SCIENCE · INVESTIGATE AND EXPLAIN

Differentiate particle angular momentum about a fixed origin

Use a cross product for a moving particle and track an origin change consistently.

Learning outcome: Compare two origins for the same assigned r,p,F. Compute both cross products directly and through the translation identities. Find a supported case with zero torque and explain why its angular momentum need not be zero. Identify which change would invalidate the fixed-origin derivative proof.

Starting recap

State the reference axis and system, distinguish time and angle integrals, and verify an accessible and supported motion branch.

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Try

At a=1,b=2, Lz=−4 kg m²/s and tau_z=−4 N m. About the shifted origin, Lprime=−6 and tauprime=−2. Directly using rprime=(0,2) reproduces both results.

Open the experiment

Predict

Can a straight-moving particle have nonzero angular momentum about a fixed point?

Your prediction and reason

Experiment

Compare two origins for the same assigned r,p,F. Compute both cross products directly and through the translation identities. Find a supported case with zero torque and explain why its angular momentum need not be zero. Identify which change would invalidate the fixed-origin derivative proof.

Compute angular momentum about two fixed origins

Worked teaching model · synthetic values · At one instant r=(x,2) m,p=(3,py) kg m/s and force(1,−2) N. Compare originO=(0,0) and fixedOprime=(1,0). Cross products in the plane; not a required circular path.

Particle x coordinate: 1 m Momentum y component: 2 kg m/s

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Instantaneous position components

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Instantaneous position components: numerical coordinates

x coordinate (m)	y coordinate (m)
0	0
1	2

Calculated values under the declared model
Quantity	Value	Unit
Angular momentum about O	-4	kg m²/s
Angular momentum about shifted Oprime	-6	kg m²/s
Torque about O	-4	N m
Torque about shifted Oprime	-2	N m
Origin-shift change in angular momentum	-2	kg m²/s

Particle angular momentum is r cross p about a specified reference point, not only a spinning-body quantity.

In the plane Lz=x py−y px and tauz=x Fy−y Fx. Differentiating r cross p about a fixed inertial origin leaves r cross F because v cross mv=0. Changing the fixed origin changes both L and torque consistently. Component-plot axes may rescale independently; infer directions from numbers, not screen angles.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Particle angular momentum is r cross p about a specified reference point, not only a spinning-body quantity.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Angular momentum is not reserved for spinning disks

A particle with position r relative to a specified origin and linear momentum p has L=r cross p. In a plane, Lz=x py−y px. A particle moving in a straight line can have nonzero angular momentum about an off-line origin. The right-hand rule gives its direction; an axis sign convention must be stated before assigning a sign.

Derive the torque relation

For a fixed inertial origin, d(r cross p)/dt=v cross p+r cross dp/dt. Since p=mv for a constant-mass particle, v cross p=0, leaving dL/dt=r cross Fnet=tau. This derivation is why the same origin is needed in the torque and angular-momentum statements. Moving-origin formulas require additional care and are outside this workbench.

Shift the origin rather than the particle

The assigned instantaneous r=(a,2) m, p=(3,b) kg m/s and F=(1,−2) N give Lz=ab−6 and tau_z=−2a−2. Shift to another fixed origin at (1,0) m: rprime=r−(1,0), so Lprime=L−b and tauprime=tau+2. Neither difference is an extra physical impulse; it is a changed reference.

Separate cancellation from absence

For multiple particles, total angular momentum is the vector sum about one common origin. Oppositely directed contributions can cancel despite nonzero individual momenta. Likewise zero instantaneous angular momentum does not imply zero torque or zero motion. The component plot may scale axes independently; read numerical coordinates rather than estimating angles from screen geometry.

Worked original case

At a=1,b=2, Lz=−4 kg m²/s and tau_z=−4 N m. About the shifted origin, Lprime=−6 and tauprime=−2. Directly using rprime=(0,2) reproduces both results.

Investigation

Public numerical transfer practice

Try a new practice case

A particle has r=(2,1) m and p=(3,4) kg m/s about a fixed origin. What is signed Lz in kg m²/s?

Numerical result in the requested units

This public, unsaved check is not a secure assessment. Explain the derivative, integral or boundary in the retained course evidence workspace.

Sources and evidence boundary

Original derivations and assigned numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax angular momentum. These are supplied or generated records, not physical measurements or performed laboratory time. Do not construct rotating equipment, release loads or conduct launch experiments. Public feedback and retained self-reports do not certify independent mastery.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Why is radius times momentum insufficient when the vectors are not perpendicular?

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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Previous: Reconstruct angular momentum from a torque history Next: Conserve angular momentum while inertia changes

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