Conservation of angular momentum — conditions: Question bank

Before you start, keep three plain-word anchors in mind:

Angular momentum $L$ = "how much spin/orbit the system carries" — for a spinning body $L = I ω$ , where $I$ (the Moment of inertia) says how spread-out the mass is, and $ω$ is the turning rate.
Torque $τ$ = "sideways twist" — the rate-changer of $L$ , just as force changes momentum. See Torque.
The master law: $\frac{d L}{d t} = τ_{e x t, n e t}$ . No external twist ⟹ no change in spin.

True or false — justify

Every answer must give the reason, never a bare yes/no.

If a system's kinetic energy is not conserved, its angular momentum cannot be conserved either.

False — they have independent conditions;

L

needs zero external torque while KE needs no work done, so a skater's

L

stays fixed while her KE jumps because her muscles do work.

Internal forces between parts of a system can change the total angular momentum.

False — internal forces come in collinear third-law pairs, giving torque

(r_{1} - r_{2}) \times f = 0

since

f

lies along

(r_{1} - r_{2})

, so they cancel. See Newton's third law.

Angular momentum, if conserved, is conserved about every possible axis.

False — it is conserved only about axes where the external torque component is zero; a body can have

τ_{e x t} = 0

about one axis and nonzero about another.

For a planet orbiting the Sun, gravity does zero net torque about the Sun.

True — gravity is a central force pointing straight at the Sun, so

r ∥ F

and

τ = r \times F = 0

The value of

L

for a given object is the same no matter which origin you pick.

False —

L = r \times p

depends on

r

, which is measured from the origin, so different origins give different

L

values.

Angular momentum is conserved even when large external forces act, provided their net torque is zero.

True — what matters is torque, not force; forces passing through the origin (or cancelling their twists) leave

L

unchanged.

A rigid body with a constant

L

must have a constant angular velocity

ω

False — if the mass redistributes and

I

changes,

ω

changes to keep

I ω

fixed, as in the skater pulling arms in.

Kepler's equal-area law is a direct consequence of angular momentum conservation.

True — constant

L

means the rate of sweeping area is constant, so equal areas are swept in equal times. See Kepler's laws.

Spot the error

Each statement below contains a flaw. Name it in one or two sentences.

"The skater speeds up when she pulls her arms in, so angular momentum increased."

Error —

L = I ω

stays constant;

ω

rises only because

I

falls, so the product is unchanged, not increased.

"Gravity pulls the planet, so it does work and changes the planet's angular momentum."

Error — the relevant quantity for changing

L

is torque, not force or work; central gravity gives zero torque, so

L

is untouched.

"We can only use conservation of

L

if there are no external forces on the system."

Error — you also may use it whenever external forces exist but their net torque is zero (e.g. central forces through the origin).

"The putty sticks to the turntable, energy is lost to heat, therefore the turntable's

L

decreased."

Error — the impact force is internal and gravity is parallel to the axis, so

L

about the axis is conserved; energy loss and

L

conservation are separate matters.

"Since

L = r \times p

, an object moving in a straight line has zero angular momentum."

Error — unless the line passes through the origin,

r

and

p

are not parallel, so

L = r \times p \neq = 0

"A wheel rolling down a ramp conserves angular momentum because it is just rolling freely."

Error — gravity produces a nonzero torque about the contact-line axis, so

L

about that axis is not conserved; choose the axis before claiming conservation.

"Internal muscular forces can't change

ω

, since internal forces don't affect the system."

Error — internal forces can change

ω

by changing

I

(moving mass); what they cannot change is the total

L

Why questions

Answer each with the underlying reason.

Why does only the external torque appear in

\frac{d L}{d t} = τ_{e x t, n e t}

Because internal third-law force pairs are collinear, their combined torque

(r_{1} - r_{2}) \times f

vanishes, leaving only external torques.

Why must we differentiate

L

to derive the conservation condition?

Conservation means

L

does not change in time, i.e.

\frac{d L}{d t} = 0

, so we must compute that derivative to see exactly when it is zero.

Why does the first term

v \times m v

vanish in the derivation?

The cross product of any vector with itself is zero because the two vectors are parallel, so

v \times v = 0

Why is

L

conserved for the skater but her rotational kinetic energy is not?

No external torque keeps

L

fixed, but her muscles do internal work to pull the mass inward, raising Rotational kinetic energy

K E = \frac{1}{2} I ω^{2}

Why does choosing the origin cleverly help when applying conservation of

L

Because

τ

depends on the origin, picking an origin where

τ_{e x t} = 0

lets you use conservation even when it fails about other points.

Why is a central force the "classic" zero-torque case?

A central force points along

r

(toward/away from the origin), so

τ = r \times F = 0

because parallel vectors have zero cross product.

Why can we write

L = m v r

for a planet only at perihelion and aphelion?

There the velocity is perpendicular to the radius, so

∣ r \times p ∣ = m v r

exactly; elsewhere

v

has a radial component and the formula needs the perpendicular part.

Why is torque called the "pace-setter" of angular momentum?

Because it equals the rate of change of

L

, playing exactly the role force plays for linear momentum in Conservation of linear momentum.

Edge cases

Boundary and degenerate scenarios — reason through each.

If the net external torque is zero but a single external force is nonzero, is

L

conserved?

Yes — conservation depends on torque, not force; a nonzero force through the origin still gives zero torque and constant

L

If a particle sits exactly at the origin (

r = 0

), what is its angular momentum?

It is zero, since

L = r \times p = 0 \times p = 0

regardless of how fast it moves.

If a body's angular velocity

ω

is zero, is its angular momentum necessarily zero?

For a rigid body spinning about a fixed axis,

L = I ω = 0

; but a translating particle can still carry orbital

L = r \times p

about an off-line origin even with no spin.

Can angular momentum be conserved in direction but not magnitude, or vice versa?

No for a truly constant

L

— conservation means the whole vector is fixed, so both magnitude and direction are constant; if either changes, some external torque acted.

What happens to

L

over one full orbit of a planet, given gravity is central?

It stays exactly constant throughout, because the torque is zero at every instant, not just on average.

If external torque is zero only along the vertical axis, what can we still conclude?

Only the vertical component

L_{z}

is conserved; the horizontal components of

L

may change since their torques are nonzero.

As the skater's moment of inertia

I \to

very small (arms tucked tight), what happens to

ω

ω \to

very large to keep

I ω

fixed, so she spins ever faster — the limiting behaviour of

I_{1} ω_{1} = I_{2} ω_{2}

Recall One-line summary of every trap here

Every trap on this page is one of four confusions: (1) conflating $L$ with KE, (2) thinking internal forces move total $L$ , (3) forgetting conservation is per-axis and origin-dependent, or (4) confusing force with torque. Fix these four and the topic is yours.

Connections

Parent topic: conditions for conservation
Torque — the twist whose absence is the whole condition.
Moment of inertia — the $I$ that trades against $ω$ .
Newton's third law — why internal torques cancel.
Central forces — the star zero-torque example.
Kepler's laws — equal-area law from constant $L$ .
Conservation of linear momentum — the translational twin.
Rotational kinetic energy — why $L$ fixed ≠ KE fixed.