Engineering Mechanics · Dynamics

Three-Dimensional Kinetics of a Rigid Body

When a body tumbles, spins, and turns all at once, planar shortcuts break down. This chapter builds the full spatial toolkit: how mass is distributed about arbitrary axes, how momentum and energy follow from that distribution, and how the equations of motion describe everything from a wobbling top to a drifting satellite.

Chapter objectives

Find the moments and products of inertia of a body about various axes.
Apply work–energy and impulse–momentum principles to a body in three-dimensional motion.
Develop and apply the equations of motion in three dimensions.
Study gyroscopic and torque-free motion.

Course Note

This is the hardest chapter in the course for most students, but only because it stacks many small ideas. Don't try to memorize the twenty-plus equations here. Learn the inertia tensor first — once you can build it and find principal axes by inspection, momentum, energy, and the Euler equations all fall out of it. Everything after the opening section applies that tensor to momentum, energy, and equations of motion.

Physical meaning

In spatial rotation, angular momentum generally is not parallel to angular velocity. The inertia tensor maps the observed rotation vector into momentum, while applied moment controls the fixed-frame rate of that momentum.

Problem-solving strategy

Classify the request as inertia, energy, impulse–momentum, instantaneous equations of motion, steady precession, or torque-free motion.
Draw the free-body diagram and declare a right-handed frame, preferably principal body axes through the mass center or fixed point.
Write angular velocity and moment components in that same frame; count unknown reactions and kinematic constraints.
Apply the tensor relation, Euler equations, energy, or angular impulse–momentum only under its stated assumptions.
Audit units, directions, support admissibility, and whether any steady- or slow-precession approximation is justified.

21.1Moments and Products of Inertia

In planar kinetics a single moment of inertia about an axis through the mass center was enough. Spatial motion is richer: describing how mass is spread out relative to a chosen coordinate system requires six inertial quantities — three moments of inertia and three products of inertia — each tied to a specific origin and axis orientation.

Moment of inertia

For a mass element, the moment of inertia about a coordinate axis is the element's mass times the square of its shortest distance to that axis. Integrating over the body gives a quantity that is always positive, since it sums positive masses times squared distances.

Moments of inertia (about x, y, z)

\[ I_{xx}=\int_m (y^2+z^2)\,dm,\quad I_{yy}=\int_m (x^2+z^2)\,dm,\quad I_{zz}=\int_m (x^2+y^2)\,dm \]

Product of inertia

A product of inertia measures mass distribution relative to a pair of orthogonal planes — the mass element times the product of its two perpendicular distances to those planes. Unlike a moment of inertia, it can be positive, negative, or zero. If either of the two defining planes is a plane of symmetry, that product of inertia vanishes, because symmetric mass pairs cancel.

Products of inertia

\[ I_{xy}=I_{yx}=\int_m xy\,dm,\quad I_{yz}=I_{zy}=\int_m yz\,dm,\quad I_{xz}=I_{zx}=\int_m xz\,dm \]

Fig. 10–1A differential mass element located by position vector r from the origin O; its coordinate distances define every moment and product of inertia. (Original redrawn diagram.)

Parallel-axis and parallel-plane theorems

When inertia is needed about axes that do not pass through the mass center G, two transfer theorems carry the values across. The parallel-axis theorem shifts moments of inertia; the parallel-plane theorem does the same for products of inertia. In both, x_G, y_G, z_G are the coordinates of G in the new frame.

Theorem · Parallel-axis

Transfer of moments of inertia

Condition: the inertia about axes through G is known and the new axes are parallel to them.

Result:

\[ I_{xx}=(I_{xx})_G+m\,(y_G^2+z_G^2) \] \[ I_{yy}=(I_{yy})_G+m\,(x_G^2+z_G^2) \] \[ I_{zz}=(I_{zz})_G+m\,(x_G^2+y_G^2) \]

Theorem · Parallel-plane

Transfer of products of inertia

Condition: the products of inertia about the central planes through G are known.

