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SCIENCE · INVESTIGATE AND EXPLAIN

Identify torsional stiffness with a known inertia increment

Use a second controlled state to break a parameter ambiguity and audit a biased period.

Learning outcome: Compare two added inertias at unbiased settings, then bias only the second period. Derive both inferred parameters and explain the sensitivity of a difference. Contrast this with a written common-gain case. Propose an independent timing or stiffness check rather than treating the known generator values as measured evidence.

Starting recap

Derive the restoring equation, retain initial conditions and distinguish exact harmonic results from approximation and calibration assumptions.

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Try

With added inertia1 kg m² and gain1, T1²=2pi² and T2²=3pi². Their difference pi² gives kappa=4 and I=2 exactly. If both periods were multiplied by1.1, the same method would still infer I=2 but kappa=4/1.21≈3.305785.

Open the experiment

Predict

What does one torsional period determine without independent inertia or stiffness information?

Experiment

Compare two added inertias at unbiased settings, then bias only the second period. Derive both inferred parameters and explain the sensitivity of a difference. Contrast this with a written common-gain case. Propose an independent timing or stiffness check rather than treating the known generator values as measured evidence.

Use a known inertia increment to identify torsional stiffness

Worked teaching model · synthetic values · Synthetic true I=2 kg m²,kappa=4 N m/rad. First period unbiased; second after known inertia addition carries selected gain. Same linear torsional stiffness, no friction or unknown attachment inertia. A separate common-clock comparison is discussed in the lesson.

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True period squared versus added inertia

period squared (s²)added inertia (kg m²)0239.47840

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.

True period squared versus added inertia: numerical coordinates
added inertia (kg m²)period squared (s²)
019.7392
0.033333320.0682
0.066666720.3972
0.120.7262
0.13333321.0552
0.16666721.3841
0.221.7131
0.23333322.0421
0.26666722.3711
0.322.7001
0.33333323.0291
0.36666723.3581
0.423.6871
0.43333324.016
0.46666724.345
0.524.674
0.53333325.003
0.56666725.332
0.625.661
0.63333325.99
0.66666726.3189
0.726.6479
0.73333326.9769
0.76666727.3059
0.827.6349
0.83333327.9639
0.86666728.2929
0.928.6219
0.93333328.9508
0.96666729.2798
129.6088
1.0333329.9378
1.0666730.2668
1.130.5958
1.1333330.9248
1.1666731.2537
1.231.5827
1.2333331.9117
1.2666732.2407
1.332.5697
1.3333332.8987
1.3666733.2277
1.433.5567
1.4333333.8856
1.4666734.2146
1.534.5436
1.5333334.8726
1.5666735.2016
1.635.5306
1.6333335.8596
1.6666736.1885
1.736.5175
1.7333336.8465
1.7666737.1755
1.837.5045
1.8333337.8335
1.8666738.1625
1.938.4915
1.9333338.8204
1.9666739.1494
239.4784
Calculated values under the declared model
QuantityValueUnit
First period4.44288s
Indicated second period5.4414s
Indicated period-squared difference9.8696s²
Inferred torsional stiffness4N m/rad
Inferred original inertia2kg m²
True torsional stiffness4N m/rad
True original inertia2kg m²

A known inertia increment adds an identifying equation, but a biased second period can corrupt the difference.

For torque=−kappa theta, I theta double-dot=−kappa theta and T²=4pi² I/kappa. Two periods with known DeltaI yield kappa=4pi² DeltaI/(T2²−T1²). One period alone identifies I/kappa, not both. If both periods share the same gain g, inferred I is unchanged but kappa is divided by g²; this control instead changes only the second period.

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

A known inertia increment adds an identifying equation, but a biased second period can corrupt the difference.

For torque=−kappa theta, I theta double-dot=−kappa theta and T²=4pi² I/kappa. Two periods with known DeltaI yield kappa=4pi² DeltaI/(T2²−T1²). One period alone identifies I/kappa, not both. If both periods share the same gain g, inferred I is unchanged but kappa is divided by g²; this control instead changes only the second period.

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

Understand

Derive the torsional oscillator

A linear torsional restoring torque is −kappa theta, where kappa has units N m/rad. Rotational dynamics gives I theta double-dot=−kappa theta, so omega=sqrt(kappa/I) and T=2pi sqrt(I/kappa). The potential energy is kappa theta²/2 and total energy is I omega_angle²/2+kappa theta²/2. Here omega_angle denotes instantaneous angular velocity, distinct from the fixed oscillation angular frequency.

Identify the one-period ambiguity

A single measured T determines I/kappa=T²/(4pi²), not I and kappa separately. Doubling both inertia and stiffness leaves the same period. Assuming one parameter while reporting the other as independently measured would overstate what the observation establishes.

Add a known independent inertia increment

With a known added coaxial inertia DeltaI and unchanged stiffness, T2²−T1²=4pi² DeltaI/kappa. Therefore kappa=4pi² DeltaI/(T2²−T1²) and I=kappa T1²/(4pi²). The added inertia is independently stipulated here; unknown attachment inertia or changed stiffness would undermine the identification.

Challenge the subtraction with a clock alternative

The generator has true I=2 kg m² and kappa=4 N m/rad. The first period is unbiased; the selected gain changes only the second. A small period-squared difference can amplify that fault. If instead both periods shared the same gain g, inferred I would remain correct while inferred kappa would be divided by g². These are two different error models; neither a good algebraic fit nor repeated readings proves a calibrated clock.

Worked original case

With added inertia1 kg m² and gain1, T1²=2pi² and T2²=3pi². Their difference pi² gives kappa=4 and I=2 exactly. If both periods were multiplied by1.1, the same method would still infer I=2 but kappa=4/1.21≈3.305785.

Investigation

Compare two added inertias at unbiased settings, then bias only the second period. Derive both inferred parameters and explain the sensitivity of a difference. Contrast this with a written common-gain case. Propose an independent timing or stiffness check rather than treating the known generator values as measured evidence.

Public numerical transfer practice

Try a new practice case

For a torsional oscillator with I=2 kg m² and kappa=8 N m/rad, what is oscillation angular frequency in rad/s?

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 harmonic motion and pendulums. 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

What does one torsional period determine without independent inertia or stiffness information?

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

Why does a second period help only when the added inertia and unchanged stiffness are justified?

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