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Coupled translational–rotational motion

A rigid body can translate through space while rotating about its center of mass. These motions are distinct but can be coupled by off-center forces, contact, strings, gears, or rolling constraints. A complete state therefore keeps linear and angular quantities separate even when an equation relates them.

Splitting rigid-body motion

For a rigid body of total mass \(M\) with center-of-mass velocity \(\mathbf V_{\mathrm{cm}}\), any point's velocity can be written as

\[\mathbf v=\mathbf V_{\mathrm{cm}}+\boldsymbol\omega\times\mathbf r'\]

Here \(\mathbf r'\) is measured from the center of mass and \(\boldsymbol\omega\) is the instantaneous angular velocity. The first term translates every point equally; the second depends on distance and direction from the center.

Kinetic energy separates correspondingly:

\[K=\frac12M V_{\mathrm{cm}}^2+\frac12I_{\mathrm{cm}}\omega^2\]

for planar motion about a center-of-mass axis. In general three dimensions, the rotational term uses the inertia tensor. There is no cross term when relative coordinates are measured from the center of mass.

Force and torque equations

Net external force controls center-of-mass translation:

\[\sum\mathbf F_{\mathrm{ext}}=M\mathbf A_{\mathrm{cm}}\]

For planar rotation about the center of mass,

\[\sum\tau_{\mathrm{cm}}=I_{\mathrm{cm}}\alpha\]

A force through the center of mass can accelerate the body without rotating it. A pure couple can rotate it without accelerating the center of mass. A general off-center force does both. Force and torque equations must be solved together when unknown contact forces appear in each.

import numpy as np

mass, radius, angle = 2.0, 0.3, np.deg2rad(20)
inertia = 0.5 * mass * radius**2
acceleration = 9.81 * np.sin(angle) / (1 + inertia / (mass * radius**2))
angular_acceleration = -acceleration / radius
print("Downhill acceleration (m/s^2):", acceleration)
print("Angular acceleration (rad/s^2):", angular_acceleration)

Rolling without slipping

For a circular body of radius \(R\) on a stationary flat surface, rolling without slipping means the point of contact has zero instantaneous velocity relative to the surface. With consistent signs, the scalar speed relation is

\[V_{\mathrm{cm}}=R|\omega|\]

Signed coordinates may instead give \(V_{\mathrm{cm}}=-R\omega\). The minus sign depends on which angular direction is positive. Memorizing a sign without declaring axes is unsafe.

Differentiating a valid no-slip relation along a straight fixed surface gives

\[a_{\mathrm{cm}}=R|\alpha|\]

in magnitude. This does not say the material contact point has zero acceleration. Its instantaneous velocity can be zero while it has centripetal and tangential acceleration.

Rolling down an incline

Consider a rigid round body rolling without slipping down an incline of angle \(\beta\). Along the slope,

\[Mg\sin\beta-f=Ma\]

With torque about the center,

\[fR=I_{\mathrm{cm}}\alpha\]

and no slip gives \(a=R\alpha\) in a matching magnitude convention. Solving yields

\[a=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)}\]

The denominator shows how rotational inertia reduces translational acceleration. A hoop with \(I=MR^2\) accelerates more slowly than a solid sphere with \(I=2MR^2/5\) under the same ideal conditions.

Static friction points up the incline in this example and supplies the torque. It need not dissipate energy because the instantaneous contact point is at rest relative to the surface. The required friction must still satisfy

\[|f|\le\mu_sN\]

If it exceeds the available static friction, the no-slip solution is invalid and sliding must be analyzed instead.

Energy view of rolling

For rolling without slip from a height change \(h\) with negligible losses,

\[Mgh=\frac12M V_{\mathrm{cm}}^2+\frac12I_{\mathrm{cm}}\omega^2\]

Substituting

\(\omega=V_{\mathrm{cm}}/R\)

reproduces the same dependence on \(I_{\mathrm{cm}}/(MR^2)\). Energy can determine speed after a height change, but it does not directly give friction force or acceleration without additional reasoning.

Contact point kinematics

The general no-slip condition compares the velocity of the body point touching the surface with the surface velocity:

\[\mathbf V_{\mathrm{cm}}+\boldsymbol\omega\times\mathbf r_{\mathrm{contact}}=\mathbf v_{\mathrm{surface}}\]

This vector equation handles moving belts and arbitrary contact orientation. Only its tangential components are constrained by no slip; the normal component is governed by nonpenetration and contact separation.

centre_velocity = np.array([2.0, 0.0, 0.0])
angular_velocity = np.array([0.0, 0.0, -2.0 / radius])
contact_offset = np.array([0.0, -radius, 0.0])
contact_velocity = centre_velocity + np.cross(angular_velocity, contact_offset)
assert np.allclose(contact_velocity, 0.0)
print("Contact velocity:", contact_velocity)

A no-slip model should check that tangential slip remains near zero and that the required friction stays within its physical limit. Simply overwriting angular velocity to satisfy the relation can hide an impossible constraint and inject energy.

