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 experimentPredict
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.
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Exact angular velocity
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Exact angular velocity: numerical coordinates
| time (s) | omega (rad/s) |
|---|---|
| 0 | 0 |
| 0.125 | 0.363522 |
| 0.25 | 0.705019 |
| 0.375 | 1.02583 |
| 0.5 | 1.3272 |
| 0.625 | 1.61031 |
| 0.75 | 1.87626 |
| 0.875 | 2.12611 |
| 1 | 2.36082 |
| 1.125 | 2.5813 |
| 1.25 | 2.78843 |
| 1.375 | 2.98301 |
| 1.5 | 3.1658 |
| 1.625 | 3.33752 |
| 1.75 | 3.49883 |
| 1.875 | 3.65037 |
| 2 | 3.79272 |
| 2.125 | 3.92646 |
| 2.25 | 4.05209 |
| 2.375 | 4.1701 |
| 2.5 | 4.28097 |
| 2.625 | 4.38512 |
| 2.75 | 4.48296 |
| 2.875 | 4.57488 |
| 3 | 4.66122 |
| 3.125 | 4.74233 |
| 3.25 | 4.81853 |
| 3.375 | 4.89011 |
| 3.5 | 4.95736 |
| 3.625 | 5.02053 |
| 3.75 | 5.07987 |
| 3.875 | 5.13562 |
| 4 | 5.18799 |
| 4.125 | 5.23719 |
| 4.25 | 5.2834 |
| 4.375 | 5.32682 |
| 4.5 | 5.3676 |
| 4.625 | 5.40592 |
| 4.75 | 5.44191 |
| 4.875 | 5.47573 |
| 5 | 5.50749 |
| Quantity | Value | Unit |
|---|---|---|
| Steady angular velocity | 6 | rad/s |
| Angular response time | 2 | s |
| Angular velocity at 2 s | 3.79272 | rad/s |
| Angular displacement at 2 s | 4.41455 | rad |
| Angular acceleration at 2 s | 1.10364 | rad/s² |
| Net torque at 2 s | 2.20728 | N 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.
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Which assumptions justify applying I alpha instead of differentiating a more general angular-momentum expression?
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