Classical-Mechanics · Unit 23 · Video 5 · Interactive Practice

Why the Height Drops Out: Angular Momentum About a Fixed Axis and the Spinning Ring

IKey Formulas

FormulaNameWhat you need
L⃗S,j=Δmj(rS,j⊥)2 ωz k^−Δmj rS,j∥rS,j⊥ ωz r^\vec{L}_{S,j} = \Delta m_j (r_{S,j}^{\perp})^2\,\omega_z\,\hat{k} - \Delta m_j\, r_{S,j}^{\parallel} r_{S,j}^{\perp}\,\omega_z\,\hat{r}One element about SS on the axisIts distance r⊥r^{\perp} from the axis and height r∥r^{\parallel} above SS
LS,zsys=IS ωz,IS=∫bodydm (rdm)2L_{S,z}^{\text{sys}} = I_S\,\omega_z, \quad I_S = \int_{\text{body}} dm\,(r_{dm})^2Axial component, whole bodyMoment of inertia about the axis — any SS on it
τS,zext=dLS,zsysdt=IS αz\tau_{S,z}^{\text{ext}} = \dfrac{dL_{S,z}^{\text{sys}}}{dt} = I_S\,\alpha_zFixed-axis equation of motionRigid body, fixed axis (ISI_S constant)
L⃗S=IS ωz k^=MR2ωz k^\vec{L}_S = I_S\,\omega_z\,\hat{k} = MR^2\omega_z\,\hat{k}Thin ring, any SS on its axisMass symmetric about the axis

Key Insight: The height r∥r^{\parallel} only tilts each element's angular momentum; it never enters the zz-component. Symmetry about the axis cancels the tilts in pairs — without it, only LS,z=IS ωzL_{S,z} = I_S\,\omega_z is guaranteed.

IIVisualization 1 — Slide SS Along the Axis

Moving SS along the axis changes the tilt of an element's angular momentum, never its zz-component.

IIIVisualization 2 — The Ring, Pair by Pair

Objects at opposite ends of a diameter have opposite tilted parts, so the ring's L⃗S\vec{L}_S lies along the axis.

IVVisualization 3 — Torque Is the Rate of Change of LS,zL_{S,z}

A constant zz-torque makes LS,z=IS ωzL_{S,z} = I_S\,\omega_z grow linearly in time, with slope τS,zext=IS αz\tau_{S,z}^{\text{ext}} = I_S\,\alpha_z.

VQuiz Questions

Problem 1 · One Element's Axial Component

Given: a rigid body turns about the fixed zz-axis with ωz=5 rad/s\omega_z = 5\ \text{rad/s}. One element has Δmj=0.20 kg\Delta m_j = 0.20\ \text{kg}, sits rS,j⊥=0.30 mr_{S,j}^{\perp} = 0.30\ \text{m} from the axis and rS,j∥=0.40 mr_{S,j}^{\parallel} = 0.40\ \text{m} above the point SS on the axis — find (LS,j)z(L_{S,j})_z.

✅ Correct! Only the distance from the axis enters: (LS,j)z=Δmj(rS,j⊥)2ωz=0.20×0.09×5=0.090 kg⋅m2/s(L_{S,j})_z = \Delta m_j (r_{S,j}^{\perp})^2\omega_z = 0.20 \times 0.09 \times 5 = 0.090\ \text{kg}\cdot\text{m}^2/\text{s}.
❌ That uses the full distance from SS, ∣r⃗S,j∣=0.50 m|\vec{r}_{S,j}| = 0.50\ \text{m}. The height r∥r^{\parallel} only produces the tilted r^\hat{r} part; the zz-component uses r⊥r^{\perp} alone.
❌ That is Δmj r∥r⊥ωz\Delta m_j\, r^{\parallel} r^{\perp}\omega_z — the size of the tilted part. It points along −r^-\hat{r}, perpendicular to the axis, so it contributes nothing to the zz-component.
❌ That is the momentum pj=Δmjr⊥ωzp_j = \Delta m_j r^{\perp}\omega_z, not the angular momentum. Crossing with r⊥r^r^{\perp}\hat{r} multiplies by r⊥r^{\perp} once more, so the distance appears squared.
Show solution

The element moves on a circle of radius r⊥r^{\perp}, so p⃗j=Δmj r⊥ωz θ^\vec{p}_j = \Delta m_j\, r^{\perp}\omega_z\,\hat{\theta}. With r⃗S,j=r⊥r^+r∥k^\vec{r}_{S,j} = r^{\perp}\hat{r} + r^{\parallel}\hat{k} and r^×θ^=k^\hat{r}\times\hat{\theta} = \hat{k}, k^×θ^=−r^\hat{k}\times\hat{\theta} = -\hat{r}:

L⃗S,j=Δmj(r⊥)2ωz k^−Δmj r∥r⊥ωz r^\vec{L}_{S,j} = \Delta m_j (r^{\perp})^2\omega_z\,\hat{k} - \Delta m_j\, r^{\parallel} r^{\perp}\omega_z\,\hat{r} (LS,j)z=(0.20)(0.30)2(5)=0.090 kg⋅m2/s(L_{S,j})_z = (0.20)(0.30)^2(5) = 0.090\ \text{kg}\cdot\text{m}^2/\text{s}

The height 0.40 m0.40\ \text{m} appears only in the tilted part, of size (0.20)(0.40)(0.30)(5)=0.12 kg⋅m2/s(0.20)(0.40)(0.30)(5) = 0.12\ \text{kg}\cdot\text{m}^2/\text{s}.

