Classical-Mechanics · Unit 23 · Video 2 · Interactive Practice

Same Orbit, Different Point: Angular Momentum Tilts Into a Cone but Keeps Its Z-Component

IKey Formulas

Formula	Name	What it says
$\vec{\mathbf{L}}_S = \vec{\mathbf{r}}_S \times \vec{\mathbf{p}}$	Angular momentum about $S$	Depends on the particle's motion and on the point $S$
$\vec{\mathbf{L}}_S = mR^2\omega_z\,\hat{\mathbf{k}} = I_S\,\vec{\boldsymbol{\omega}}$	About the centre of the orbit	One fixed arrow along the axis; $I_S = mR^2$
$\vec{\mathbf{L}}_S = mR^2\omega_z\,\hat{\mathbf{k}} - hmR\omega_z\,\hat{\mathbf{r}}$	About a point a height $h$ below the centre	Tilts off the axis, away from the particle
$\|\vec{\mathbf{L}}_S\| = mR\omega_z\sqrt{h^2+R^2}, \quad \tan\phi = \dfrac{R}{h}$	Magnitude and tilt	Constant magnitude; rises at $\phi$ above the horizontal

Key Insight: The height enters only through $(h\,\hat{\mathbf{k}}) \times \vec{\mathbf{p}}$ , and a vector product with $\hat{\mathbf{k}}$ has no $\hat{\mathbf{k}}$ -component. So $L_{S,z} = mR^2\omega_z$ for every point $S$ on the axis: moving $S$ along the axis changes the direction of $\vec{\mathbf{L}}_S$ , and whether it is constant in time, but never its $z$ -component.

IIThe Vector Product, Term by Term

Splitting $\vec{\mathbf{r}}_S$ into $R\,\hat{\mathbf{r}} + h\,\hat{\mathbf{k}}$ splits $\vec{\mathbf{L}}_S$ into a vertical part and a horizontal part.

Step 1

Step 1 — Position and momentum

\vec{\mathbf{r}}_S = R\,\hat{\mathbf{r}} + h\,\hat{\mathbf{k}}

\vec{\mathbf{p}} = mR\omega_z\,\hat{\boldsymbol{\theta}}

Here

\hat{\mathbf{r}}

points right,

\hat{\mathbf{k}}

up,

\hat{\boldsymbol{\theta}}

into the page.

Step 2 — First term

R\,\hat{\mathbf{r}} \times mR\omega_z\,\hat{\boldsymbol{\theta}} = mR^2\omega_z\,\hat{\mathbf{k}}

since

\hat{\mathbf{r}} \times \hat{\boldsymbol{\theta}} = \hat{\mathbf{k}}

— the same as about the centre.

Step 3 — Second term

h\,\hat{\mathbf{k}} \times mR\omega_z\,\hat{\boldsymbol{\theta}} = -hmR\omega_z\,\hat{\mathbf{r}}

since

\hat{\boldsymbol{\theta}} \times \hat{\mathbf{k}} = \hat{\mathbf{r}}

, so

\hat{\mathbf{k}} \times \hat{\boldsymbol{\theta}} = -\hat{\mathbf{r}}

Step 4 — Sum and magnitude

\vec{\mathbf{L}}_S = mR^2\omega_z\,\hat{\mathbf{k}} - hmR\omega_z\,\hat{\mathbf{r}}

|\vec{\mathbf{L}}_S| = mR\omega_z\sqrt{h^2+R^2}

Step 5 — Direction

-\frac{L_{S,z}}{L_{S,r}} = \frac{mR^2\omega_z}{hmR\omega_z} = \frac{R}{h} = \tan\phi

\vec{\mathbf{L}}_S

rises at

\phi

above the horizontal, on the side opposite the particle.

IIIOne Orbit, a Cone of Arrows

About the centre $\vec{\mathbf{L}}_S$ never moves; about a point below, it turns once per orbit around a cone.

Reference point S Particle angle θ (°) 30

IVSlide S Along the Axis

Wherever $S$ sits on the axis, the vertical part of $\vec{\mathbf{L}}_S$ is $mR^2\omega_z$ ; only the horizontal part changes.

Height of the orbit above S, h (m) 1.4

💡 For a single particle only the centre gives $\vec{\mathbf{L}}_S \parallel \vec{\boldsymbol{\omega}}$ ; points above the orbit tilt $\vec{\mathbf{L}}_S$ toward the particle's side, along $+\hat{\mathbf{r}}$ .

VQuiz Questions

Problem 1 · About the Centre

Given: A particle of mass $m = 0.20\ \text{kg}$ moves counterclockwise (seen from above) in a horizontal circle of radius $R = 0.50\ \text{m}$ with $\omega_z = 4.0\ \text{rad/s}$ . Find the $z$ -component of its angular momentum about the centre of the circle.

✅ Correct!