Result:

\[ I_{xy}=(I_{xy})_G+m\,x_G y_G,\quad I_{yz}=(I_{yz})_G+m\,y_G z_G,\quad I_{zx}=(I_{zx})_G+m\,z_G x_G \]

Course Note

The single most common mistake on this section: forgetting that the parallel-plane terms (\(m\,x_G y_G\)) carry a sign. Moments of inertia only ever add positively, but a product of inertia can drop when you transfer it. Always write the coordinates of \(G\) with their signs before plugging in.

The inertia tensor

The six independent quantities assemble into a single 3×3 array — the inertia tensor — that completely characterizes the body's mass distribution for a given origin and orientation. The off-diagonal entries carry minus signs, a convention inherited from the angular-momentum derivation.

Inertia tensor

\[ \mathbf{I}=\begin{pmatrix} I_{xx} & -I_{xy} & -I_{xz}\\ -I_{yx} & I_{yy} & -I_{yz}\\ -I_{zx} & -I_{zy} & I_{zz} \end{pmatrix} \]

Principal axes & principal moments of inertia

For any point there is an orientation of axes that makes all three products of inertia zero. The tensor then becomes diagonal, and its three diagonal entries are the principal moments of inertia \(I_x, I_y, I_z\). One is the maximum and one the minimum of the body's moment of inertia. If two of the three coordinate planes are planes of symmetry, those axes are already principal — no calculation needed.

Moment of inertia about an arbitrary axis

Once the full tensor is known for the x, y, z axes, the moment of inertia about any line through the origin follows from the direction cosines \(u_x, u_y, u_z\) of that line. This is the inertia transformation equation.

Inertia about axis Oa

\[ I_{Oa}=I_{xx}u_x^2+I_{yy}u_y^2+I_{zz}u_z^2-2I_{xy}u_x u_y-2I_{yz}u_y u_z-2I_{zx}u_z u_x \]

21.2Angular Momentum

Angular momentum links the inertia tensor to motion. For a rigid body of angular velocity ω, the angular momentum about a chosen reference point comes from integrating the moment of each element's linear momentum. The reference point dictates which simplified form applies.

Three reference points

About a fixed point O in the body, the velocity of that point is zero and the angular momentum reduces to a clean integral. About the mass center G, the first-moment term vanishes by definition of G. About an arbitrary point A, the angular momentum is the moment of the linear momentum \(m\mathbf{v}_G\) about A, added vectorially to \(\mathbf{H}_G\).

Angular momentum — special cases

\[ \mathbf{H}_O=\int_m \boldsymbol{\rho}_O\times(\boldsymbol{\omega}\times\boldsymbol{\rho}_O)\,dm \qquad \text{(fixed point)} \] \[ \mathbf{H}_G=\int_m \boldsymbol{\rho}_G\times(\boldsymbol{\omega}\times\boldsymbol{\rho}_G)\,dm \qquad \text{(mass center)} \] \[ \mathbf{H}_A=\boldsymbol{\rho}_{G/A}\times m\mathbf{v}_G+\mathbf{H}_G \qquad \text{(arbitrary point)} \]

Fig. 10–2A rigid body spinning with angular velocity ω, referenced to an arbitrary point A offset from the mass center G. (Original redrawn diagram.)

Rectangular components of H

Expanding the cross products in x, y, z components and recognizing the inertia integrals gives the scalar components of angular momentum directly in terms of the tensor.

Components of H (general axes)

\[ \begin{aligned} H_x &= I_{xx}\omega_x-I_{xy}\omega_y-I_{xz}\omega_z\\ H_y &= -I_{yx}\omega_x+I_{yy}\omega_y-I_{yz}\omega_z\\ H_z &= -I_{zx}\omega_x-I_{zy}\omega_y+I_{zz}\omega_z \end{aligned} \]

Components of H (principal axes)

\[ H_x=I_x\omega_x,\qquad H_y=I_y\omega_y,\qquad H_z=I_z\omega_z \]

Principle of impulse and momentum

With angular momentum formulated, the impulse–momentum principle handles problems of force, velocity, and time. In three dimensions there are six scalar equations: three for linear impulse and momentum along x, y, z, and three for angular impulse and momentum about those axes.

