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Angular Momentum

Conservation of angular momentum, cross products, and applications

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Angular Momentum

Definition

For a particle: L⃗=r⃗×p⃗=m(r⃗×v⃗)\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})

Magnitude: L=rmvsin⁡θ=r⊥mv=mrv⊥L = rmv\sin\theta = r_{\perp}mv = mrv_{\perp}

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

Relationship to Torque

τ⃗=dL⃗dt\vec{\tau} = \frac{d\vec{L}}{dt}

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

For fixed axis with constant II: τ=Idωdt=Iα\tau = I\frac{d\omega}{dt} = I\alpha

Conservation of Angular Momentum

When τ⃗ext=0\vec{\tau}_{ext} = 0:

L⃗=constant\vec{L} = \text{constant}

Iiωi=IfωfI_i\omega_i = I_f\omega_f

Example: Figure Skater

Skater pulls arms in, reducing moment of inertia:

Iiωi=IfωfI_i\omega_i = I_f\omega_f

If If<IiI_f < I_i, then ωf>ωi\omega_f > \omega_i (spins faster)

Energy changes: KEi=12Iiωi2,KEf=12Ifωf2KE_i = \frac{1}{2}I_i\omega_i^2, \quad KE_f = \frac{1}{2}I_f\omega_f^2

KEf=12If(IiIfωi)2=IiIf⋅12Iiωi2=IiIfKEiKE_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

L⃗total=∑iL⃗i=∑ir⃗i×p⃗i\vec{L}_{total} = \sum_i \vec{L}_i = \sum_i \vec{r}_i \times \vec{p}_i

About center of mass: L⃗=L⃗cm+L⃗orbital\vec{L} = \vec{L}_{cm} + \vec{L}_{orbital}

where:

  • L⃗cm=Icmω⃗\vec{L}_{cm} = I_{cm}\vec{\omega} (spin)
  • L⃗orbital=r⃗cm×Mv⃗cm\vec{L}_{orbital} = \vec{r}_{cm} \times M\vec{v}_{cm} (orbital)

Central Force Motion

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

τ⃗=r⃗×F⃗=0\vec{\tau} = \vec{r} \times \vec{F} = 0

(because F⃗\vec{F} parallel to r⃗\vec{r})

Therefore: L⃗\vec{L} = constant

Consequences:

  1. Motion confined to a plane
  2. Areal velocity constant (Kepler's second law)
  3. r2θ˙=Lmr^2\dot{\theta} = \frac{L}{m} = constant

Areal Velocity

Area swept out per unit time:

dAdt=12r2dθdt=12r2ω=L2m\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 τ⃗ext=0\vec{\tau}_{ext} = 0 about some point, then L⃗\vec{L} conserved about that point.

Example: Putty Ball Hitting Rod

Ball of mass mm, speed vv hits rod of length LL, mass MM at distance dd from pivot.

Before: Li=mvdL_i = mvd (ball's angular momentum)

After: Lf=ItotalωL_f = I_{total}\omega

where Itotal=13ML2+md2I_{total} = \frac{1}{3}ML^2 + md^2 (rod + stuck ball)

mvd=(13ML2+md2)ωmvd = \left(\frac{1}{3}ML^2 + md^2\right)\omega

ω=mvd13ML2+md2\omega = \frac{mvd}{\frac{1}{3}ML^2 + md^2}

Precession

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

Gravitational torque: τ=mgrsin⁡θ\tau = mgr\sin\theta

This causes precession (axis rotates) with angular velocity:

Ω=τLsin⁡θ=mgrL=mgrIω\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 τ⃗\vec{\tau} causes change: ΔL⃗=τ⃗Δt\Delta\vec{L} = \vec{\tau}\Delta t

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

Angular Impulse

∫t1t2τ⃗ dt=ΔL⃗\int_{t_1}^{t_2} \vec{\tau} \, dt = \Delta\vec{L}

Analog of linear impulse ∫F⃗ dt=Δp⃗\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: Li=LfL_i = L_f I1ω1i+I2ω2i=(I1+I2)ωfI_1\omega_{1i} + I_2\omega_{2i} = (I_1 + I_2)\omega_f

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

5.0=0.8ωf5.0 = 0.8\omega_f

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

(b) Angular momenta:

Initial: Li=(0.5)(10)=5.0 kg⋅m2/sL_i = (0.5)(10) = \boxed{5.0 \text{ kg·m}^2\text{/s}}

