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Angular momentum $\vec{L} = \mathbf{I}\vec{\omega}$, moments and products of inertia, principal axes as eigenvectors, the parallel-axis theorem, and dynamic imbalance.
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By the end of this lesson you will be able to compute a body's inertia tensor, find its principal axes and moments, and predict when spinning it needs a torque.
From Physics C you know the moment of inertia $I = \int r_\perp^2\,dm$ about a fixed axis, $L = I\omega$ and $T = \tfrac{1}{2}I\omega^2$, and the parallel-axis theorem. From linear algebra you know matrices, eigenvalues and eigenvectors, and that a symmetric matrix has perpendicular eigenvectors. This lesson drops the fixed axis: a body free to turn about any axis needs its moment of inertia as a matrix.
| Term | What it means |
|---|---|
| Inertia tensor | The symmetric matrix $\mathbf{I}$ with $\vec{L} = \mathbf{I}\vec{\omega}$. |
| Moments of inertia | The diagonal entries, such as $I_{xx} = \sum m(y^2 + z^2)$. |
| Products of inertia | The off-diagonal entries, such as $I_{xy} = -\sum mxy$. |
| Principal axes | Three perpendicular axes, the eigenvectors of $\mathbf{I}$, about which $\vec{L}$ is parallel to $\vec{\omega}$. |
| Principal moments | The eigenvalues $\lambda_1, \lambda_2, \lambda_3$ of $\mathbf{I}$. |
| Dynamic imbalance | Nonzero products of inertia about a spin axis, which make the axis wobble. |
| Symmetric top | A body with two equal principal moments, such as a cylinder or a spinning top. |
For a rigid body turning with angular velocity $\vec{\omega}$ about a fixed point, each particle moves at $\vec{\omega} \times \vec{r}$, and the angular momentum is $\vec{L} = \sum m\vec{r} \times (\vec{\omega} \times \vec{r})$. Expanding the triple product shows $\vec{L}$ is a linear function of $\vec{\omega}$:
$$\vec{L} = \mathbf{I}\vec{\omega}, \qquad I_{ij} = \sum m\left(r^2\delta_{ij} - r_ir_j\right).$$
The diagonal entries are the familiar moments of inertia, $I_{xx} = \sum m(y^2 + z^2)$; the off-diagonal products of inertia, $I_{xy} = -\sum mxy$, measure lopsidedness. Because of them, $\vec{L}$ is in general not parallel to $\vec{\omega}$.
The inertia tensor is symmetric, so it has three real eigenvalues and three perpendicular eigenvectors: the principal axes. About a principal axis, $\mathbf{I}\vec{\omega} = \lambda\vec{\omega}$, so $\vec{L} = \lambda\vec{\omega}$ and the body can spin steadily with no torque. In principal axes the tensor is diagonal, and the kinetic energy is $T = \tfrac{1}{2}(\lambda_1\omega_1^2 + \lambda_2\omega_2^2 + \lambda_3\omega_3^2)$.
Another way: picture
Spin a dumbbell about an axis through its middle but tilted to the bar. Each weight circles the axis, and its angular momentum points perpendicular to the bar, not along the axis. As the dumbbell turns, that tilted angular momentum turns with it, and something must twist the axle to make that happen. Tilt the axis to lie along or across the bar and the twisting stops: those are principal axes.
Another way: steps
The vector identity $\vec{r} \times (\vec{\omega} \times \vec{r}) = r^2\vec{\omega} - (\vec{r} \cdot \vec{\omega})\vec{r}$ gives, for one particle, $\vec{L} = m(r^2\vec{\omega} - (\vec{r} \cdot \vec{\omega})\vec{r})$. Its $x$ component is $m((y^2 + z^2)\omega_x - xy\,\omega_y - xz\,\omega_z)$, which is the first row of $\mathbf{I}\vec{\omega}$ with the entries above. Summing over particles, or integrating over a continuous body, gives the full tensor.
The word tensor signals that $\mathbf{I}$ is a geometric object, like a vector, whose components change in a definite way when the axes are rotated: $\mathbf{I}' = R\mathbf{I}R^T$. The principal axes are the special orientation in which the matrix is diagonal. Finding them is exactly diagonalizing a symmetric matrix, which is why the eigenvalue problem from linear algebra appears here.
Any axis of rotational symmetry is a principal axis, and so is any axis perpendicular to a plane of mirror symmetry. For a uniform rectangular block the principal axes through the center are parallel to the edges; for a cylinder, its axis and any two perpendicular diameters.