Pulley and string coupling

An inextensible string that does not slip on a pulley links linear string acceleration to angular acceleration:

\[a=R\alpha\]

Different tensions on the two sides give net torque

\[(T_1-T_2)R=I\alpha\]

For a massive pulley, the tensions generally differ. Assuming one common tension while also retaining pulley inertia is inconsistent unless another feature justifies it. The translational equations for attached masses, the rotational pulley equation, and the string constraint form one coupled system.

Sliding and energy dissipation

If the contact point slips, the rolling constraint does not apply. Kinetic friction may change both center velocity and angular velocity, often pushing them toward the no-slip relation. Mechanical energy is converted to thermal energy at a rate related to friction and relative slip speed.

The direction of friction opposes relative motion at contact, not necessarily the motion of the center of mass. A spinning ball placed on a surface can experience friction in a direction that initially speeds up its center while slowing its spin.

Numerical constraint handling

An unconstrained linear and angular update can drift away from no slip. Directly projecting state back onto the constraint can control drift but may alter momentum or energy. A constraint impulse method solves for forces or impulses that enforce contact while respecting mass and inertia.

velocity, omega = 2.0, 0.0
slip = velocity + radius * omega
impulse = -slip / (1 / mass + radius**2 / inertia)
velocity += impulse / mass
omega += radius * impulse / inertia
print("Impulse needed for no slip (N s):", impulse)
assert np.isclose(velocity + radius * omega, 0.0)
# A real contact must also have enough friction to supply this impulse.

Reduce the time step and track slip residual, energy balance, linear momentum change, and angular momentum change about a useful point. Constraint satisfaction alone does not prove that the dynamics are correct.

Three-dimensional limits

General rigid-body motion needs an orientation representation, inertia tensor, and vector angular momentum. Multiple contacts can make constraints redundant or inconsistent. Deformable tires, rolling resistance, and changing contact patches are not captured by a single rigid radius and ideal point contact.

The scalar rolling relation remains valuable for wheels, disks, and spheres on simple surfaces, but its geometry and friction conditions must be named.

Reliability checks

Coupled translation and rotation are linked through forces, torques, and kinematic constraints, not by identifying linear and angular variables. Keeping both parts of the state visible makes rolling, pulley, and off-center-force problems consistent and checkable.

A worked Python notebook

A uniform solid disk rolls down an incline from rest. The question for this calculation is precise: how does rolling acceleration change when we change incline angle, 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.

Static friction must be sufficient to maintain rolling without slipping. 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 0.785398 rad. 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 incline angle, expressed in rad. The output is rolling acceleration, expressed in m/s^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.

\[f(q) = 6.54 \sin{\left(q \right)}\]

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 0.349066 rad, the expression gives 2.23681 m/s^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("2*9.81*sin(q)/3", locals={"q": q})
reference_input = 0.3490658503988659
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "rad")
print("Predicted rolling acceleration:", reference_output, "m/s^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 m/s^2. The midpoint, at 0.392699 rad, gives 2.50275 m/s^2. The upper endpoint gives 4.62448 m/s^2. Read these together with the assumptions: they describe different choices of incline angle 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, 0.39269908169872414, 0.7853981633974483])
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} rad -> {output_value:.6g} m/s^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 incline angle; the vertical axis shows rolling acceleration. 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, 0.7853981633974483, 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("Incline angle (rad)")
ax.set_ylabel("Rolling acceleration (m/s^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 rolling acceleration change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 0.349066 rad is 6.14559 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.

\[f'(q) = 6.54 \cos{\left(q \right)}\]

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 = 0.3529928412158531
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, "m/s^2")
print("Full expression:", full_prediction, "m/s^2")
print("Approximation difference:", full_prediction - linear_prediction, "m/s^2")

Here the changed input is 0.352993 rad. The tangent estimate is 2.26095 m/s^2, while the full expression gives 2.26093 m/s^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 = 0.7853981633974483
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 incline angle inside the stated interval, predict whether rolling acceleration 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: Static friction must be sufficient to maintain rolling without slipping. 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 m/s^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(0.7853981633974483) - evaluate(0.0))
for points in [101, 201, 401]:
    grid = np.linspace(0.0, 0.7853981633974483, 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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