Problem 2 · Move the Reference Point

Given: the same element as Problem 1 (Δmj=0.20 kg\Delta m_j = 0.20\ \text{kg}, r⊥=0.30 mr^{\perp} = 0.30\ \text{m}, ωz=5 rad/s\omega_z = 5\ \text{rad/s}), but now angular momentum is taken about a point S′S' on the axis 1.20 m1.20\ \text{m} below the element's level — find the zz-component and the size of the part perpendicular to the axis.

What is (LS′,j)z(L_{S',j})_z?

What is the size of the perpendicular (tilted) part?

✅ Correct! The axial part stays 0.090 kg⋅m2/s0.090\ \text{kg}\cdot\text{m}^2/\text{s} about every point on the axis; only the tilt grows, to 0.20×1.20×0.30×5=0.36 kg⋅m2/s0.20 \times 1.20 \times 0.30 \times 5 = 0.36\ \text{kg}\cdot\text{m}^2/\text{s}.
❌ That puts the full distance ∣r⃗S′,j∣2=0.09+1.44|\vec{r}_{S',j}|^2 = 0.09 + 1.44 into the zz-component. The height never enters (LS′,j)z=Δmj(r⊥)2ωz(L_{S',j})_z = \Delta m_j (r^{\perp})^2\omega_z.
❌ The zz-component does not scale with the height. Tripling r∥r^{\parallel} (from 0.400.40 to 1.20 m1.20\ \text{m}) triples only the tilted part.
❌ Check the zz-component. It is Δmj(r⊥)2ωz\Delta m_j (r^{\perp})^2\omega_z, with no height in it.
❌ That is the tilt about the old point, 0.40 m0.40\ \text{m} below. The tilted part is Δmj r∥r⊥ωz\Delta m_j\, r^{\parallel} r^{\perp}\omega_z, so it grows with the new height r∥=1.20 mr^{\parallel} = 1.20\ \text{m}.
❌ Moving SS along the axis does not remove the zz-component. The element still moves on its circle of radius r⊥r^{\perp}, so (LS′,j)z=Δmj(r⊥)2ωz(L_{S',j})_z = \Delta m_j (r^{\perp})^2\omega_z is the same as about SS.
❌ That is the zz-component, Δmj(r⊥)2ωz\Delta m_j (r^{\perp})^2\omega_z. The perpendicular part is a different term, Δmj r∥r⊥ωz\Delta m_j\, r^{\parallel} r^{\perp}\omega_z, with r∥=1.20 mr^{\parallel} = 1.20\ \text{m}.
❌ That is Δmj r∥r⊥=(0.20)(1.20)(0.30)\Delta m_j\, r^{\parallel} r^{\perp} = (0.20)(1.20)(0.30) without ωz\omega_z. The momentum is Δmj r⊥ωz\Delta m_j\, r^{\perp}\omega_z, so multiply by ωz=5 rad/s\omega_z = 5\ \text{rad/s}.
❌ Check the tilted part. Its size is Δmj r∥r⊥ωz\Delta m_j\, r^{\parallel} r^{\perp}\omega_z, with r∥=1.20 mr^{\parallel} = 1.20\ \text{m} now.
Show solution

About S′S': r⊥=0.30 mr^{\perp} = 0.30\ \text{m} (unchanged) and r∥=1.20 mr^{\parallel} = 1.20\ \text{m}.

(LS′,j)z=Δmj(r⊥)2ωz=(0.20)(0.30)2(5)=0.090 kg⋅m2/s(L_{S',j})_z = \Delta m_j (r^{\perp})^2\omega_z = (0.20)(0.30)^2(5) = 0.090\ \text{kg}\cdot\text{m}^2/\text{s} ∣(L⃗S′,j)⊥∣=Δmj r∥r⊥ωz=(0.20)(1.20)(0.30)(5)=0.36 kg⋅m2/s\left|(\vec{L}_{S',j})_{\perp}\right| = \Delta m_j\, r^{\parallel} r^{\perp}\omega_z = (0.20)(1.20)(0.30)(5) = 0.36\ \text{kg}\cdot\text{m}^2/\text{s}

Moving SS down the axis tilts L⃗\vec{L} further from the axis — from arctan⁡(0.12/0.090)≈53°\arctan(0.12/0.090) \approx 53\degree to arctan⁡(0.36/0.090)≈76°\arctan(0.36/0.090) \approx 76\degree — while its projection on the axis stays fixed.

Problem 3 · The Spinning Ring

Given: a thin ring of mass M=0.40 kgM = 0.40\ \text{kg} and radius R=0.25 mR = 0.25\ \text{m} lies horizontally, centred on the zz-axis, a height h=0.60 mh = 0.60\ \text{m} above the point SS on the axis, and spins with ωz=12 rad/s\omega_z = 12\ \text{rad/s} — find L⃗S\vec{L}_S.