Impulse–momentum (3-D)

\[ m(\mathbf{v}_G)_1+\int_{t_1}^{t_2}\mathbf{F}\,dt=m(\mathbf{v}_G)_2 \] \[ (\mathbf{H}_O)_1+\int_{t_1}^{t_2}\mathbf{M}_O\,dt=(\mathbf{H}_O)_2 \]

21.3Kinetic Energy

To use work–energy methods in three dimensions, the kinetic energy of the body is needed. As with momentum, the general expression simplifies dramatically when the reference point is fixed or is the mass center, and further still when the axes are principal axes of inertia.

Two clean forms

About a fixed point O, all kinetic energy is rotational. About the mass center G, energy splits into a translational part \(\tfrac{1}{2}mv_G^2\) plus the rotational part. Using principal axes removes the product-of-inertia cross terms entirely.

Kinetic energy — fixed point O

\[ T=\tfrac{1}{2}I_x\omega_x^2+\tfrac{1}{2}I_y\omega_y^2+\tfrac{1}{2}I_z\omega_z^2 \]

Kinetic energy — mass center G

\[ T=\tfrac{1}{2}mv_G^2+\tfrac{1}{2}I_x\omega_x^2+\tfrac{1}{2}I_y\omega_y^2+\tfrac{1}{2}I_z\omega_z^2 \]

Principle of work and energy

\[ T_1+\textstyle\sum U_{1\text{-}2}=T_2 \]

21.4Equations of Motion

The motion of a body in space is governed by translational equations for the mass center and rotational equations about G or a fixed point O. The rotational set depends on how the chosen x, y, z axes themselves move.

Translational equations

\[ \textstyle\sum F_x=m(a_G)_x,\quad \sum F_y=m(a_G)_y,\quad \sum F_z=m(a_G)_z \]

Why the axes' motion matters

When inertia components are computed about axes that rotate with angular velocity Ω (different from the body's ω), the time derivative of angular momentum must include the rotation of the frame: \(\sum\mathbf{M}=(\dot{\mathbf{H}})_{xyz}+\boldsymbol{\Omega}\times\mathbf{H}\). The smart choice of Ω keeps the moments and products of inertia constant, which is what makes the equations solvable.

Equations · Euler (Ω = ω)

Euler equations of motion

Condition: the x, y, z axes are fixed in and move with the body, and are chosen as principal axes of inertia (so all products of inertia vanish).

Result:

\[ \begin{aligned} \textstyle\sum M_x &= I_x\dot\omega_x-(I_y-I_z)\,\omega_y\omega_z\\ \textstyle\sum M_y &= I_y\dot\omega_y-(I_z-I_x)\,\omega_z\omega_x\\ \textstyle\sum M_z &= I_z\dot\omega_z-(I_x-I_y)\,\omega_x\omega_y \end{aligned} \]

Named for Leonhard Euler. They apply only for moments summed about point O or G.

Equations · Ω ≠ ω

Rotational equations for a separately-rotating frame

Condition: the axes rotate with Ω different from the body's spin ω (useful for symmetric spinning bodies like tops and gyroscopes), chosen as principal axes.

Result:

\[ \begin{aligned} \textstyle\sum M_x &= I_x\dot\omega_x-I_y\Omega_z\omega_y+I_z\Omega_y\omega_z\\ \textstyle\sum M_y &= I_y\dot\omega_y-I_z\Omega_x\omega_z+I_x\Omega_z\omega_x\\ \textstyle\sum M_z &= I_z\dot\omega_z-I_x\Omega_y\omega_x+I_y\Omega_x\omega_y \end{aligned} \]

Course Note

Three exam-favorite cases for the frame: Ω = 0 (axes translate only — almost never useful because the inertias become time-dependent), Ω = ω (axes glued to the body — gives the Euler equations), and Ω ≠ ω (axes follow the precession, not the spin — the gyroscope case). Decide which frame you're using before you write a single moment equation, and label it on your free-body diagram.

