Classical-Mechanics · Unit 23 · Video 1 · Interactive Practice

Momentum Times Moment Arm: The Angular Momentum of a Particle About a Point

IKey Formulas

FormulaNameWhat it says
L⃗S=r⃗S×p⃗,p⃗=mv⃗\vec{\mathbf{L}}_S = \vec{\mathbf{r}}_S \times \vec{\mathbf{p}},\qquad \vec{\mathbf{p}} = m\vec{\mathbf{v}}Angular momentum about SSPosition vector from SS to the particle first, momentum second; units kg⋅m2⋅s−1=N⋅m⋅s=J⋅s\text{kg}\cdot\text{m}^2\cdot\text{s}^{-1} = \text{N}\cdot\text{m}\cdot\text{s} = \text{J}\cdot\text{s}
∣L⃗S∣=∣r⃗S∣ ∣p⃗∣sin⁡θ=rS⊥ ∣p⃗∣=∣r⃗S∣ pS⊥|\vec{\mathbf{L}}_S| = |\vec{\mathbf{r}}_S|\,|\vec{\mathbf{p}}|\sin\theta = r_S^\perp\,|\vec{\mathbf{p}}| = |\vec{\mathbf{r}}_S|\,p_S^\perpMagnitude, two readingsθ\theta is the angle between r⃗S\vec{\mathbf{r}}_S and p⃗\vec{\mathbf{p}}, 0≤θ≤π0 \le \theta \le \pi; moment arm times momentum, or distance times the perpendicular piece of momentum
L⃗S=(x py−y px) k^\vec{\mathbf{L}}_S = (x\,p_y - y\,p_x)\,\hat{\mathbf{k}}Motion in the xx-yy planeFrom i^×j^=k^\hat{\mathbf{i}}\times\hat{\mathbf{j}} = \hat{\mathbf{k}}, j^×i^=−k^\hat{\mathbf{j}}\times\hat{\mathbf{i}} = -\hat{\mathbf{k}}; a counterclockwise shorter arc from r⃗S\vec{\mathbf{r}}_S to p⃗\vec{\mathbf{p}} gives +k^+\hat{\mathbf{k}}, clockwise gives −k^-\hat{\mathbf{k}}
τ⃗S=dL⃗Sdt,LS,z=IS ωz\vec{\boldsymbol{\tau}}_S = \dfrac{d\vec{\mathbf{L}}_S}{dt},\qquad L_{S,z} = I_S\,\omega_zWhere the chapter is goingThe rotational analog of F⃗ext=dp⃗sys/dt\vec{\mathbf{F}}^{\text{ext}} = d\vec{\mathbf{p}}_{\text{sys}}/dt; for a rigid body about a fixed axis it recovers τS,z=IS αz\tau_{S,z} = I_S\,\alpha_z

Key Insight: Angular momentum is always about a named point. A particle moving in a straight line has L⃗S≠0⃗\vec{\mathbf{L}}_S \neq \vec{\mathbf{0}} about any point its path misses — and exactly 0⃗\vec{\mathbf{0}} about every point on its path.

IIMoment Arm or Perpendicular Momentum

Both readings of ∣r⃗S∣ ∣p⃗∣sin⁡θ|\vec{\mathbf{r}}_S|\,|\vec{\mathbf{p}}|\sin\theta reproduce the magnitude of x py−y pxx\,p_y - y\,p_x for any position and momentum.

IIIA Straight Line Still Carries Angular Momentum

At constant velocity ∣r⃗S∣|\vec{\mathbf{r}}_S| and θ\theta keep changing, yet L⃗S\vec{\mathbf{L}}_S never changes — for every choice of SS.

💡 Challenge: find a point SS about which the same particle has L⃗S=+12 kg⋅m2⋅s−1 k^\vec{\mathbf{L}}_S = +12\ \text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}\ \hat{\mathbf{k}} — twice the example's magnitude, with the opposite sense.