When two principal moments are equal, the body is a symmetric top, and every axis in the plane of the equal pair is principal. A cube about its center has all three equal, $\tfrac{1}{6}Ma^2$, so every axis through its center is principal, and a cube spins about a corner-to-corner diagonal as smoothly as about an edge direction. About a corner the symmetry is lower: the diagonal through the corner is principal, with moment $\tfrac{1}{6}Ma^2$, and every axis perpendicular to it has $\tfrac{11}{12}Ma^2$.
Moving the reference point from the center of mass by $\vec{d}$ changes each entry of the tensor by the entry for a point mass $M$ at $\vec{d}$:
$$I_{ij} = I_{ij}^{\text{cm}} + M\left(d^2\delta_{ij} - d_id_j\right).$$
For the diagonal entries this is the familiar $I = I_{\text{cm}} + Md_\perp^2$. The off-diagonal terms show that shifting the reference point can create products of inertia where there were none, which is why a body spinning about a point other than its center of mass usually needs a torque. It also means the moment of inertia about any axis is least when the axis passes through the center of mass.
The rotational kinetic energy is $T = \tfrac{1}{2}\vec{\omega} \cdot \vec{L} = \tfrac{1}{2}\vec{\omega} \cdot \mathbf{I}\vec{\omega}$, a quadratic form. About an axis along the unit vector $\hat{n}$, the moment of inertia is $I_n = \hat{n} \cdot \mathbf{I}\hat{n}$, and in principal axes $I_n = \lambda_1n_1^2 + \lambda_2n_2^2 + \lambda_3n_3^2$.
Plotting the points $\hat{n}/\sqrt{I_n}$ for every direction gives the inertia ellipsoid, whose axes are the principal axes. The largest principal moment gives the shortest axis. This picture is how Louis Poinsot described the motion of a free rigid body in 1834: the ellipsoid rolls without slipping on a fixed plane, which is the geometry behind the next lesson's result about stable and unstable rotation.
A body spinning about a fixed axis that is not principal has angular momentum off the axis, turning with the body at the spin rate. The rate of change of that component, $\omega^2|I_{xz}|$, must be supplied by the bearings as a torque that rotates with the wheel: a vibration at the spin frequency. This is dynamic imbalance, distinct from static imbalance, which is a center of mass off the axis.
Tire shops correct it by spinning each wheel on a balancing machine, which measures the rotating force at each rim, and clipping small weights at two points so that both the center of mass and the products of inertia are zero. Turbines, hard drives and spacecraft are balanced the same way, and a satellite's spin axis is designed to be a principal axis so that it can spin for years with no torque at all.
Checking an answer. All principal moments must be positive, and each must be no larger than the sum of the other two, the triangle inequality for moments of inertia. Their sum, the trace, must equal the sum of the diagonal entries in any axes. Principal axes must come out perpendicular. And an axis of symmetry must turn out principal.
The minus sign in $I_{xy} = -\sum mxy$ is a convention, but a useful one. With it, $\vec{L} = \mathbf{I}\vec{\omega}$ holds as an ordinary matrix product, and the tensor is positive definite: $\hat{n} \cdot \mathbf{I}\hat{n} > 0$ for every direction. Some engineering texts define the products without the sign and put it in the matrix instead; the physics is the same, but mixing conventions produces wrong principal axes.
A quick way to remember the sign: mass lying along the line $y = x$ in the first and third quadrants makes $\sum mxy$ positive and so $I_{xy}$ negative. Spinning such a body about $z$ is fine, but spinning it about $x$ pulls its angular momentum toward the line of mass, and the negative product of inertia is what records that pull.
For a body that lies entirely in the $xy$ plane, every particle has $z = 0$. Then $I_{xx} = \sum my^2$, $I_{yy} = \sum mx^2$ and $I_{zz} = \sum m(x^2 + y^2) = I_{xx} + I_{yy}$: the moment about the axis perpendicular to the plane is the sum of the two in-plane moments. This perpendicular-axis theorem holds only for flat bodies, but it saves work often. A thin uniform disk has $I_{zz} = \tfrac{1}{2}MR^2$ about its axis, and by symmetry its two in-plane moments are equal, so each is $\tfrac{1}{4}MR^2$.
For a flat body, $I_{xz}$ and $I_{yz}$ are zero, so the $z$ axis is always principal, and the other two principal axes lie in the plane, found from the $2 \times 2$ block with $I_{xx}$, $I_{yy}$ and $I_{xy}$. That is why most of the worked examples here reduce to two-by-two eigenvalue problems: the plane of a flat body does the work that symmetry does for a solid one.