L_{S,z} = I_S\,\omega_z = mR^2\omega_z = (0.20)(0.50)^2(4.0) = 0.20\ \text{kg}\cdot\text{m}^2/\text{s}

❌ Not quite.

0.40\ \text{kg}\cdot\text{m/s}

is the momentum

p = mR\omega_z

; the angular momentum is

R

times that,

Rp

❌ Not quite. A point particle has

I_S = mR^2

exactly — the factor

\tfrac12

belongs to a solid disc, not a single particle.

❌ Close, but…

0.050\ \text{kg}\cdot\text{m}^2

I_S = mR^2

; multiply by

\omega_z

to get

L_{S,z}

Show solution

About the centre, $\vec{\mathbf{r}}_S = R\,\hat{\mathbf{r}}$ and $\vec{\mathbf{p}} = mR\omega_z\,\hat{\boldsymbol{\theta}}$ are perpendicular and $\hat{\mathbf{r}} \times \hat{\boldsymbol{\theta}} = \hat{\mathbf{k}}$ , so

\vec{\mathbf{L}}_S = mR^2\omega_z\,\hat{\mathbf{k}} = (0.20)(0.50)^2(4.0)\,\hat{\mathbf{k}} = 0.20\,\hat{\mathbf{k}}\ \text{kg}\cdot\text{m}^2/\text{s}.

The arrow points along $+\hat{\mathbf{k}}$ and is the same at every instant.

Problem 2 · A Point Below the Orbit

Given: The same particle ( $m = 0.20\ \text{kg}$ , $R = 0.50\ \text{m}$ , $\omega_z = 4.0\ \text{rad/s}$ ), now with the reference point $S$ on the axis a distance $h = 1.2\ \text{m}$ below the centre of the orbit, so $|\vec{\mathbf{r}}_S| = 1.3\ \text{m}$ . Find $L_{S,z}$ .

✅ Correct! Moving

S

along the axis adds only a horizontal term, so

L_{S,z} = mR^2\omega_z = 0.20\ \text{kg}\cdot\text{m}^2/\text{s}

, exactly as about the centre.

❌ Not quite.

hmR\omega_z = 0.48

is the horizontal part, along

-\hat{\mathbf{r}}

; the vertical part comes from

R\,\hat{\mathbf{r}} \times \vec{\mathbf{p}}

❌ Not quite.

|\vec{\mathbf{r}}_S|\,mv = 0.52

is the full magnitude

|\vec{\mathbf{L}}_S|

; the arrow is tilted, so its

z

-component is smaller.

❌ Not quite.

0.077 = 0.20 \times \tfrac{0.50}{1.3}

projects the

z

-component a second time.

L_{S,z} = mR^2\omega_z

with no factor of

h

|\vec{\mathbf{r}}_S|

Show solution

With $\vec{\mathbf{r}}_S = R\,\hat{\mathbf{r}} + h\,\hat{\mathbf{k}}$ and $\vec{\mathbf{p}} = mR\omega_z\,\hat{\boldsymbol{\theta}}$ :

\vec{\mathbf{L}}_S = mR^2\omega_z\,\hat{\mathbf{k}} - hmR\omega_z\,\hat{\mathbf{r}}.

The term $(h\,\hat{\mathbf{k}}) \times \vec{\mathbf{p}}$ is perpendicular to $\hat{\mathbf{k}}$ , so only the first term has a $z$ -component:

L_{S,z} = mR^2\omega_z = (0.20)(0.50)^2(4.0) = 0.20\ \text{kg}\cdot\text{m}^2/\text{s}.

Problem 3 · Magnitude and Tilt

Given: The setup of Problem 2 ( $m = 0.20\ \text{kg}$ , $R = 0.50\ \text{m}$ , $\omega_z = 4.0\ \text{rad/s}$ , $h = 1.2\ \text{m}$ ). Find $|\vec{\mathbf{L}}_S|$ and the angle $\phi$ at which $\vec{\mathbf{L}}_S$ rises above the horizontal.

What is $|\vec{\mathbf{L}}_S|$ ?

What is $\phi$ ?

✅ Correct!

|\vec{\mathbf{L}}_S| = |\vec{\mathbf{r}}_S|\,mv = (1.3)(0.40) = 0.52\ \text{kg}\cdot\text{m}^2/\text{s}

, rising at

\phi = \arctan(R/h) \approx 22.6\degree

on the side of the axis opposite the particle.

❌ Not quite.

0.20 + 0.48

adds the two parts as numbers, but they are perpendicular — combine them with Pythagoras.

❌ Not quite.

0.48 = hmR\omega_z

is only the horizontal part along

-\hat{\mathbf{r}}

; the vertical part

0.20

adds to the magnitude as well, by Pythagoras.

❌ Not quite.

0.20 = mR^2\omega_z

is only

L_{S,z}

; about a point below the orbit there is also a horizontal part

0.48

, so

|\vec{\mathbf{L}}_S|

is larger.