IVInto or Out of the Page

With r⃗S\vec{\mathbf{r}}_S and p⃗\vec{\mathbf{p}} placed tail to tail, the sense of the shorter arc between them fixes the sign of LS,zL_{S,z}.

VQuiz Questions

Problem 1 · Components of the Cross Product

Given: A particle of mass 1.5 kg1.5\ \text{kg} passes through (4.0 m, 1.0 m)(4.0\ \text{m},\ 1.0\ \text{m}) with velocity v⃗=2.0 m⋅s−1 j^\vec{\mathbf{v}} = 2.0\ \text{m}\cdot\text{s}^{-1}\ \hat{\mathbf{j}}. Find its angular momentum about the origin SS.

✅ Correct! x py−y px=(4.0)(3.0)−(1.0)(0)=12x\,p_y - y\,p_x = (4.0)(3.0) - (1.0)(0) = 12, and the shorter arc from r⃗S\vec{\mathbf{r}}_S to p⃗\vec{\mathbf{p}} turns counterclockwise.
❌ Check the order. That is p⃗×r⃗S\vec{\mathbf{p}}\times\vec{\mathbf{r}}_S. The position vector comes first, as for torque: L⃗S=r⃗S×p⃗\vec{\mathbf{L}}_S = \vec{\mathbf{r}}_S \times \vec{\mathbf{p}}.
❌ Not quite. That is r⃗S×v⃗\vec{\mathbf{r}}_S \times \vec{\mathbf{v}}. Angular momentum uses the momentum p⃗=mv⃗=3.0 kg⋅m⋅s−1 j^\vec{\mathbf{p}} = m\vec{\mathbf{v}} = 3.0\ \text{kg}\cdot\text{m}\cdot\text{s}^{-1}\ \hat{\mathbf{j}}.
❌ Missing the sin⁡θ\sin\theta. ∣r⃗S∣ ∣p⃗∣=17⋅3.0≈12.4|\vec{\mathbf{r}}_S|\,|\vec{\mathbf{p}}| = \sqrt{17}\cdot 3.0 \approx 12.4 counts the piece of r⃗S\vec{\mathbf{r}}_S along p⃗\vec{\mathbf{p}}, which contributes nothing.
❌ Not quite. Form p⃗=mv⃗\vec{\mathbf{p}} = m\vec{\mathbf{v}} first, then expand r⃗S×p⃗\vec{\mathbf{r}}_S\times\vec{\mathbf{p}} term by term.
Show solution

Momentum: p⃗=mv⃗=(1.5 kg)(2.0 m⋅s−1) j^=3.0 kg⋅m⋅s−1 j^\vec{\mathbf{p}} = m\vec{\mathbf{v}} = (1.5\ \text{kg})(2.0\ \text{m}\cdot\text{s}^{-1})\,\hat{\mathbf{j}} = 3.0\ \text{kg}\cdot\text{m}\cdot\text{s}^{-1}\ \hat{\mathbf{j}}. Position from SS: r⃗S=(4.0 i^+1.0 j^) m\vec{\mathbf{r}}_S = (4.0\,\hat{\mathbf{i}} + 1.0\,\hat{\mathbf{j}})\ \text{m}.

L⃗S=(4.0 i^+1.0 j^)×3.0 j^=12 (i^×j^)+3.0 (j^×j^)=+12 kg⋅m2⋅s−1 k^\vec{\mathbf{L}}_S = (4.0\,\hat{\mathbf{i}} + 1.0\,\hat{\mathbf{j}}) \times 3.0\,\hat{\mathbf{j}} = 12\,(\hat{\mathbf{i}}\times\hat{\mathbf{j}}) + 3.0\,(\hat{\mathbf{j}}\times\hat{\mathbf{j}}) = +12\ \text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}\ \hat{\mathbf{k}}

Check with the moment arm: the line of motion is x=4.0 mx = 4.0\ \text{m}, so rS⊥=4.0 mr_S^\perp = 4.0\ \text{m} and ∣L⃗S∣=(4.0)(3.0)=12|\vec{\mathbf{L}}_S| = (4.0)(3.0) = 12. The particle moves up on the right of SS — counterclockwise, out of the page.