21.5Gyroscopic Motion

A body symmetric about a spin axis and rotating about a fixed point — a top or a gyroscope — is best described by the three Euler angles. Their time rates carry physical names that recur throughout the rest of the chapter.

Euler angles: precession, nutation, spin

The orientation of the body is reached by three ordered rotations: \(\phi\) about the vertical axis, then \(\theta\) about the line of nodes, then \(\psi\) about the spin axis. Their rates are the precession \(\dot\phi\), the nutation \(\dot\theta\), and the spin \(\dot\psi\). The order \(\phi,\theta,\psi\) must be kept, since finite rotations don't add as vectors — but the differential rotations do.

Fig. 10–3A spinning top: spin \(\dot\psi\) about its own axis, precession \(\dot\phi\) about the vertical, and nutation \(\dot\theta\) tilting the axis. (Original redrawn diagram.)

Substituting the Euler-angle components of ω and the frame velocity Ω into the rotational equations yields the general gyroscopic equations. For the common case of steady precession — constant \(\theta\), \(\dot\phi\), and \(\dot\psi\) — they collapse to a single working equation.

Steady precession (general θ)

\[ \textstyle\sum M_x=-I\dot\phi^2\sin\theta\cos\theta+I_z\dot\phi\sin\theta\,(\dot\phi\cos\theta+\dot\psi) \] \[ \textstyle\sum M_y=0,\qquad \sum M_z=0 \]

Gyroscopic moment at θ = 90°

\[ \textstyle\sum M_x=I_z\,\Omega_y\,\omega_z \]

The gyroscopic effect

A spinning rotor whose axis is horizontal seems like it should topple under gravity — yet it doesn't. Instead its weight moment is balanced by the rate of change of its angular momentum as the axis precesses. The applied moment swings the angular-momentum vector toward itself rather than tipping the rotor. This is why a gyroscope in gimbals holds a fixed orientation and serves as a compass or stabilizer.

21.6Torque-Free Motion

When gravity is the only external force, no moment acts about the mass center, so the angular momentum stays constant. This is the regime of planets, satellites, and projectiles. Assuming an axisymmetric body (\(I_{xx}=I_{yy}=I\), \(I_{zz}=I_z\)), the motion settles into a steady spin plus a steady precession.

Constant angular momentum sets everything

Choosing the inertial Z axis along the fixed \(\mathbf{H}_G\), the tilt angle \(\theta\) between the spin axis and \(\mathbf{H}_G\) stays constant. The precession and spin rates are then both constant and fixed by the inertias and the magnitude of \(\mathbf{H}_G\).

Torque-free motion (axisymmetric body)

\[ \theta=\text{constant},\qquad \dot\phi=\frac{H_G}{I},\qquad \dot\psi=\frac{I-I_z}{I\,I_z}\,H_G\cos\theta \]

Spin–precession relation

\[ \dot\psi=\frac{I-I_z}{I_z}\,\dot\phi\cos\theta \]

Regular vs. retrograde precession

The body-cone / space-cone picture distinguishes two regimes. When \(I>I_z\) the spin is positive and the body cone rolls on the outside of the space cone — regular precession. When \(Iretrograde precession.

Fig. 10–4Body cone rolling on the fixed space cone; the contact line is the instantaneous axis of rotation, along which ω points. (Original redrawn diagram.)

Course Note

Watch the sign of \(I-I_z\). For a long thin body (football, rocket) the transverse inertia I is larger, so \(I>I_z\) and you get regular precession. For a flat disk-like body (a frisbee, a flat satellite) \(I_z>I\) and the precession is retrograde. Identifying the body shape first tells you which way the cones roll before you compute anything.

