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SCIENCE · INVESTIGATE AND EXPLAIN

Integrate a rotational torque law with its initial condition

Extend fixed-inertia torque dynamics to a bounded velocity-dependent resistance model.

Learning outcome: At b=1 compare initial omega 0=0,6,8. Derive the rate and angle, retain net torque and explain the different approach directions. Then double b and predict the steady value and response time before evaluating.

Starting recap

Declare the fixed axis, integrate the actual mass or load distribution, and test constraint feasibility before interpreting a numerical result.

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Try

b=1,omega 0=0 gives omega(2)≈3.79272 and theta(2)≈4.41455. Starting at omega 0=8 instead approaches 6 from above.

Open the experiment

Predict

Does zero net torque require a fixed-inertia body to stop rotating?

Experiment

At b=1 compare initial omega 0=0,6,8. Derive the rate and angle, retain net torque and explain the different approach directions. Then double b and predict the steady value and response time before evaluating.

Integrate a viscous angular initial-value model

Worked teaching model · synthetic values · Fixed-axis rigid body with I=2 kg m², applied torque 6 N m and resistive torque −b omega. Constant I and coefficients, no bearing threshold. Evaluate angle/omega at 2 s; optional rotational ODE extension beyond constant-torque cases.

Reset keeps your prediction, captures and practice history.

Exact angular velocity

omega (rad/s)time (s)055.507490

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.

Exact angular velocity: numerical coordinates
time (s)omega (rad/s)
00
0.1250.363522
0.250.705019
0.3751.02583
0.51.3272
0.6251.61031
0.751.87626
0.8752.12611
12.36082
1.1252.5813
1.252.78843
1.3752.98301
1.53.1658
1.6253.33752
1.753.49883
1.8753.65037
23.79272
2.1253.92646
2.254.05209
2.3754.1701
2.54.28097
2.6254.38512
2.754.48296
2.8754.57488
34.66122
3.1254.74233
3.254.81853
3.3754.89011
3.54.95736
3.6255.02053
3.755.07987
3.8755.13562
45.18799
4.1255.23719
4.255.2834
4.3755.32682
4.55.3676
4.6255.40592
4.755.44191
4.8755.47573
55.50749
Calculated values under the declared model
QuantityValueUnit
Steady angular velocity6rad/s
Angular response time2s
Angular velocity at 2 s3.79272rad/s
Angular displacement at 2 s4.41455rad
Angular acceleration at 2 s1.10364rad/s²
Net torque at 2 s2.20728N m

Constant rotational inertia turns the torque law into a first-order angular initial-value equation.

Solve I dω/dt=6−bω with its initial condition, then integrate ω for angle. The steady value can be approached from above or below; exact initial equilibrium stays there. This optional ODE extension assumes the same fixed axis and resistance law throughout.

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

Constant rotational inertia turns the torque law into a first-order angular initial-value equation.

Solve I dω/dt=6−bω with its initial condition, then integrate ω for angle. The steady value can be approached from above or below; exact initial equilibrium stays there. This optional ODE extension assumes the same fixed axis and resistance law throughout.

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

Understand

Begin with the core rotational equation

For a rigid body about a fixed axis with constant I, net torque is I alpha=I dω/dt. A torque changes angular velocity; a nonzero angular velocity does not by itself require nonzero net torque. Inertia depends on the mass distribution and axis, not on the value of angular speed in this model.

Declare optional differential-equation enrichment

The following exact viscous-resistance solution is optional calculus enrichment of the rotational force law. The workbench stipulates I=2 kg m², drive torque 6 N m and resistance−bω. The simple law is not claimed to model every bearing, static friction threshold or actual apparatus.

Solve for angular velocity

The ODE is dω/dt=(6−bω)/2. Define steady angular velocity ωinfinity=6/b and response time tau=2/b. Separation or direct substitution gives ω=ωinfinity+(ω0−ωinfinity)exp(−t/tau). The initial condition determines approach from below, approach from above or exact steady rotation. A terminal value is not a universal upper bound on the initial speed.

Integrate again for angle

With theta 0=0, theta=ωinfinity t+(ω0−ωinfinity)tau(1−exp(−t/tau)). The angle has a long-time linear term even though angular acceleration decays to zero. For b=1,ω0=0,t=2, omega=6(1−e⁻¹)≈3.79272 rad/s and theta=12e⁻¹≈4.41455 rad. Differentiate theta to verify the velocity and its initial value.

Test an unchanged equilibrium state

For b=1 andω0=6, drive and resistance balance at every instant. Angular velocity remains 6, net torque and alpha are zero, and theta at 2 s is 12 rad. This is rotational equilibrium with motion, not static equilibrium. A changed I, axis or friction law would require a different model, not reuse of the same exponential by habit.

Worked original case

b=1,omega 0=0 gives omega(2)≈3.79272 and theta(2)≈4.41455. Starting at omega 0=8 instead approaches 6 from above.

Investigation

At b=1 compare initial omega 0=0,6,8. Derive the rate and angle, retain net torque and explain the different approach directions. Then double b and predict the steady value and response time before evaluating.

Public numerical transfer practice

Try a new practice case

For drive torque 8 N m and viscous coefficient 2 N m s/rad, what is steady angular velocity 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 teaching and numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax fixed-axis rotation. Supplied/generated evidence is not physical measurement, a load rating or laboratory participation. No rotating machinery, hanging weights or mechanical construction is requested. Public feedback and retained self-reports do not certify independent learner mastery.

Apply

Does zero net torque require a fixed-inertia body to stop rotating?

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

Which assumptions justify applying I alpha instead of differentiating a more general angular-momentum expression?

Continue

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