Final: Lf=(0.8)(6.25)=5.0 kg⋅m2/sL_f = (0.8)(6.25) = \boxed{5.0 \text{ kg·m}^2\text{/s}} ✓

(c) Energy lost:

Initial rotational KE: KEi=12I1ω1i2=12(0.5)(10)2=25 JKE_i = \frac{1}{2}I_1\omega_{1i}^2 = \frac{1}{2}(0.5)(10)^2 = 25 \text{ J}

Final rotational KE: KEf=12(I1+I2)ωf2=12(0.8)(6.25)2KE_f = \frac{1}{2}(I_1 + I_2)\omega_f^2 = \frac{1}{2}(0.8)(6.25)^2

KEf=0.4(39.1)=15.6 JKE_f = 0.4(39.1) = 15.6 \text{ J}

ΔKE=25−15.6\Delta KE = 25 - 15.6

ΔKE=9.4 J lost\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:

Iplatform=MR22=(100)(2.0)22=200 kg⋅m2I_{platform} = \frac{MR^2}{2} = \frac{(100)(2.0)^2}{2} = 200 \text{ kg·m}^2

Student at edge: Istudent,i=mR2=(60)(2.0)2=240I_{student,i} = mR^2 = (60)(2.0)^2 = 240 kg·m²

Total initial: Ii=200+240=440 kg⋅m2I_i = 200 + 240 = 440 \text{ kg·m}^2

L=Iiω0=(440)(0.5)L = I_i\omega_0 = (440)(0.5)

L=220 kg⋅m2/s\boxed{L = 220 \text{ kg·m}^2\text{/s}}

(b) Final angular velocity:

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

If=Iplatform=200 kg⋅m2I_f = I_{platform} = 200 \text{ kg·m}^2

Conservation of angular momentum: Li=LfL_i = L_f 440(0.5)=200ωf440(0.5) = 200\omega_f

ωf=220200\omega_f = \frac{220}{200}

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

(c) Change in KE:

KEi=12Iiω02=12(440)(0.5)2=55 JKE_i = \frac{1}{2}I_i\omega_0^2 = \frac{1}{2}(440)(0.5)^2 = 55 \text{ J}

KEf=12Ifωf2=12(200)(1.1)2=121 JKE_f = \frac{1}{2}I_f\omega_f^2 = \frac{1}{2}(200)(1.1)^2 = 121 \text{ J}

ΔKE=121−55\Delta KE = 121 - 55

ΔKE=+66 J\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 v⃗=3i^+4j^\vec{v} = 3\hat{i} + 4\hat{j} m/s at position r⃗=2i^+1j^\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 τ⃗=5k^\vec{\tau} = 5\hat{k} N·m acts, find dL⃗\vec{L}/dt.

💡 Show Solution

Given:

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

(a) Angular momentum vector:

L⃗=r⃗×p⃗=r⃗×(mv⃗)\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})

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

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

Using i^×i^=0\hat{i} \times \hat{i} = 0, i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×i^=−k^\hat{j} \times \hat{i} = -\hat{k}, j^×j^=0\hat{j} \times \hat{j} = 0:

L⃗=(2)(2)k^+(1)(1.5)(−k^)\vec{L} = (2)(2)\hat{k} + (1)(1.5)(-\hat{k})

L⃗=4k^−1.5k^\vec{L} = 4\hat{k} - 1.5\hat{k}

L⃗=2.5k^ kg⋅m2/s\boxed{\vec{L} = 2.5\hat{k} \text{ kg·m}^2\text{/s}}

(b) Magnitude:

∣L⃗∣=2.5 kg⋅m2/s|\vec{L}| = \boxed{2.5 \text{ kg·m}^2\text{/s}}

(c) Rate of change:

dL⃗dt=τ⃗\frac{d\vec{L}}{dt} = \vec{\tau}

dL⃗dt=5k^ kg⋅m2/s2\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: L⃗(t)=2.5k^+5tk^=(2.5+5t)k^\vec{L}(t) = 2.5\hat{k} + 5t\hat{k} = (2.5 + 5t)\hat{k}

Explain using:
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❓ Frequently Asked Questions

What is Angular Momentum?▾
Conservation of angular momentum, cross products, and applications
How can I study Angular Momentum effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Angular Momentum is part of the AP Physics C: Mechanics course on Study Mondo, specifically in the Rotational Motion section. You can explore the full course for more related topics and practice resources.
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Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.