Problem 2 · Which Distance Is the Moment Arm?

Given: A particle moves along the line y=2.0 my = 2.0\ \text{m} with momentum p⃗=5.0 kg⋅m⋅s−1 i^\vec{\mathbf{p}} = 5.0\ \text{kg}\cdot\text{m}\cdot\text{s}^{-1}\ \hat{\mathbf{i}}. At this instant it is at x=3.0 mx = 3.0\ \text{m}. Find ∣L⃗S∣|\vec{\mathbf{L}}_S| about the origin and its direction.

✅ Correct! The path misses SS by 2.0 m2.0\ \text{m}, so ∣L⃗S∣=(2.0)(5.0)=10|\vec{\mathbf{L}}_S| = (2.0)(5.0) = 10, and moving right above SS is clockwise.
❌ Missing the sin⁡θ\sin\theta. ∣r⃗S∣ ∣p⃗∣=13⋅5.0≈18|\vec{\mathbf{r}}_S|\,|\vec{\mathbf{p}}| = \sqrt{13}\cdot 5.0 \approx 18 uses the full distance, not the perpendicular distance to the line of motion.
❌ Wrong distance. x=3.0 mx = 3.0\ \text{m} is measured along the line of motion. The moment arm is the perpendicular distance from SS to that line.
❌ Check the direction. Tail to tail at SS, r⃗S\vec{\mathbf{r}}_S points up-right and p⃗\vec{\mathbf{p}} points right: the shorter arc turns clockwise, giving −k^-\hat{\mathbf{k}}.
❌ Not quite. Drop a perpendicular from SS to the line of motion.
Show solution

The line of motion is y=2.0 my = 2.0\ \text{m}, so the perpendicular from SS meets it at (0, 2.0 m)(0,\ 2.0\ \text{m}): rS⊥=2.0 mr_S^\perp = 2.0\ \text{m}.

∣L⃗S∣=rS⊥ ∣p⃗∣=(2.0 m)(5.0 kg⋅m⋅s−1)=10 kg⋅m2⋅s−1|\vec{\mathbf{L}}_S| = r_S^\perp\,|\vec{\mathbf{p}}| = (2.0\ \text{m})(5.0\ \text{kg}\cdot\text{m}\cdot\text{s}^{-1}) = 10\ \text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}

Components confirm it and give the direction: LS,z=x py−y px=(3.0)(0)−(2.0)(5.0)=−10L_{S,z} = x\,p_y - y\,p_x = (3.0)(0) - (2.0)(5.0) = -10, so L⃗S=−10 kg⋅m2⋅s−1 k^\vec{\mathbf{L}}_S = -10\ \text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}\ \hat{\mathbf{k}}, into the page. The answer does not depend on xx: every point of the path has the same moment arm.

Problem 3 · Worked Example, New Numbers

Given: A 3.0 kg3.0\ \text{kg} particle moves with constant velocity v⃗=(2.0 i^+1.0 j^) m⋅s−1\vec{\mathbf{v}} = (2.0\,\hat{\mathbf{i}} + 1.0\,\hat{\mathbf{j}})\ \text{m}\cdot\text{s}^{-1} and passes through (1.0 m, 2.0 m)(1.0\ \text{m},\ 2.0\ \text{m}). Find L⃗S\vec{\mathbf{L}}_S about the origin, then the moment arm rS⊥r_S^\perp.