✅ Correct! The tilts cancel in diametric pairs, so L⃗S=MR2ωz k^=0.40×0.0625×12 k^=0.30 k^ kg⋅m2/s\vec{L}_S = MR^2\omega_z\,\hat{k} = 0.40 \times 0.0625 \times 12\,\hat{k} = 0.30\,\hat{k}\ \text{kg}\cdot\text{m}^2/\text{s} — the same about every point on the axis.
❌ That uses M(R2+h2)M(R^2 + h^2). The moment of inertia measures distance from the axis, not from SS: every piece of the ring is RR from the axis, so IS=MR2I_S = MR^2.
❌ That treats the whole ring like one object at r^\hat{r}. Each object's tilted part −h mR ωz r^i-h\,mR\,\omega_z\,\hat{r}_i is cancelled by its partner's, since r^2=−r^1\hat{r}_2 = -\hat{r}_1: nothing is left perpendicular to the axis.
❌ 12MR2\tfrac{1}{2}MR^2 is a uniform disk. All of a thin ring's mass sits at distance RR, so IS=∫dm R2=MR2I_S = \int dm\,R^2 = MR^2.
Show solution

Split the ring into diametric pairs of objects of mass mm. For one pair,

L⃗S pair=2mR2ωz k^−h mR ωz(r^1+r^2)=2mR2ωz k^\vec{L}_S^{\,\text{pair}} = 2mR^2\omega_z\,\hat{k} - h\,mR\,\omega_z(\hat{r}_1 + \hat{r}_2) = 2mR^2\omega_z\,\hat{k}

because r^2=−r^1\hat{r}_2 = -\hat{r}_1. Summing over the pairs, the masses add to MM:

L⃗S=MR2ωz k^=(0.40)(0.25)2(12) k^=0.30 k^ kg⋅m2/s\vec{L}_S = MR^2\omega_z\,\hat{k} = (0.40)(0.25)^2(12)\,\hat{k} = 0.30\,\hat{k}\ \text{kg}\cdot\text{m}^2/\text{s}

The height h=0.60 mh = 0.60\ \text{m} is a distractor: it dropped out with the tilts.

Problem 4 · Spinning Up on a Fixed Axle

Given: a thin ring of mass M=1.5 kgM = 1.5\ \text{kg} and radius R=0.40 mR = 0.40\ \text{m} turns on a fixed vertical axle through its centre. Starting from rest, a constant external torque brings its angular momentum about a point SS on the axle, 0.50 m0.50\ \text{m} below the ring, to LS,z=2.4 kg⋅m2/sL_{S,z} = 2.4\ \text{kg}\cdot\text{m}^2/\text{s} in 3.0 s3.0\ \text{s} — find the angular acceleration αz\alpha_z.

✅ Correct! τS,zext=ΔLS,z/Δt=0.80 N⋅m\tau_{S,z}^{\text{ext}} = \Delta L_{S,z}/\Delta t = 0.80\ \text{N}\cdot\text{m} and IS=MR2=0.24 kg⋅m2I_S = MR^2 = 0.24\ \text{kg}\cdot\text{m}^2, so αz=0.80/0.24=3.3 rad/s2\alpha_z = 0.80/0.24 = 3.3\ \text{rad/s}^2.
❌ That uses I=M(R2+h2)I = M(R^2 + h^2) with the 0.50 m0.50\ \text{m} drop to SS. ISI_S depends on the axis alone — the position of SS along it never enters.
❌ That is the final angular velocity, ωz=LS,z/IS=10 rad/s\omega_z = L_{S,z}/I_S = 10\ \text{rad/s}. Divide by the 3.0 s3.0\ \text{s} it took to get there.
❌ That uses IS=12MR2I_S = \tfrac{1}{2}MR^2, the disk value. A thin ring has all its mass at RR: IS=MR2I_S = MR^2.
Show solution

The torque is constant, so LS,zL_{S,z} grows linearly from 00:

τS,zext=dLS,zdt=2.43.0=0.80 N⋅m\tau_{S,z}^{\text{ext}} = \frac{dL_{S,z}}{dt} = \frac{2.4}{3.0} = 0.80\ \text{N}\cdot\text{m}

With IS=MR2=(1.5)(0.40)2=0.24 kg⋅m2I_S = MR^2 = (1.5)(0.40)^2 = 0.24\ \text{kg}\cdot\text{m}^2, the same for every point on the axle:

αz=τS,zextIS=0.800.24=103≈3.3 rad/s2\alpha_z = \frac{\tau_{S,z}^{\text{ext}}}{I_S} = \frac{0.80}{0.24} = \frac{10}{3} \approx 3.3\ \text{rad/s}^2

Check: ωz=αzt=103(3.0)=10 rad/s\omega_z = \alpha_z t = \tfrac{10}{3}(3.0) = 10\ \text{rad/s} and IS ωz=0.24×10=2.4 kg⋅m2/sI_S\,\omega_z = 0.24 \times 10 = 2.4\ \text{kg}\cdot\text{m}^2/\text{s}.

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