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Rotational dynamics simulation

A numerical rotational model advances angular state from torque. The calculation follows the same broad pattern as translational motion: torque determines angular acceleration, angular acceleration changes angular velocity, and angular velocity changes orientation. The result is trustworthy only when the fixed-axis assumptions and numerical errors are made explicit.

Fixed-axis physical model

For a rigid body constrained to rotate about one fixed axis with constant moment of inertia \(I\), the rotational equation of motion is

\[\tau_{\mathrm{net}}=I\alpha=I\frac{d\omega}{dt}\]

Angular velocity and angle are connected by

\[\omega=\frac{d\theta}{dt}\]

Together, these form a first-order state system:

\[\dot\theta=\omega,\qquad \dot\omega=\frac{\tau_{\mathrm{net}}(\theta,\omega,t)}{I}\]

The angle is in radians, angular velocity in radians per second, torque in newton metres, and inertia in kilogram metres squared. Their ratio has units of radians per second squared, conventionally written even though radians are dimensionless in SI.

inertia, torque = 0.5, 3.0
angle, omega = 0.0, 0.0
alpha = torque / inertia
print(f"Angular acceleration: {alpha:.1f} rad/s^2")

Constant torque as an exact fixture

For constant torque and constant inertia, angular acceleration is constant:

\[\alpha=\frac{\tau}{I}\]

The exact solution from initial angle \(\theta_0\) and angular velocity \(\omega_0\) is

\[\omega(t)=\omega_0+\frac{\tau}{I}t\] \[\theta(t)=\theta_0+\omega_0t+\frac12\frac{\tau}{I}t^2\]

This case tests signs, units, update order, and convergence. Doubling torque doubles angular acceleration; doubling inertia halves it. Reversing torque reverses acceleration. Zero torque leaves angular velocity constant rather than forcing it to zero.

Torque models

A constant motor torque is only one possibility. A torsional spring produces restoring torque

\[\tau_s=-\kappa(\theta-\theta_{\mathrm{eq}})\]

where \(\kappa\) has units of newton metres per radian. Linear rotational damping is often approximated by

\[\tau_d=-b\omega\]

with damping coefficient \(b\). An applied drive may depend explicitly on time. A combined model could be

\[I\ddot\theta=-\kappa(\theta-\theta_{\mathrm{eq}})-b\dot\theta+\tau_{\mathrm{drive}}(t)\]

Every term must use the same positive rotation convention. Friction and damping transfer mechanical energy; they are not reasons to expect constant rotational energy.

A basic discrete update

Forward Euler evaluates derivatives at the old state:

\[\theta_{n+1}=\theta_n+\omega_n\Delta t,\qquad \omega_{n+1}=\omega_n+\alpha_n\Delta t\]

It is simple but can add artificial energy to an undamped torsional oscillator. A semi-implicit Euler update changes angular velocity before angle:

\[\omega_{n+1}=\omega_n+\alpha_n\Delta t,\qquad \theta_{n+1}=\theta_n+\omega_{n+1}\Delta t\]

This method is still first order, but often behaves better for conservative oscillatory motion. The order of two short lines changes the numerical method and its energy behavior.

dt = 0.01
omega_next = omega + alpha * dt
angle_next = angle + omega_next * dt
print(f"After one step: {angle_next:.6f} rad, {omega_next:.3f} rad/s")

Wrapping the angle belongs only to a derived presentation value. The physical update must retain completed turns and must not create a discontinuity each time the orientation crosses zero.

Energy diagnostics

Fixed-axis rotational kinetic energy is

\[K_{\mathrm{rot}}=\frac12I\omega^2\]\n+ For the torsional spring, choose potential

\[U=\frac12\kappa(\theta-\theta_{\mathrm{eq}})^2\]

Without damping or drive, the total

\(E=K_{\mathrm{rot}}+U\)

should remain constant in the continuous model. With damping,

\[\frac{dE}{dt}=-b\omega^2\]

and a drive supplies power \(P=\tau_{\mathrm{drive}}\omega\). The energy ledger should distinguish expected transfer from numerical drift.

Angular momentum diagnostics

For the fixed-axis constant-inertia model,

\[L=I\omega\]

and external torque impulse gives

\[L_f-L_i=\int_{t_i}^{t_f}\tau_{\mathrm{ext}}\,dt\]

With zero external torque, angular momentum should remain constant. If inertia changes because mass moves, angular velocity can change even with constant angular momentum; the simple constant-inertia state equation must then be replaced by the more general derivative of angular momentum.