A tire and wheel are never perfectly uniform: the valve stem, the tire's splice and small variations in rubber all shift mass. If the heavy spots are on opposite sides and at different distances along the axle, the center of mass can be on the axle while the product of inertia $I_{xz}$ is not zero. At highway speed, about $90$ rad/s, even $30$ g spots $0.2$ m out and $0.1$ m apart need a rotating torque of about $5$ N m from the bearings, which the driver feels through the steering wheel as a shimmy.
A balancing machine at a tire shop spins the wheel and measures the forces at two planes, the inner and outer rim. From them it computes both the static and the dynamic imbalance, and tells the technician where to clip weights on each rim. Making the spin axis a principal axis through the center of mass removes the vibration entirely.
Many satellites and probes are spin-stabilized: they spin about one axis so that their angular momentum holds their orientation in space. Pioneer 10 and 11, built by NASA's Ames Research Center in California, spun about the axis of their big dish antennas to keep them pointed at the Earth all the way to Jupiter and beyond.
For this to work the spin axis must be a principal axis, or the spacecraft will wobble, and engineers balance spacecraft on spin tables before launch, just as tires are balanced, adding small masses until the products of inertia vanish. The choice of which principal axis matters too, for a reason the next lesson explains: spin about the axis of largest moment is the stable choice once a spacecraft loses any energy, as Explorer 1 discovered in 1958.
For rotation about a fixed axis in Physics C, $L = I\omega$ was a number along the axis, and it is natural to think $\vec{L}$ always points along $\vec{\omega}$. It does only about a principal axis. About any other axis the products of inertia tip $\vec{L}$ away, the tipped part rotates with the body, and a torque is needed just to keep the axis still. That is the shimmy of an unbalanced wheel.
A second error is to think a body's principal axes are where its mass is. They depend on the whole distribution and the reference point, and moving the reference point can change them.
Masses $m$ sit at $(a, 0, b)$ and $(-a, 0, -b)$ and spin about $z$. Find $I_{zz}$.
$I_{zz} = \sum m(x^2 + y^2) = 2ma^2$
Squared distance from the $z$ axis.
Find the product of inertia $I_{xz}$.
$I_{xz} = -\sum mxz = -(mab + mab) = -2mab$
Both have $xz = ab$.
Find the angular momentum for $\vec{\omega} = \omega\hat{z}$.
$\vec{L} = \omega(I_{xz}, I_{yz}, I_{zz}) = \omega(-2mab, 0, 2ma^2)$
The third column of the tensor.
Compare its direction with the axis.
$\tan\beta = \dfrac{|L_x|}{L_z} = \dfrac{b}{a}$
Tilted from $z$ toward the bar, perpendicular to it.
Find the torque needed to keep the axis fixed.
$\Gamma = \omega|L_x| = 2mab\,\omega^2$
The off-axis part of $\vec{L}$ turns with the body at rate $\omega$.
A flat body has $\mathbf{I} = \begin{pmatrix} 5 & -2 \\ -2 & 2 \end{pmatrix}$ kg m² in its plane. Write the characteristic equation.
$(5 - \lambda)(2 - \lambda) - 4 = 0$
$\det(\mathbf{I} - \lambda\mathbf{1}) = 0$.
Expand the determinant.
$\lambda^2 - 7\lambda + 6 = 0$
Multiply out and collect.
Solve for the principal moments.
$\lambda = 1 \text{ or } 6\ \text{kg m}^2$
Factor as $(\lambda - 1)(\lambda - 6)$.
Find the axis for $\lambda = 6$.
$\begin{pmatrix} -1 & -2 \\ -2 & -4 \end{pmatrix}\vec{e} = 0 \Rightarrow \vec{e} = (2, -1)/\sqrt{5}$
Solve $(\mathbf{I} - 6)\vec{e} = 0$.
Find the axis for $\lambda = 1$.
$\vec{e} = (1, 2)/\sqrt{5}$
Perpendicular to the first, as it must be.
Check the trace.
$1 + 6 = 5 + 2$
The sum of the principal moments equals the sum of the diagonal.
A uniform cube of mass $M$ and side $a$ has axes along its edges from a corner. Find $I_{xx}$.
$I_{xx} = \dfrac{M}{a^3}\displaystyle\int_0^a\int_0^a\int_0^a(y^2 + z^2)\,dx\,dy\,dz = \tfrac{2}{3}Ma^2$
$\int_0^a y^2\,dy = a^3/3$, twice.
Find the product of inertia.
$I_{xy} = -\dfrac{M}{a^3}\displaystyle\int xy\,dV = -\tfrac{1}{4}Ma^2$
$\int_0^a x\,dx = a^2/2$, twice.
Write the tensor in units of $Ma^2/12$.