Angular momentum about S

Moment arm

✅ Correct! L⃗S=−9.0 k^\vec{\mathbf{L}}_S = -9.0\,\hat{\mathbf{k}} and rS⊥=∣L⃗S∣/∣p⃗∣=9/45=3/5 mr_S^\perp = |\vec{\mathbf{L}}_S|/|\vec{\mathbf{p}}| = 9/\sqrt{45} = 3/\sqrt{5}\ \text{m}.
❌ Sign slip. j^×i^=−k^\hat{\mathbf{j}}\times\hat{\mathbf{i}} = -\hat{\mathbf{k}}, so the y pxy\,p_x term is subtracted: 3−123 - 12, not 3+123 + 12.
❌ Check the order. That is p⃗×r⃗S\vec{\mathbf{p}}\times\vec{\mathbf{r}}_S; the position vector comes first.
❌ Not quite. That is r⃗S×v⃗\vec{\mathbf{r}}_S\times\vec{\mathbf{v}} — multiply by m=3.0 kgm = 3.0\ \text{kg} to get the momentum first.
❌ Check the cross product. Expand the four products of (1.0 i^+2.0 j^)×(6.0 i^+3.0 j^)(1.0\,\hat{\mathbf{i}} + 2.0\,\hat{\mathbf{j}})\times(6.0\,\hat{\mathbf{i}} + 3.0\,\hat{\mathbf{j}}).
❌ That is ∣r⃗S∣|\vec{\mathbf{r}}_S|. The moment arm is the perpendicular distance from SS to the line of motion, ∣r⃗S∣sin⁡θ|\vec{\mathbf{r}}_S|\sin\theta, and here θ≠90°\theta \neq 90\degree.
❌ That distance is measured along the yy-axis. The path crosses the yy-axis 1.5 m1.5\ \text{m} above SS (equivalently ∣L⃗S∣/px=9/6.0|\vec{\mathbf{L}}_S|/p_x = 9/6.0), but the yy-axis is not perpendicular to the path. The moment arm is shorter: 1.5sin⁡α1.5\sin\alpha with sin⁡α=2/5\sin\alpha = 2/\sqrt{5}.
❌ Divide by the whole momentum. 9/3.0=3.09/3.0 = 3.0 divides ∣L⃗S∣|\vec{\mathbf{L}}_S| by pyp_y alone (or by the mass). The moment arm is ∣L⃗S∣/∣p⃗∣|\vec{\mathbf{L}}_S|/|\vec{\mathbf{p}}| with ∣p⃗∣=6.02+3.02=35|\vec{\mathbf{p}}| = \sqrt{6.0^2 + 3.0^2} = 3\sqrt{5}.
❌ Check the moment arm. Use rS⊥=∣L⃗S∣/∣p⃗∣r_S^\perp = |\vec{\mathbf{L}}_S| / |\vec{\mathbf{p}}|, with ∣p⃗∣=6.02+3.02|\vec{\mathbf{p}}| = \sqrt{6.0^2 + 3.0^2}.
Show solution

r⃗S=(1.0 i^+2.0 j^) m\vec{\mathbf{r}}_S = (1.0\,\hat{\mathbf{i}} + 2.0\,\hat{\mathbf{j}})\ \text{m} and p⃗=mv⃗=(6.0 i^+3.0 j^) kg⋅m⋅s−1\vec{\mathbf{p}} = m\vec{\mathbf{v}} = (6.0\,\hat{\mathbf{i}} + 3.0\,\hat{\mathbf{j}})\ \text{kg}\cdot\text{m}\cdot\text{s}^{-1}. The four products:

1.0 i^×6.0 i^=0⃗,1.0 i^×3.0 j^=3 k^,2.0 j^×6.0 i^=−12 k^,2.0 j^×3.0 j^=0⃗1.0\,\hat{\mathbf{i}}\times 6.0\,\hat{\mathbf{i}} = \vec{\mathbf{0}},\quad 1.0\,\hat{\mathbf{i}}\times 3.0\,\hat{\mathbf{j}} = 3\,\hat{\mathbf{k}},\quad 2.0\,\hat{\mathbf{j}}\times 6.0\,\hat{\mathbf{i}} = -12\,\hat{\mathbf{k}},\quad 2.0\,\hat{\mathbf{j}}\times 3.0\,\hat{\mathbf{j}} = \vec{\mathbf{0}} L⃗S=(3−12) k^=−9.0 kg⋅m2⋅s−1 k^(into the page)\vec{\mathbf{L}}_S = (3 - 12)\,\hat{\mathbf{k}} = -9.0\ \text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}\ \hat{\mathbf{k}}\quad\text{(into the page)}