Choosing output times

The internal step used for accuracy and the interval used to record results serve different purposes. Saving every tiny step may be expensive, while saving too sparsely can hide oscillation, wrap crossings, or peak torque. Integrate with a justified step and sample output at declared times without feeding rounded output back into the calculation.

When comparing two step sizes, interpolate or arrange both runs to report at the same physical times. Comparing step number 100 from different step sizes compares different times and says nothing about convergence.

Events and discontinuous torque

Stops, clutches, impacts, and dry friction can change torque discontinuously. A step that crosses an event without locating it can overshoot an angle limit or apply torque for the wrong duration. Detect the event, estimate its time within the step, apply its impulse or new constraint, and continue from a consistent state.

Static friction is not simply a constant torque opposite angular velocity at zero speed. It can take whatever value up to a limit is required to prevent motion. A sign function at exactly zero can chatter numerically and needs an explicit sticking rule.

Step-size and method checks

Run the same initial-value problem to a fixed final time using decreasing step sizes. Compare unwrapped angle, angular velocity, phase, energy balance, and event times. Results should approach a stable answer. A smooth-looking orientation is not evidence of convergence.

duration = 2.0
for steps in [100, 200, 400]:
    angle, omega = 0.0, 0.0
    dt = duration / steps
    for _ in range(steps):
        omega += alpha * dt
        angle += omega * dt
    angle_error = angle - 0.5 * alpha * duration**2
    impulse_residual = inertia * omega - torque * duration
    print(steps, angle_error, impulse_residual)

Use exact constant-torque motion, undamped torsional oscillation, and pure damping as separate fixtures. One case cannot expose every sign and integration error.

Limits of the scalar model

The equation \(\tau=I\alpha\) assumes one constrained axis and appropriate constant inertia. A free three-dimensional rigid body needs orientation, vector angular momentum, body-to-world transformations, and an inertia tensor. Torque and angular acceleration need not be parallel.

A flexible rotor also has vibration modes, and a high-speed rotor may require bearing, aerodynamic, and relativistic corrections depending on scale. A correct fixed-axis calculation should be presented as that model, not as universal rigid-body dynamics.

Reliability checks

A numerical rotational model is most useful when its state, force law, and error checks remain visible without any animation. Analytic fixtures and conservation ledgers reveal whether a plausible rotational trace actually follows the intended physics.

A worked Python notebook

A constant torque of 3 newton metres acts on inertia 0.5 kilogram metres squared. The question for this calculation is precise: how does angle from rest change when we change elapsed time, while keeping the other choices fixed? The earlier sections explain the physical relationship. Here we turn one part of that relationship into a calculation that you can run, inspect, and change in Google Colab.

The rotor has a fixed axis and constant inertia. Keep that sentence beside the results. A formula can return a number outside its useful range, and a plotting library will usually draw that number without asking whether the physical assumptions still hold. The selected interval is from 0 to 2 s. It gives us a specific domain to discuss instead of leaving the graph's range to a software default.

Use the notebook link above the lesson, run the setup cell, and then run the cells in order. The code below is also included there. NumPy handles the arrays of numerical values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these libraries or notebook cells are unfamiliar, start with Your first physics notebook.

State the relationship before choosing a graph

In this small example, q stands for elapsed time, expressed in s. The output is angle from rest, expressed in rad. The equation below uses the numerical values in those units. Any fixed coefficients therefore carry the units needed to make the relation consistent. A value supplied in another unit must be converted before it enters this expression.

\[f(q) = 3 q^{2}\]

This is a function of one changing input. It is not a statement that the physical system has only one variable. We deliberately hold the other inputs fixed so that one cause of variation is visible. Changing several inputs at once would make it harder to tell which change produced the output difference. Write down the held-fixed choices before trying a different experiment.

At 1 s, the expression gives 3 rad. This is our reference calculation. It is a prediction of the stated example, not a new measurement. Its displayed digits help us compare implementations. They should not be read as a claim that physical inputs are known to six significant figures.

import numpy as np
import sympy as sp
import matplotlib.pyplot as plt

q = sp.Symbol("q", real=True)
expression = sp.sympify("3*q**2", locals={"q": q})
reference_input = 1.0
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "s")
print("Predicted angle from rest:", reference_output, "rad")

sp.Symbol declares a mathematical symbol. It does not assign a numerical measurement to that symbol. subs then replaces the symbol with our chosen input, and float converts the resulting symbolic number into a standard Python numerical value. Keeping these steps separate lets us inspect the equation before numerical substitution. It also prevents a later plot from becoming the only surviving record of the model.

