Recognize angular momentum in straight-line motion
Angular momentum is specified relative to a reference point or axis. A rigid body about a fixed principal rotation axis has planar angular momentum L=Iω. A moving point particle has magnitude mvr sinφ, where r runs from the reference point to the particle and φ is the angle between that radius and its velocity. The same expression is momentum times perpendicular distance to the line of motion.
Learning outcome: Compare offsets +1, 0 and −1 m at speed 3 m/s. Draw the actual horizontal paths separately from the angular-momentum plot. Explain the unchanged linear momentum, changing reference-dependent angular momentum and zero torque in each case.
Starting recap
Review axis-specific inertia, force application locations, torque-angle versus torque-time area, energy conservation and system boundaries.
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Try
At v=3 m/s and y=+1 m, the 2 kg particle has linear momentum +6 kg m/s and angular momentum −6 kg m²/s about the origin. At x=0 its radius is perpendicular to motion; at x=√3 m the radius is 2 m and the angle is 30°, giving the same magnitude 2×3×2×½=6.
Open the experimentPredict
If the particle is farther from the reference point later along the same straight path, must its angular momentum increase?
Experiment
Compare offsets +1, 0 and −1 m at speed 3 m/s. Draw the actual horizontal paths separately from the angular-momentum plot. Explain the unchanged linear momentum, changing reference-dependent angular momentum and zero torque in each case.
Compare straight-line motion about a chosen reference
Worked teaching model · synthetic values · A2 kg particle travels along the horizontal line y=b at constant speed a. The reference is fixed at the origin of an inertial frame. Signed planar angular momentum uses counterclockwise tendency as positive; no three-dimensional vector direction is required.
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Spatial path, not the angular-momentum graph
The origin is fixed. The particle moves right on y=1 m at 3 m/s. Its perpendicular line offset is 1 m; signed angular momentum is -6 kg m²/s.
The dashed horizontal line is the trajectory. The separate graph below plots angular momentum against position; its vertical axis is not spatial height.
Axes rescale with settings; compare numerical values, not screen angles or apparent lengths. Solid/points show the selected model relationship or observations; the dashed comparison is explained below.
| Quantity | Value | Unit |
|---|---|---|
| Particle linear momentum | 6 | kg m/s |
| Signed angular momentum about origin | -6 | kg m²/s |
| Perpendicular distance to line of motion | 1 | m |
| Torque about origin | 0 | N m |
Accessible plot coordinates
| particle horizontal position (m) | signed angular momentum (kg m²/s) |
|---|---|
| -3 | -6 |
| -2 | -6 |
| -1 | -6 |
| 0 | -6 |
| 1 | -6 |
| 2 | -6 |
| 3 | -6 |
Straight-line motion can have nonzero angular momentum about a chosen reference point.
The magnitude mvr sinφ equals momentum times perpendicular line offset. Here signed L=−mvy is constant even as distance r and angle φ change. Moving the reference onto the particle’s line makes this angular momentum zero without changing linear momentum.
Straight-line motion can have nonzero angular momentum about a chosen reference point.
The magnitude mvr sinφ equals momentum times perpendicular line offset. Here signed L=−mvy is constant even as distance r and angle φ change. Moving the reference onto the particle’s line makes this angular momentum zero without changing linear momentum.
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Public practice checks a specific claim. It does not establish independent mastery.
Why would using mr times speed without the angle factor fail for a straight-line trajectory?
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