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Moment of inertia by integration

Moment of inertia measures how mass is distributed relative to a chosen rotation axis. It is rotational inertia: for the same torque, a body with more moment of inertia has less angular acceleration. Mass farther from the axis matters more because perpendicular distance is squared.

From particles to a continuous body

For point masses, the scalar moment of inertia about a fixed axis is

\[I=\sum_{i=1}^{N}m_i r_{\perp i}^2\]

Here \(r_{\perp i}\) is the shortest distance from mass \(m_i\) to the axis, not necessarily the distance to the origin. The SI unit is kilogram metre squared.

For a continuous body, divide the mass into small elements and take the limit:

\[I=\int r_\perp^2\,dm\]

This integral is a definition, not a universal formula for one shape. The expression for \(dm\) depends on the body's dimension and density.

import numpy as np

masses = np.array([1.0, 2.0, 1.0])
positions = np.array([[1, 0, 0], [0, 2, 0], [-1, 0, 0]])
axis = np.array([0.0, 0.0, 1.0])
perpendicular = positions - (positions @ axis)[:, None] * axis
inertia = np.sum(masses * np.sum(perpendicular**2, axis=1))
print(f"Inertia about z axis: {inertia:.3f} kg m^2")

The result is never negative for ordinary positive mass. It is zero only in the ideal case where all mass lies on the axis.

Choosing the mass element

For a thin wire with linear density \(\lambda\),

\[dm=\lambda\,ds\]

For a thin lamina with surface density \(\sigma\),

\[dm=\sigma\,dA\]

For a solid with volume density \(\rho\),

\[dm=\rho\,dV\]

Density may vary with position. Before integrating inertia, verify that integrating the same mass element alone gives the stated total mass:

\[M=\int dm\]

This catches missing Jacobian factors and incorrect density normalization.

Uniform slender rod about its center

Take a uniform rod of length \(L\) and mass \(M\) along the \(x\), rotating about a perpendicular axis through its midpoint. Linear density is \(\lambda=M/L\), and \(dm=\lambda\,dx\). The perpendicular distance is \(|x|\), so

\[I=\int_{-L/2}^{L/2}x^2\frac{M}{L}\,dx=\frac{1}{12}ML^2\]

The squared distance makes contributions from both halves positive. Integrating from zero to \(L\) without shifting the coordinate would instead describe an axis through one end and give

\[I_{\mathrm{end}}=\frac13ML^2\]

Axis location is part of the answer.

Uniform disk and ring

For a uniform thin disk of radius \(R\) rotating about its central perpendicular axis, use concentric rings. Surface density is

\[\sigma=\frac{M}{\pi R^2}\]

A ring of radius \(r\) and thickness \(dr\) has mass

\[dm=\sigma(2\pi r\,dr)\]

Therefore,

\[I=\int_0^R r^2\sigma 2\pi r\,dr=\frac12MR^2\]

A thin hoop has essentially all mass at radius \(R\), giving \(I=MR^2\). The hoop has twice the disk's inertia for equal mass and outer radius because more mass lies far from the axis.

Parallel-axis theorem

If an axis is parallel to an axis through the center of mass and separated by distance \(d\), then

\[I=I_{\mathrm{cm}}+Md^2\]

The theorem requires parallel axes and the known reference axis to pass through the center of mass. Applying it between two arbitrary parallel axes by inserting their separation can be wrong.

For the rod, shifting the center axis by \(d=L/2\) gives

\[I_{\mathrm{end}}=\frac1{12}ML^2+M\left(\frac L2\right)^2=\frac13ML^2\]

This independently confirms the direct integration.

Perpendicular-axis theorem

For a planar lamina lying in the \(xy\), with three axes through the same point,

\[I_z=I_x+I_y\]

This follows from \(r_z^2=x^2+y^2\). It applies to flat bodies, not arbitrary three-dimensional solids. It also does not replace the parallel-axis theorem; the hypotheses differ.

Composite bodies and removed material

Moments of inertia about one common axis add:

\[I_{\mathrm{total}}=\sum_k I_k\]

Shift each component's tabulated center-of-mass inertia to the common axis before adding. A hole may be treated algebraically as negative removed mass, but this is only a subtraction technique. The remaining physical body still has positive mass and nonnegative inertia.

Avoid overlap in a decomposition. Counting the same volume in two components doubles both mass and inertia contributions there.

Relation to rotational energy and dynamics

For a rigid body rotating about a fixed axis at angular speed \(\omega\),

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

and, under the corresponding fixed-axis assumptions,

\[\tau_{\mathrm{net}}=I\alpha\]

Moment of inertia depends on mass distribution and axis, not on angular speed. A deformable body can change its inertia by moving mass; angular velocity can then change even without external torque because angular momentum, rather than angular velocity, is the conserved quantity.

