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Angular momentum conservation

Angular momentum describes rotational motion relative to a chosen origin. It remains constant for a system when the net external torque about that origin is zero. Internal forces can exchange angular momentum among parts while the system total stays fixed.

Angular momentum of a particle

For a particle at position \(\mathbf r\) relative to origin \(O\) with linear momentum \(\mathbf p\), angular momentum about that origin is

\[\mathbf L_O=\mathbf r\times\mathbf p\]

Its magnitude is

\[L_O=rp\sin\phi=r_\perp p\]

where \(\phi\) is the included angle and \(r_\perp\) is the shortest distance from the origin to the particle's line of motion. A particle moving directly toward the origin has zero angular momentum about that origin even while its linear momentum is nonzero.

Direction follows the right-hand rule for the cross product. Swapping the product order reverses the direction and is incorrect for the definition.

Torque changes angular momentum

For a particle,

\[\boldsymbol\tau_O=\mathbf r\times\mathbf F\]

and, for a fixed origin in an inertial frame,

\[\boldsymbol\tau_O=\frac{d\mathbf L_O}{dt}\]

Integrating gives the angular impulse relation

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

Therefore, zero net external torque makes angular momentum constant. Zero force is not required: a central force can be nonzero while its torque about the force center is zero.

import numpy as np

position = np.array([2.0, 0.0, 0.0])
momentum = np.array([0.0, 3.0, 0.0])
initial_l = np.cross(position, momentum)
external_torque = np.array([0.0, 0.0, 2.0])
final_l = initial_l + external_torque * 0.5
print("Expected final angular momentum (kg m^2/s):", final_l)

A system of particles

Total angular momentum about one common origin is

\[\mathbf L_O=\sum_i\mathbf r_i\times\mathbf p_i\]

When internal forces are equal, opposite, and directed along the line joining each pair, their internal torques cancel. The system balance then contains only external torque:

\[\frac{d\mathbf L_O}{dt}=\boldsymbol\tau_{O,\mathrm{ext}}\]

The system boundary decides which torques are external. A skater pulling in their arms can treat forces between arms and body as internal. A hand pushing a turntable transfers angular momentum from the person and Earth into a turntable-only system.

Not every equal-and-opposite pair automatically has zero internal torque. The pair must also be central, or internal spin and field angular momentum must be included in a more complete model.

Dependence on origin

Angular momentum generally changes when the origin changes. For a system with total linear momentum \(\mathbf P\), shifting the origin by constant vector \(\mathbf a\) gives

\[\mathbf L_{O'}=\mathbf L_O-\mathbf a\times\mathbf P\]

If total linear momentum is zero, angular momentum is independent of a fixed origin shift. Otherwise, every conservation claim must name its origin. A moving origin or accelerating frame introduces further terms and cannot be handled by this simple shift alone.

Orbital and spin parts

For a system of particles, total angular momentum can be split into center-of-mass orbital motion and motion relative to the center:

\[\mathbf L_O=\mathbf R_{\mathrm{cm}}\times M\mathbf V_{\mathrm{cm}}+\mathbf L_{\mathrm{cm}}\]

The first term is angular momentum of total mass moving at the center of mass about the chosen origin. The second includes rotation and internal particle motion about the center. Calling these "orbital" and "spin" is useful in rigid-body mechanics, but the separation depends on the reference point.

Fixed-axis rigid body

For a rigid body rotating about a fixed principal axis,

\[L=I\omega\]

where \(I\) is measured about that axis. This scalar relation is a special case. With zero external torque and changing inertia,

\[I_i\omega_i=I_f\omega_f\]

If a skater reduces inertia by pulling mass closer to the axis, angular speed increases. No external torque is required for the internal rearrangement, but internal forces do work, so rotational kinetic energy need not remain constant.

Since

\[K_{\mathrm{rot}}=\frac12I\omega^2=\frac{L^2}{2I}\]

decreasing inertia at constant angular momentum increases kinetic energy. Chemical energy from the skater's muscles supplies the increase. Conservation of angular momentum does not imply conservation of rotational kinetic energy.

Central-force motion

For a force parallel to the radius from a fixed center,

\[\boldsymbol\tau=\mathbf r\times\mathbf F=\mathbf0\]

so angular momentum about the center is conserved. The motion lies in a plane perpendicular to the constant angular-momentum vector. The areal velocity is

\[\frac{dA}{dt}=\frac{L}{2m}\]

and is constant. This is the mechanical basis of equal areas in equal times for ideal two-body gravitational motion.