Integrated applications

EXAMPLE 1Inertia about an inspection axis

A balanced instrument package has principal mass moments \(I_x=0.120\), \(I_y=0.200\), and \(I_z=0.320\ \mathrm{kg\cdot m^2}\) at its mass center. Find its moment of inertia about the line through the center directed from the origin toward the point \((2,1,2)\).

Recognition strategy

The target is an inertia about a specified spatial line, not a motion response. Principal data and a direction vector cue the quadratic inertia transformation without products of inertia.

Given information

Principal moments are \(0.120, 0.200, 0.320\ \mathrm{kg\cdot m^2}\).
The axis direction is proportional to \(2\mathbf i+\mathbf j+2\mathbf k\); the unknown is \(I_u\).

Model and interpretation

Treat the package as rigid, place a right-handed principal frame at its mass center, and normalize the line direction. Because these are principal axes, the inertia tensor is diagonal and no transfer theorem is needed.

Step-by-step solution

1. Normalize. The direction magnitude is \(\sqrt{2^2+1^2+2^2}=3\), so \(\mathbf u=(2\mathbf i+\mathbf j+2\mathbf k)/3\).

2. Formulate. For a unit axis through the tensor origin, \(I_u=I_xu_x^2+I_yu_y^2+I_zu_z^2\) because every product of inertia vanishes.

3. Substitute. Insert the direction cosines: \(I_u=[0.120(4)+0.200(1)+0.320(4)]/9\ \mathrm{kg\cdot m^2}\).

4. Resolve. The numerator is \(1.960\ \mathrm{kg\cdot m^2}\), hence \(I_u=0.21778\ \mathrm{kg\cdot m^2}\), rounded to four significant figures.

Verification

Units: Each squared direction cosine is dimensionless, so the weighted sum retains the required inertia unit \(\mathrm{kg\cdot m^2}\).

Signs and direction: Changing the axis sense from \(\mathbf u\) to \(-\mathbf u\) leaves every squared component unchanged, as a physical axis requires.

Plausibility: The result lies between the minimum and maximum principal moments, \(0.120

Final result: \(\boxed{I_u=0.2178\ \mathrm{kg\cdot m^2}}\).

Engineering interpretation

The inspection axis is appreciably harder to accelerate than the minimum-inertia axis but easier than the symmetry direction with the largest inertia; fixture torque estimates should use this transformed value.

EXAMPLE 2Instantaneous control moment for a service module

A service module has principal moments \((I_x,I_y,I_z)=(0.24,0.36,0.50)\ \mathrm{kg\cdot m^2}\). At one instant its body-frame angular velocity is \((4,-2,6)\ \mathrm{rad/s}\) and the desired component rates are \((1.5,0.5,-1.0)\ \mathrm{rad/s^2}\). Determine the external moment about its mass center.

Recognition strategy

Forces are not requested and the data are instantaneous angular velocity, angular acceleration, and principal inertias. Euler’s body-axis equations expose the gyroscopic coupling that a componentwise \(I\alpha\) calculation would miss.

Given information

A right-handed principal body frame is used with positive components as listed.
The mass center is the moment reference; translation is irrelevant to this rotational calculation.

Model and interpretation

Model the module as rigid with constant inertia and take moments about its mass center. The listed accelerations are body-frame component derivatives \(\dot\omega_x,\dot\omega_y,\dot\omega_z\), so Euler’s equations apply directly.

Step-by-step solution

1. Select equations. Write \(M_x=I_x\dot\omega_x-(I_y-I_z)\omega_y\omega_z\), with cyclic companion equations for \(y\) and \(z\).

2. Compute x. Substitution gives \(M_x=0.24(1.5)-[0.36-0.50](-2)(6)=-1.32\ \mathrm{N\cdot m}\).

3. Compute y. The second equation gives \(M_y=0.36(0.5)-[0.50-0.24](6)(4)=-6.06\ \mathrm{N\cdot m}\).

4. Compute z. Finally, \(M_z=0.50(-1)-[0.24-0.36](4)(-2)=-1.46\ \mathrm{N\cdot m}\).

Verification

Units: Moment of inertia times angular acceleration has units \(\mathrm{kg\cdot m^2/s^2}=\mathrm{N\cdot m}\); coupling terms match.

