

m = 60 kg (student)
M = 100 kg, R = 2.0 m (platform)
I_platform = MR²/2
ω₀ = 0.5 rad/s

(a) Initial angular momentum:

$I_{platform} = \frac{MR^2}{2} = \frac{(100)(2.0)^2}{2} = 200 \text{ kg·m}^2$

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.

3Problem 3hard

❓ Question:

A particle (mass m = 0.5 kg) moves with velocity $\vec{v} = 3\hat{i} + 4\hat{j}$ m/s at position $\vec{r} = 2\hat{i} + 1\hat{j}$ m. Find: (a) the angular momentum vector about the origin, (b) the magnitude of angular momentum, and (c) if a torque $\vec{\tau} = 5\hat{k}$ N·m acts, find d $\vec{L}$ /dt.

💡 Show Solution

Given:

m = 0.5 kg
$\vec{v} = 3\hat{i} + 4\hat{j}$ m/s
$\vec{r} = 2\hat{i} + 1\hat{j}$ m

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

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