❌ Check the magnitude. The two parts

0.20\,\hat{\mathbf{k}}

and

-0.48\,\hat{\mathbf{r}}

are perpendicular; the magnitude is the square root of the sum of their squares.

❌ Close, but…

67.4\degree

is the angle above the horizontal of

\vec{\mathbf{r}}_S

, not of

\vec{\mathbf{L}}_S

\vec{\mathbf{L}}_S \perp \vec{\mathbf{r}}_S

, so

\vec{\mathbf{L}}_S

tilts from the horizontal by the angle

\vec{\mathbf{r}}_S

makes with the vertical.

❌ Check the angle.

\tan\phi = \dfrac{L_{S,z}}{|L_{S,r}|} = \dfrac{0.20}{0.48} = \dfrac{R}{h}

— neither part is zero, so

\vec{\mathbf{L}}_S

is neither horizontal nor vertical.

Show solution

The parts are $L_{S,z} = mR^2\omega_z = 0.20$ and $L_{S,r} = -hmR\omega_z = -(1.2)(0.20)(0.50)(4.0) = -0.48$ (both in $\text{kg}\cdot\text{m}^2/\text{s}$ ). They are perpendicular, so

|\vec{\mathbf{L}}_S| = \sqrt{0.20^2 + 0.48^2} = \sqrt{0.2704} = 0.52\ \text{kg}\cdot\text{m}^2/\text{s},

which matches $mR\omega_z\sqrt{h^2+R^2} = (0.40)(1.3) = 0.52$ . For the direction,

\tan\phi = -\frac{L_{S,z}}{L_{S,r}} = \frac{0.20}{0.48} = \frac{R}{h} = \frac{0.50}{1.2} \quad\Rightarrow\quad \phi \approx 22.6\degree.

Problem 4 · A Point Above the Orbit

Given: The same particle orbits in the plane $z = 0$ with $\omega_z = 4.0\ \text{rad/s}$ (counterclockwise from above). The reference point is now $S = (0,\,0,\,1.2\ \text{m})$ , above the orbit, and at this instant the particle is at $(0,\,0.50\ \text{m},\,0)$ . Find $\vec{\mathbf{L}}_S$ in $\text{kg}\cdot\text{m}^2/\text{s}$ .

✅ Correct! With

S

above,

h = -1.2\ \text{m}

-hmR\omega_z\,\hat{\mathbf{r}}

, so the horizontal part points along

+\hat{\mathbf{r}} = +\hat{\mathbf{j}}

, toward the particle's side;

L_{S,z}

is still

0.20

❌ Not quite. That is the answer for

S

1.2 m below the orbit. From above,

\vec{\mathbf{r}}_S

has a

-1.2\,\hat{\mathbf{k}}

leg, which flips the sign of the horizontal part.

❌ Not quite. The

z

-component does not depend on where

S

sits on the axis:

R\,\hat{\mathbf{r}} \times \vec{\mathbf{p}}

still gives

+mR^2\omega_z\,\hat{\mathbf{k}}

❌ Close, but…

\hat{\mathbf{r}}

points from the axis to the particle. At

(0,\,0.50,\,0)

that is

\hat{\mathbf{j}}

, not

\hat{\mathbf{i}}

Show solution

Velocity: $\vec{\mathbf{v}} = \vec{\boldsymbol{\omega}} \times \vec{\mathbf{r}} = 4.0\,\hat{\mathbf{k}} \times 0.50\,\hat{\mathbf{j}} = -2.0\,\hat{\mathbf{i}}\ \text{m/s}$ , so $\vec{\mathbf{p}} = -0.40\,\hat{\mathbf{i}}\ \text{kg}\cdot\text{m/s}$ .

Position from $S$ : $\vec{\mathbf{r}}_S = 0.50\,\hat{\mathbf{j}} - 1.2\,\hat{\mathbf{k}}$ . Term by term, using $\hat{\mathbf{j}} \times \hat{\mathbf{i}} = -\hat{\mathbf{k}}$ and $\hat{\mathbf{k}} \times \hat{\mathbf{i}} = \hat{\mathbf{j}}$ :

\vec{\mathbf{L}}_S = (0.50\,\hat{\mathbf{j}}) \times (-0.40\,\hat{\mathbf{i}}) + (-1.2\,\hat{\mathbf{k}}) \times (-0.40\,\hat{\mathbf{i}}) = 0.20\,\hat{\mathbf{k}} + 0.48\,\hat{\mathbf{j}}.

This is $mR^2\omega_z\,\hat{\mathbf{k}} - hmR\omega_z\,\hat{\mathbf{r}}$ with $h = -1.2\ \text{m}$ and $\hat{\mathbf{r}} = \hat{\mathbf{j}}$ : the arrow leans toward the particle, and its $z$ -component is unchanged.

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