SCIENCE · INVESTIGATE AND EXPLAIN

Reconcile generalized force with a first-branch energy integral

A final mechanics packet links a 2 kg translating mass to a rotor with inertia 0.02 kg m² at effective radius 0.1 m. The ideal rigid coupling imposes ω = v/R and can transmit forces of either sign. It is explicitly bilateral, not a cord that would fail when asked to push. No construction or real loading is requested.

Learning outcome: Derive force/torque effective inertia and independently integrate net work. Retain one accelerating, one decelerating and one unreachable requested state, preserving actual versus requested coordinates.

Starting recap

Reconcile independent equations and evidence, distinguish identifiable combinations from parameters, and retain branch, calibration and observation limitations.

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Try

At G = 1 and x = 4 m, net work is 4 J and speed sqrt(2) m/s, but acceleration is −0.75 m/s². Forward motion and negative acceleration coexist. At x = 5, drive work and resistance conversion are both 5 J and the kinetic energy is zero.

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Predict

With G = 1, a request for x = 6 m on the first forward branch should:

Your prediction and reason

Experiment

Derive force/torque effective inertia and independently integrate net work. Retain one accelerating, one decelerating and one unreachable requested state, preserving actual versus requested coordinates.

Integrate a constrained translation-rotation branch

Worked teaching model · synthetic values · Hypothetical 2 kg translating mass rigidly coupled to rotor I=.02 kg m² at effective radius 0.1 m, so omega=v/.1. Bilateral ideal linkage can transmit either signed force; it is not a pull-only cord. Drive F=gain(6−2x) N, resisting force 1 N during the first forward-moving branch, initially rest at x = 0. Stop at the first turning point; later friction/reversal dynamics are not modeled. No construction.

Requested forward endpoint: 2 m Drive-force multiplier: 1 ratio

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Accumulated net work on the first forward branch

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: integrated drive minus resisting work (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Accumulated net work on the first forward branch

integrated drive minus resisting work

forward position (m)	available kinetic energy (J)
0	0
0.0625	0.3085938
0.125	0.609375
0.1875	0.9023438
0.25	1.1875
0.3125	1.464844
0.375	1.734375
0.4375	1.996094
0.5	2.25
0.5625	2.496094
0.625	2.734375
0.6875	2.964844
0.75	3.1875
0.8125	3.402344
0.875	3.609375
0.9375	3.808594
1	4
1.0625	4.183594
1.125	4.359375
1.1875	4.527344
1.25	4.6875
1.3125	4.839844
1.375	4.984375
1.4375	5.121094
1.5	5.25
1.5625	5.371094
1.625	5.484375
1.6875	5.589844
1.75	5.6875
1.8125	5.777344
1.875	5.859375
1.9375	5.933594
2	6
2.0625	6.058594
2.125	6.109375
2.1875	6.152344
2.25	6.1875
2.3125	6.214844
2.375	6.234375
2.4375	6.246094
2.5	6.25
2.5625	6.246094
2.625	6.234375
2.6875	6.214844
2.75	6.1875
2.8125	6.152344
2.875	6.109375
2.9375	6.058594
3	6
3.0625	5.933594
3.125	5.859375
3.1875	5.777344
3.25	5.6875
3.3125	5.589844
3.375	5.484375
3.4375	5.371094
3.5	5.25
3.5625	5.121094
3.625	4.984375
3.6875	4.839844
3.75	4.6875
3.8125	4.527344
3.875	4.359375
3.9375	4.183594
4	4
4.0625	3.808594
4.125	3.609375
4.1875	3.402344
4.25	3.1875
4.3125	2.964844
4.375	2.734375
4.4375	2.496094
4.5	2.25
4.5625	1.996094
4.625	1.734375
4.6875	1.464844
4.75	1.1875
4.8125	0.9023438
4.875	0.609375
4.9375	0.3085938
5	0

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Requested endpoint on first forward branch	1	flag
First forward turning position	5	m
Reported first-branch position	2	m
Effective translating inertia	4	kg
Signed drive work to reported position	8	J
Forward-branch friction conversion	2	J
Kinetic energy at reported position	6	J
Speed at reported position	1.732051	m/s
Translational kinetic energy	3	J
Rotational kinetic energy	3	J
Acceleration immediately before any turn	0.25	m/s²
Signed linkage force accelerating rotor equivalent mass	0.5	N

The requested position is reachable on the disclosed first forward branch.

The ideal bilateral linkage gives kinetic energy (m+I/R²)v²/2=2v². Net generalized work is integral from 0 to x of [gain(6−2s)−1]ds=(6gain−1)x−gain x². Its positive zero is 6−1/gain. Differentiating the kinetic relation yields acceleration=[gain(6−2x)−1]/4 while v>0. At the known turning endpoint speed is exactly zero; subsequent contact/friction behavior is not inferred. Linkage force can be negative because this stipulated coupling is bilateral, unlike a cord.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

The requested position is reachable on the disclosed first forward branch.

Observation to attach to a capture

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

Understand

Reconcile generalized force with a first-branch energy integral

Total kinetic energy is mv²/2 + I(v/R)²/2 = 2v², equivalent to a generalized translating inertia of 4 kg. The supplied drive force is G(6 − 2x) N, while a 1 N resistance opposes the first forward motion. Starting from rest, the net work integral gives K(x) = (6G − 1)x − Gx².

At G = 1 and x = 2 m, drive work is 8 J and resistance converts 2 J, leaving 6 J kinetic energy. Speed is sqrt(3) m/s; translation and rotation each hold 3 J. Differentiating the energy relation along the forward motion gives a = [G(6 − 2x) − 1]/4, so at that point acceleration is 0.25 m/s².

The positive zero of K is xturn = 6 − 1/G. At G = 1 it is 5 m. A request for x = 6 m is outside the first forward branch: it cannot be accepted by taking the square root of negative energy or by clipping only the speed while retaining the impossible position. The display reports the actual turning boundary with zero speed. Later reversal and friction switching require new dynamics, not an extrapolation of this branch.

A negative instantaneous acceleration before the turn is admissible: the rigid linkage can transmit a reversed force while the system still moves forward. The rotor’s equivalent coupling force is (I/R²)a = 2a. That signed force would require a separate feasibility test for a pull-only cord, illustrating why a kinematic constraint must be tied to an actual force capability.

Changed-case check

Predict before reading the comparison. At G = 1 and x = 4 m, net work is 4 J and speed sqrt(2) m/s, but acceleration is −0.75 m/s². Forward motion and negative acceleration coexist. At x = 5, drive work and resistance conversion are both 5 J and the kinetic energy is zero.

Retained investigation

Sources and evidence limits

Primary conceptual reference checked 30 October 2026: OpenStax University Physics Volume 1. These original derived packets synthesize the course; no source exercise is reproduced. They are supplied mathematical evidence, not observed apparatus or independent practical certification. Preserve the first report, critique, revision and actual reviewer-frozen unfamiliar assignment in the course evidence workflow. A missing human review or unfamiliar response remains missing evidence.

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Which conclusion would change if the linkage were a tension-only cord?

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