$\mathbf{I} = \dfrac{Ma^2}{12}\begin{pmatrix} 8 & -3 & -3 \\ -3 & 8 & -3 \\ -3 & -3 & 8 \end{pmatrix}$
By symmetry all diagonal entries match, as do all products.
Apply it to the diagonal direction.
$\mathbf{I}(1, 1, 1) = \dfrac{Ma^2}{12}(2, 2, 2)$
Each row sums to $8 - 3 - 3 = 2$.
Read off the moment about the diagonal.
$\lambda_1 = \tfrac{2}{12}Ma^2 = \tfrac{1}{6}Ma^2$
The diagonal is a principal axis.
Find the other two from the trace.
$\lambda_2 = \lambda_3 = \dfrac{24 - 2}{2} \cdot \dfrac{Ma^2}{12} = \tfrac{11}{12}Ma^2$
The trace is $24 \cdot Ma^2/12$, and the two remaining moments are equal by symmetry.
Write the parallel-axis theorem.
$I = I_{\text{cm}} + Md^2$
Shift from the center of mass.
Substitute the values.
$I = 4 + 2 \times 3^2$
SI units.
Evaluate the moment.
A rigid body spins at $15$ rad/s about the $z$ axis, and its inertia tensor has a nonzero product of inertia $I_{xz}$. Which statement about its angular momentum is true?
Complete the worked solution: a uniform cube of mass $M$ and side $a$ has $Ma^2 = 48$ kg m². Using axes along its edges from one corner, find $I_{xx}$, the product $I_{xy}$, and the moment about the diagonal through that corner, all in kg m².
Integrate the squared distance from the $x$ edge.
$I_{xx} = \rho\displaystyle\int_0^a\int_0^a\int_0^a(y^2 + z^2)\,dx\,dy\,dz = \tfrac{2}{3}Ma^2 =$ i
Each squared coordinate averages to $a^2/3$.
Integrate minus the product of coordinates.
$I_{xy} = -\rho\displaystyle\int xy\,dV = -\tfrac{1}{4}Ma^2 =$ p
Each coordinate averages to $a/2$.
Apply the tensor to the diagonal direction.
$\mathbf{I}(1, 1, 1) = (I_{xx} + 2I_{xy})(1, 1, 1) \Rightarrow \lambda =$ d
By symmetry the diagonal is a principal axis.
Check the diagonal's moment against intuition.
$\text{smallest of the three principal moments}$
The mass is closest, on average, to the long diagonal.
Match each quantity to its expression.
| $\sum m(y^2 + z^2)$ | $-\sum mxy$ | $\mathbf{I}\vec{\omega} = \lambda\vec{\omega}$ | $\tfrac{1}{2}\vec{\omega} \cdot \mathbf{I}\vec{\omega}$ | |
|---|---|---|---|---|
| moment of inertia | ||||
| product of inertia | ||||
| principal axis condition | ||||
| kinetic energy |
Two particles of mass $4$ kg sit at $(2, 2, 0)$ m and $(-2, -2, 0)$ m. Fill in $I_{xx}$, $I_{yy}$ and $I_{xy}$ of the pair, in kg m².
| value | |
|---|---|
| $I_{xx}$ (kg m²) | |
| $I_{yy}$ (kg m²) | |
| $I_{xy}$ (kg m²) |
A body of mass $4$ kg has moment of inertia $3$ kg m² about an axis through its center of mass. Write its moment of inertia about a parallel axis a distance $d$ away, in kg m², as a formula in $d$ (meters).
Answer:
In the $xy$ plane a flat body's inertia tensor is $\begin{pmatrix} 8 & -3 \\ -3 & 8 \end{pmatrix}$ kg m². What is its larger principal moment of inertia in that plane, in kg m²?
Answer: kg m² principal moment
At a tire shop in Ohio, a wheel has two unbalanced spots of $47$ g each, $0.20$ m from the axle, on opposite sides of the wheel and $0.10$ m apart along the axle. The wheel spins at $59$ rad/s. What torque must the axle's bearings supply, in N m, to keep it turning about the axle?
Answer: N m on the bearings
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A body of mass $5$ kg has moment of inertia $4$ kg m² about an axis through its center of mass. Write its moment of inertia about a parallel axis a distance $d$ away, in kg m², as a formula in $d$ (meters).
Answer:
You can use the inertia tensor. Explain to someone why a wheel can wobble even when its center of mass sits exactly on the axle.
19. Your turn: a body has $I_{cm} = 4$ kg m² and mass $2$ kg. Find its moment about a parallel axis $3$ m away., step 3
$I = 22\ \text{kg m}^2$
Always larger than about the center of mass.