Fit rotational inertia without hiding resistance or readout bias
The drive torque on a rotating body need not equal its net torque. In this fixed-axis model, an opposing torque q acts while the body rotates positively, so τdrive−q=Iα. For unchanged I and constant q, plotting drive torque against independently known angular acceleration gives slope I and intercept q.
Learning outcome: Vary resistance with zero fault, then vary the one-reading fault at fixed resistance. Retain slope, intercept and held-out residual. Propose a third condition that would distinguish changing resistance from a shared acceleration scale error.
Starting recap
Declare the axis, angular units, recipient and valid contact or material branch. Distinguish supplied shape factors from independently measured inertia.
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Try
At q=0 and δ=0.4 N m, the fit gives inertia 2.2 kg m² and intercept−0.2 N m. The sign is a diagnostic warning about the assumed passive-resistance interpretation, not a physical negative-friction conclusion.
Open the experimentPredict
For drive torque τ=Iα+q with constant q, the slope of τ versus α is:
Experiment
Vary resistance with zero fault, then vary the one-reading fault at fixed resistance. Retain slope, intercept and held-out residual. Propose a third condition that would distinguish changing resistance from a shared acceleration scale error.
Infer inertia and resistance from a drive-torque record
Worked teaching model · synthetic values · True fixed-axis inertia 2 kg m². At positive angular velocity, resistance is a constant opposing torque. Drive readings at independently exact angular accelerations 1 and 3 rad/s² include an extra fault only in the second; midpoint reading at 2 is held out. No real bearing law is established.
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Drive torque fit with one held-out state
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: fitted drive relation (solid).2: fitting readings (filled markers).3: held-out reading (filled markers). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Drive torque fit with one held-out state
fitted drive relation
| angular acceleration (rad/s²) | indicated drive torque (N m) |
|---|---|
| 0 | 0.9 |
| 3 | 7.2 |
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.
fitting readings
| angular acceleration (rad/s²) | indicated drive torque (N m) |
|---|---|
| 1 | 3 |
| 3 | 7.2 |
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.
held-out reading
| angular acceleration (rad/s²) | indicated drive torque (N m) |
|---|---|
| 2 | 5 |
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.
| Quantity | Value | Unit |
|---|---|---|
| Drive reading at alpha 1 | 3 | N m |
| Drive reading at alpha 3 | 7.2 | N m |
| Inferred inertia | 2.1 | kg m² |
| Inferred opposing torque | 0.9 | N m |
| Held-out reading at alpha 2 | 5 | N m |
| Held-out residual | -0.1 | N m |
The slope estimates inertia while the intercept represents opposing torque only under the stated model.
Drive torque minus resistance equals I alpha. Differencing two readings cancels constant resistance, but a fault confined to one reading alters both slope and intercept. Negative inferred opposing torque in a faulty packet is a warning about assumptions, not proof that a passive bearing supplies energy. Fixed axis and unchanged distribution are necessary for this constant-I relation.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
The slope estimates inertia while the intercept represents opposing torque only under the stated model.
Drive torque minus resistance equals I alpha. Differencing two readings cancels constant resistance, but a fault confined to one reading alters both slope and intercept. Negative inferred opposing torque in a faulty packet is a warning about assumptions, not proof that a passive bearing supplies energy. Fixed axis and unchanged distribution are necessary for this constant-I relation.
Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.
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Why does an exact two-point fit not validate the model that made those two points sufficient?
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