Compare a few cases you can explain

Before drawing hundreds of points, calculate three. The lower endpoint gives 0 rad. The midpoint, at 1 s, gives 3 rad. The upper endpoint gives 12 rad. Read these together with the assumptions: they describe different choices of elapsed time within the same example, with the other inputs held fixed.

These three values are a useful first check, but they do not establish the shape between them. Two endpoints can be equal even when the curve bends or oscillates in between. Likewise, a large change between endpoints does not tell us whether the relationship is linear. We need the equation and intermediate samples to answer those questions.

evaluate = sp.lambdify(q, expression, modules="numpy")
inputs = np.array([0.0, 1.0, 2.0])
outputs = np.asarray(evaluate(inputs), dtype=float)
outputs = np.broadcast_to(outputs, inputs.shape)
for input_value, output_value in zip(inputs, outputs):
    print(f"{input_value:.6g} s -> {output_value:.6g} rad")
assert np.all(np.isfinite(outputs))

lambdify turns the symbolic expression into a numerical function. We explicitly select NumPy as its numerical backend so it can accept an array. The same expression now has two useful forms: one for algebra and one for calculating many cases. The assertion checks that the chosen inputs produce finite outputs. It would catch an infinity or an undefined numerical result, but a finite value can still be physically inappropriate.

Make a graph that answers the question

The horizontal axis shows elapsed time; the vertical axis shows angle from rest. Every plotted point is a prediction of the same expression. We mark the reference case separately so that a reader can connect the earlier printed value to its place on the curve. The axis labels include units because a shape without a scale cannot tell us the size of a physical effect.

input_grid = np.linspace(0.0, 2.0, 301)
output_grid = np.asarray(evaluate(input_grid), dtype=float)
output_grid = np.broadcast_to(output_grid, input_grid.shape)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(input_grid, output_grid, label="Stated model")
ax.scatter([reference_input], [reference_output], label="Reference case", zorder=3)
ax.set_xlabel("Elapsed time (s)")
ax.set_ylabel("Angle from rest (rad)")
ax.legend()
ax.grid(alpha=0.25)
fig.tight_layout()
plt.show()

Look for a sign change, a turning point, or a region where the curve becomes steep. Connect each feature to the expression rather than to the colour of the line. If the curve has a minimum or maximum, changing the input in opposite directions may have similar effects on the output near that point. If it is nearly straight over a short interval, a local linear approximation may be useful even when the complete expression is nonlinear.

Increasing the number of samples makes the line smoother and may reveal features that a coarse grid missed. It does not improve the physical inputs or enlarge the model's validity range. For quantities that can take only discrete values, use the line as a guide to the relationship and interpret only the allowed input values. A plotting function does not enforce such restrictions for you.

Measure the local sensitivity

A derivative asks a more specific question than the whole graph: how quickly does angle from rest change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 1 s is 6 in those ratio units. Locally, the output increases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = 6 q\]

For a small input change, multiply the derivative by that change to estimate the output change. Then compare that estimate with a fresh evaluation of the full expression. The distinction between these two calculations matters: the derivative gives a tangent approximation, while substitution evaluates the chosen expression at the new point.

slope_expression = sp.diff(expression, q)
slope_at_reference = float(slope_expression.subs(q, reference_input))
changed_input = 1.01
linear_prediction = reference_output + slope_at_reference * (
    changed_input - reference_input
)
full_prediction = float(evaluate(changed_input))
print("Symbolic derivative:", slope_expression)
print("Tangent prediction:", linear_prediction, "rad")
print("Full expression:", full_prediction, "rad")
print("Approximation difference:", full_prediction - linear_prediction, "rad")

Here the changed input is 1.01 s. The tangent estimate is 3.06 rad, while the full expression gives 3.0603 rad. If the two values are close, it means the chosen input change is small enough for this local approximation at this point. It does not mean the entire model is linear. If the slope is zero, a quadratic or higher-order term may supply the leading change instead.

This is also the beginning of uncertainty propagation. If the input has a small stated standard uncertainty, the magnitude of the slope times that uncertainty estimates one contribution to output uncertainty. Do not automatically use the same rule for a wide interval, a discontinuity, a turning point with zero first derivative, or several correlated uncertain inputs. Those situations need more of the model than one slope.