Moment arm: ∣p⃗∣=36+9=45=35 kg⋅m⋅s−1|\vec{\mathbf{p}}| = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}\ \text{kg}\cdot\text{m}\cdot\text{s}^{-1}, so

rS⊥=∣L⃗S∣∣p⃗∣=935=35 m≈1.34 mr_S^\perp = \frac{|\vec{\mathbf{L}}_S|}{|\vec{\mathbf{p}}|} = \frac{9}{3\sqrt{5}} = \frac{3}{\sqrt{5}}\ \text{m} \approx 1.34\ \text{m}

Geometric check: the path has slope 12\tfrac{1}{2} through (1,2)(1, 2), so it crosses the yy-axis at 1.5 m1.5\ \text{m}; that segment meets the path at angle α\alpha with sin⁡α=2/5\sin\alpha = 2/\sqrt{5}, giving rS⊥=1.5⋅2/5=3/5 mr_S^\perp = 1.5 \cdot 2/\sqrt{5} = 3/\sqrt{5}\ \text{m}.

Problem 4 · Name the Point First

Given: The video's particle: p⃗=(6.0 i^+6.0 j^) kg⋅m⋅s−1\vec{\mathbf{p}} = (6.0\,\hat{\mathbf{i}} + 6.0\,\hat{\mathbf{j}})\ \text{kg}\cdot\text{m}\cdot\text{s}^{-1}, moving along the line y=x+1 my = x + 1\ \text{m} through (2.0 m, 3.0 m)(2.0\ \text{m},\ 3.0\ \text{m}). About which point is L⃗=+6 kg⋅m2⋅s−1 k^\vec{\mathbf{L}} = +6\ \text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}\ \hat{\mathbf{k}}?

✅ Correct! (−1,1)(-1, 1) sits a moment arm of 12 m\tfrac{1}{\sqrt{2}}\ \text{m} on the other side of the path from the origin, so the sense flips.
❌ That is the worked example. About the origin L⃗S=−6 k^\vec{\mathbf{L}}_S = -6\,\hat{\mathbf{k}} — the right magnitude, the opposite direction.
❌ That point is on the path. 1=0+11 = 0 + 1, so r⃗\vec{\mathbf{r}} is parallel to p⃗\vec{\mathbf{p}} and the angular momentum about it is 0⃗\vec{\mathbf{0}}.
❌ Not quite. (2,1)(2, 1) is on the same side as the origin but twice as far from the path: L⃗=−12 k^\vec{\mathbf{L}} = -12\,\hat{\mathbf{k}}.
❌ Not quite. Compute Lz=(x−xP) py−(y−yP) pxL_z = (x - x_P)\,p_y - (y - y_P)\,p_x for each candidate point PP.
Show solution

About a point P=(xP,yP)P = (x_P, y_P) the position vector is (2−xP) i^+(3−yP) j^(2 - x_P)\,\hat{\mathbf{i}} + (3 - y_P)\,\hat{\mathbf{j}}, so

LP,z=6(2−xP)−6(3−yP)=6 (yP−xP−1)L_{P,z} = 6(2 - x_P) - 6(3 - y_P) = 6\,(y_P - x_P - 1)
  • (0,0)(0, 0): 6(0−0−1)=−66(0 - 0 - 1) = -6
  • (−1,1)(-1, 1): 6(1+1−1)=+66(1 + 1 - 1) = +6 ✓
  • (0,1)(0, 1): 6(1−0−1)=06(1 - 0 - 1) = 0 — on the path
  • (2,1)(2, 1): 6(1−2−1)=−126(1 - 2 - 1) = -12

The factor yP−xP−1y_P - x_P - 1 is 2\sqrt{2} times the signed distance from PP to the path, so LP,zL_{P,z} is the moment arm times ∣p⃗∣=62|\vec{\mathbf{p}}| = 6\sqrt{2}, with the sign set by which side of the path PP is on. The same particle has different angular momenta about different points.

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