Signs and direction: All three negative components mean the required moment points opposite the corresponding positive body axes; no sign is discarded as a magnitude.

Plausibility: Direct \(I\dot\omega\) would predict small moments, while the large y coupling is credible because both \(\omega_x\) and \(\omega_z\) are substantial.

Final result: \(\boxed{\mathbf M_G=(-1.32\mathbf i-6.06\mathbf j-1.46\mathbf k)\ \mathrm{N\cdot m}}\).

Engineering interpretation

The dominant actuator demand is about the negative body y-axis and is caused mainly by gyroscopic coupling, not the requested y acceleration. Real sizing must add disturbances, saturation margins, and flexible-mode limits.

EXAMPLE 3Precession estimate for a camera rotor

A 12.0 kg camera rotor has its center of mass 0.180 m from a frictionless pivot. Its axial mass moment is \(I_z=0.420\ \mathrm{kg\cdot m^2}\), and it spins at \(180\ \mathrm{rad/s}\) with a horizontal axis. Estimate the steady gravitational precession rate and assess the slow-precession assumption.

Recognition strategy

A rapidly spinning axisymmetric rotor, gravity moment, horizontal spin axis, and requested steady rate cue the slow-precession gyroscope balance \(mgd\approx I_z\omega_s\Omega\).

Given information

Use \(g=9.81\ \mathrm{m/s^2}\), positive precession defined by the right-hand rule that balances the gravity moment.
The pivot is ideal and fixed; bearing friction, nutation, and rotor imbalance are neglected.

Model and interpretation

Choose the pivot as moment center so ideal support reactions vanish from the moment equation. Treat spin as constant and use the slow branch only after comparing the resulting precession rate with the spin rate.

Step-by-step solution

1. Find gravity moment. For the horizontal shaft, the gravity lever arm is perpendicular, so \(M_g=mgd=12.0(9.81)(0.180)=21.1896\ \mathrm{N\cdot m}\).

2. Write gyro balance. Steady slow precession requires the angular-momentum direction to turn at rate \(\Omega\), giving \(M_g\approx \Omega H_s=\Omega I_z\omega_s\).

3. Calculate rate. Thus \(\Omega=21.1896/[0.420(180)]=0.28029\ \mathrm{rad/s}\), or about \(2.68\ \mathrm{rev/min}\).

4. Check applicability. The ratio \(\Omega/\omega_s=0.00156\) is far below unity, so neglected quadratic precession terms are small for this estimate.

Verification

Units: Dividing \(\mathrm{N\cdot m}\) by \(\mathrm{kg\cdot m^2/s}\) yields \(\mathrm{s^{-1}}=\mathrm{rad/s}\), as required.

Signs and direction: The precession sense must make \(d\mathbf H/dt\) parallel to the gravity moment; reversing rotor spin reverses the precession direction.

Plausibility: A precession rate roughly six hundred times smaller than spin is consistent with the declared slow-precession approximation and a rapidly spinning rotor.

Final result: \(\boxed{\Omega=0.280\ \mathrm{rad/s}\approx2.68\ \mathrm{rev/min}}\), with \(\Omega/\omega_s=1.56\times10^{-3}\).

Engineering interpretation

The rotor should precess slowly rather than simply fall in this ideal model. A real camera mount still needs bearing-load, startup, imbalance, damping, clearance, and fail-safe reviews before operation.

Chapter recap

Choose principal axes whenever possible, but transform the inertia tensor when the requested axis differs.
Separate kinematics from kinetics: angular velocity describes motion; moments determine the fixed-frame change of angular momentum.
Use Euler equations for instantaneous moment–acceleration questions and gyroscope relations only for their stated symmetry and steadiness assumptions.
Before moving on, be able to draw the spatial free-body diagram, declare the frame, compute coupling terms, and audit approximation and support admissibility.