Check the derivative with numerical differences

We can estimate the same slope from values on either side of the reference point. Subtract the lower output from the upper output and divide by twice the input step. Parentheses are essential: the entire output difference must be divided by the entire input interval. Without them, Python can execute a different mathematical expression while still producing a plausible number.

The calculation below uses three decreasing step sizes. Each estimate is compared with the symbolic derivative, so the reference is independent of the numerical difference formula. Near an endpoint, it is acceptable here to evaluate a small extension only because the displayed mathematical expression is smooth there; such a check does not assert that extrapolated values describe additional physical cases.

input_span = 2.0
for fraction in [1e-2, 1e-3, 1e-4]:
    step = input_span * fraction
    upper = float(evaluate(reference_input + step))
    lower = float(evaluate(reference_input - step))
    numerical_slope = (upper - lower) / (2 * step)
    difference = numerical_slope - slope_at_reference
    print(f"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}")

For a smooth function, reducing the step often reduces the approximation error at first. Very small steps can eventually make subtraction of nearly equal floating-point values less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare the observed behaviour over a useful range of steps instead of choosing the smallest number the computer can represent.

An absolute slope difference is useful when the expected slope is near zero. A relative difference would divide by a very small reference and could look enormous even when the absolute discrepancy is tiny. Away from zero, a relative difference can help compare quantities with very different scales. The tolerance should reflect the question and the numerical method, rather than a universal decimal threshold copied into every test.

Change the example and preserve the reasoning

Make one copy of the reference calculation before editing it. Choose a second value of elapsed time inside the stated interval, predict whether angle from rest will rise or fall, and then evaluate the expression. Place the prediction and the result in a text cell. If they disagree, identify whether the mistake was in the sign, the unit conversion, the algebra, or the assumption about how the curve behaves.

Next choose a shorter plotting interval around the reference point. Compare the local curve with its tangent approximation. Then widen the interval again. The difference shows why a useful local estimate need not describe the whole range. Keep the axis labels and numerical scale visible when comparing the plots; automatic axis limits can make two different effects appear equally large.

Finally, return to the physical limit stated at the start: The rotor has a fixed axis and constant inertia. Describe one change to the setup that would violate that limit. State what new input, equation, or observation would be needed before trusting an answer in the changed situation. This last step connects the Python experiment to the physics instead of treating successful execution as the only criterion for a useful result.

Before saving, restart the runtime and run all cells in order. A result that depends on an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep unrounded values for later calculations, round only the reported output, and label any new data as measured, assumed, or generated. The graph and printed values should remain traceable to the same expression and the same choices of inputs.

Recover the output change by integrating the slope

There is another way to check that the expression and its derivative describe the same relationship. Add up the small output changes predicted by the slope across the chosen input interval. In the continuous limit, this integral equals the output at the upper endpoint minus the output at the lower endpoint. The result has units of rad: the input unit in the integration cancels the input unit in the denominator of the slope.

This statement is about the mathematics of this particular relationship. It does not assign a new physical meaning to an arbitrary area under a graph. For example, an area under force versus distance represents work, while an area under a different pair of quantities may represent something else or have no standard name. Always carry the axis units through the calculation before naming an integral.

slope_function = sp.lambdify(q, slope_expression, modules="numpy")
endpoint_change = float(evaluate(2.0) - evaluate(0.0))
for points in [101, 201, 401]:
    grid = np.linspace(0.0, 2.0, points)
    slopes = np.asarray(slope_function(grid), dtype=float)
    slopes = np.broadcast_to(slopes, grid.shape)
    integrated_change = np.trapezoid(slopes, grid)
    print(
        points,
        "integrated change:",
        integrated_change,
        "endpoint change:",
        endpoint_change,
    )

The trapezoidal rule joins neighbouring slope samples with straight segments and adds their signed areas. A negative contribution remains negative; replacing every slope by its magnitude would calculate a different quantity. Increasing the sample count provides a practical refinement check. If the values do not settle toward the endpoint difference, first inspect the interval, singularities, and array ordering.

Do not conclude that two agreeing calculations prove the model is physically complete. Both use the same selected relationship, so an omitted effect would be omitted from both. Their agreement is evidence that the numerical tools implement that relationship consistently. Comparing with an independent experiment or a more appropriate physical model is a separate test.

Keep this comparison in the notebook when you change the expression. It is particularly useful after adding a coefficient, changing a sign convention, or converting an input unit. A code cell that produces a smooth graph can conceal these mistakes; a second calculation of the same output change gives you another place to notice them.

Sources

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