Angular momentum in collisions

During a brief collision, external torque impulse may be negligible about a wisely chosen origin, making total angular momentum approximately conserved. Linear momentum and angular momentum provide independent constraints. An off-center impact can convert translational motion into spin.

For a body colliding with a fixed pivot, the pivot force can have a large impulse but zero torque about the pivot because its lever arm is zero. Angular momentum about that pivot may be conserved during the impact even though linear momentum of the body is not. Kinetic energy still requires a separate collision model.

General three-dimensional rigid body

In three dimensions,

\[\mathbf L_{\mathrm{cm}}=\mathbf I\boldsymbol\omega\]

where \(\mathbf I\) is a matrix. Angular momentum need not point along angular velocity except when rotation is about a principal axis or the inertia is isotropic. Conservation fixes the angular-momentum vector in an inertial frame, but the body can change orientation relative to it.

The scalar shortcut \(L=I\omega\) cannot be applied independently to arbitrary coordinate components unless the tensor is diagonal in the relevant frame and its motion is handled consistently.

Numerical conservation ledger

For computed motion, test the angular impulse balance:

\[\mathbf r_L=\mathbf L_f-\mathbf L_i-\sum_n\boldsymbol\tau_{\mathrm{ext},n}\Delta t_n\]

Calculate every term about the same origin and in the same frame. Use both an absolute tolerance and a scale based on individual angular momenta and accumulated impulse. A relative error divided by a theoretically zero initial total is undefined or misleading.

positions = np.array([[2, 0, 0], [-1, 0, 0]])
momenta = np.array([[0, 3, 0], [0, -2, 0]])
spins = np.array([[0, 0, 1], [0, 0, 0.5]])
total = np.sum(np.cross(positions, momenta) + spins, axis=0)
print("Total orbital plus spin angular momentum:", total)

Test a torque-free rigid rotor, a central-force orbit, a constant external torque, and an internal shape change. Reduce the time step and compare at identical physical times. Angular momentum conservation alone cannot prove forces, positions, or energy are correct, so retain independent fixtures.

Reliability checks

Angular momentum conservation is a vector balance tied to an origin and system. When those choices are explicit, it explains orbital area laws, changing spin rates, and off-center impacts while providing a precise check on rotational calculations.

A worked Python notebook

Keep axial angular momentum at 2 kilogram metres squared per second while changing inertia. The question for this calculation is precise: how does angular velocity change when we change moment of inertia, 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.

External torque is zero, but internal work may change rotational kinetic energy. 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 0.1 to 1 kg m^2. 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 moment of inertia, expressed in kg m^2. The output is angular velocity, expressed in rad/s. 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.

\[f(q) = \frac{2}{q}\]

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 0.5 kg m^2, the expression gives 4 rad/s. 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("2/q", locals={"q": q})
reference_input = 0.5
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "kg m^2")
print("Predicted angular velocity:", reference_output, "rad/s")

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 20 rad/s. The midpoint, at 0.55 kg m^2, gives 3.63636 rad/s. The upper endpoint gives 2 rad/s. Read these together with the assumptions: they describe different choices of moment of inertia 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([0.1, 0.55, 1.0])
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} kg m^2 -> {output_value:.6g} rad/s")
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 moment of inertia; the vertical axis shows angular velocity. 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(0.1, 1.0, 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("Moment of inertia (kg m^2)")
ax.set_ylabel("Angular velocity (rad/s)")
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 angular velocity change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 0.5 kg m^2 is -8 in those ratio units. Locally, the output decreases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = - \frac{2}{q^{2}}\]

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 = 0.5045
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/s")
print("Full expression:", full_prediction, "rad/s")
print("Approximation difference:", full_prediction - linear_prediction, "rad/s")

Here the changed input is 0.5045 kg m^2. The tangent estimate is 3.964 rad/s, while the full expression gives 3.96432 rad/s. 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 = 0.9
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 moment of inertia inside the stated interval, predict whether angular velocity 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: External torque is zero, but internal work may change rotational kinetic energy. 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/s: 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(1.0) - evaluate(0.1))
for points in [101, 201, 401]:
    grid = np.linspace(0.1, 1.0, 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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