Numerical quadrature

For measured density or irregular geometry, approximate the integral with cells:

\[I\approx\sum_c \rho(\mathbf r_c)\,\Delta V_c\,r_{\perp c}^2\]

Use cell centroids and actual cell volumes. Refine the mesh and compare both total mass and inertia. Inertia is especially sensitive to outer regions because of the squared distance, so a mesh accurate for volume alone may still underresolve boundary geometry.

mass, length = 3.0, 2.0
for cells in [10, 100, 1000]:
    edges = np.linspace(-length / 2, length / 2, cells + 1)
    centres = (edges[:-1] + edges[1:]) / 2
    estimate = np.sum((mass / cells) * centres**2)
    print(cells, estimate, "exact:", mass * length**2 / 12)

Simple bounds help. If all mass lies no farther than \(R_{\max}\) from the axis, then

\[0\le I\le MR_{\max}^2\]

A result outside this range signals a mass, distance, or unit error.

Three-dimensional inertia

One scalar describes resistance only about one specified axis. General rigid-body rotation uses the inertia tensor. The moment about unit axis \(\hat{\mathbf n}\) is

\[I_{\hat n}=\hat{\mathbf n}^{\mathsf T}\mathbf I\hat{\mathbf n}\]

Off-diagonal products of inertia describe coupling between axes. Principal axes diagonalize the tensor. Dividing an arbitrary torque vector by one scalar moment of inertia is not a valid general three-dimensional update.

Choosing coordinates efficiently

The best integration coordinates follow the body's symmetry and make perpendicular distance easy to express. Cartesian slices suit rods and rectangular solids. Cylindrical shells or rings suit disks, cylinders, and bodies symmetric around an axis. Spherical shells suit some radial density distributions. A coordinate system does not change the physical inertia, but it changes the volume element, bounds, and algebra. Forgetting the radial factor in a cylindrical or spherical volume element produces both the wrong mass and the wrong inertia.

Before evaluating a difficult integral, sketch the body, axis, and one representative mass element. Mark the shortest distance from the element to the axis. Write the density and differential measure separately, then multiply them to form the mass element. Use symmetry to predict which coordinates should cancel and estimate whether the answer should be closer to a hoop-like or center-concentrated distribution. This physical estimate helps detect a formally neat integral built from the wrong geometry.

Dimensional scaling supplies another check. If every length of a geometrically similar body is multiplied by a factor while total mass is held fixed, its moment of inertia scales with the square of that factor. If density rather than total mass is held fixed, mass also changes with size, so the overall scaling includes the dimension of the body. State which quantity is fixed before comparing differently sized objects.

Reliability checks

Moment of inertia is a weighted description of mass geometry. Integration makes that geometry explicit, while standard theorems and numerical bounds provide independent ways to catch an incorrect axis, density, or distance.

A worked Python notebook

Compare uniform rods of fixed mass 3 kilograms and different lengths. The question for this calculation is precise: how does central moment of inertia change when we change rod length, 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 axis passes through the centre and is perpendicular to the rod; another axis has another 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 0.5 to 3 m. 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 rod length, expressed in m. The output is central moment of inertia, expressed in kg m^2. 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.

\[f(q) = \frac{q^{2}}{4}\]

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 2 m, the expression gives 1 kg m^2. 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**2/12", locals={"q": q})
reference_input = 2.0
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "m")
print("Predicted central moment of inertia:", reference_output, "kg m^2")

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 0.0625 kg m^2. The midpoint, at 1.75 m, gives 0.765625 kg m^2. The upper endpoint gives 2.25 kg m^2. Read these together with the assumptions: they describe different choices of rod length 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([0.5, 1.75, 3.0])
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} m -> {output_value:.6g} kg m^2")
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 rod length; the vertical axis shows central moment of inertia. 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(0.5, 3.0, 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("Rod length (m)")
ax.set_ylabel("Central moment of inertia (kg m^2)")
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 central moment of inertia change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 2 m is 1 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.

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

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 = 2.0125
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, "kg m^2")
print("Full expression:", full_prediction, "kg m^2")
print("Approximation difference:", full_prediction - linear_prediction, "kg m^2")

Here the changed input is 2.0125 m. The tangent estimate is 1.0125 kg m^2, while the full expression gives 1.01254 kg m^2. 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 = 2.5
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 rod length inside the stated interval, predict whether central moment of inertia 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 axis passes through the centre and is perpendicular to the rod; another axis has another 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 kg m^2: 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(3.0) - evaluate(0.5))
for points in [101, 201, 401]:
    grid = np.linspace(0.5, 3.0, 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.

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