Lesson 6 · 35 min

The Inertia Tensor and Angular Momentum

In the plane, angular momentum is \(I\omega\) and points along the axis. In three dimensions it generally does not: spin a tilted rod about a vertical shaft and its angular momentum leans away from the shaft. Six numbers, arranged as a matrix, tell you exactly how.

Learning objectives

Angular momentum of a rotating body

Take a rigid body turning with angular velocity \(\wvec = \omega_x\ihat + \omega_y\jhat + \omega_z\khat\) about a fixed point \(O\) (or about its center of mass \(G\)). A mass element at \(\rvec\) has velocity \(\wvec \times \rvec\), so the body's angular momentum about \(O\) is

\[ \Hvec_O = \int \rvec \times (\wvec \times \rvec)\,dm. \]
Expanding the cross products

With \(\rvec = x\ihat + y\jhat + z\khat\), the vector identity \(\rvec \times (\wvec \times \rvec) = \wvec\,(\rvec\cdot\rvec) - \rvec\,(\rvec\cdot\wvec)\) gives, for the \(x\)-component,

\[ \omega_x(x^2 + y^2 + z^2) - x(x\omega_x + y\omega_y + z\omega_z) = (y^2 + z^2)\,\omega_x - xy\,\omega_y - xz\,\omega_z. \]

Integrate over the body:

\[ H_x = \left(\int (y^2 + z^2)\,dm\right)\omega_x - \left(\int xy\,dm\right)\omega_y - \left(\int xz\,dm\right)\omega_z = I_{xx}\,\omega_x - I_{xy}\,\omega_y - I_{xz}\,\omega_z \]

and \(H_y\), \(H_z\) follow by cycling the letters.

Each component of \(\Hvec\) mixes all three components of \(\wvec\), with the moments on the diagonal and the products, with minus signs, off it. That is a matrix product:

The inertia tensor and angular momentum

\[ \begin{Bmatrix} H_x \\ H_y \\ H_z \end{Bmatrix} = \begin{bmatrix} I_{xx} & -I_{xy} & -I_{xz} \\ -I_{yx} & I_{yy} & -I_{yz} \\ -I_{zx} & -I_{zy} & I_{zz} \end{bmatrix} \begin{Bmatrix} \omega_x \\ \omega_y \\ \omega_z \end{Bmatrix}, \qquad \Hvec_O = \Imat_O\,\wvec \]

The matrix \(\Imat_O\) is the inertia tensor about \(O\). It is symmetric (\(I_{yx} = I_{xy}\), …), so six numbers define it. It holds about a fixed point \(O\) or about \(G\).

Plane rotation is the special case \(\wvec = \omega\khat\) with \(I_{xz} = I_{yz} = 0\): then \(\Hvec = I_{zz}\,\omega\,\khat\), the familiar \(H = I\omega\).

Why \(\Hvec\) is not parallel to \(\wvec\)

Spin a body about the \(z\)-axis, \(\wvec = \omega\khat\). The matrix product picks out the third column of \(\Imat\):

\[ \Hvec = -I_{xz}\,\omega\,\ihat - I_{yz}\,\omega\,\jhat + I_{zz}\,\omega\,\khat. \]

If \(I_{xz}\) or \(I_{yz}\) is not zero, \(\Hvec\) has components perpendicular to the spin axis. As the body turns, those components turn with it, so \(\Hvec\) sweeps out a cone around the shaft. A changing angular momentum needs a moment, \(\sum\Mvec_O = \dot{\Hvec}_O\), and the only things that can supply it are the bearings. That is a dynamic imbalance: a rotor that shakes its bearings even when its center of mass is exactly on the shaft.

Figure 6.1 A body fixed to a vertical shaft (the \(z\)-axis) and spinning about it. The green arrow is \(\colW{\wvec}\), along the shaft; the red arrow is \(\colH{\Hvec_O} = \Imat_O\wvec\). Tilt the rod or the disk: \(\Hvec_O\) leans away from the shaft, and when the body spins it sweeps a cone, traced by the dashed circle. At \(0^\circ\) and \(90^\circ\) the products \(I_{xz}, I_{yz}\) vanish and \(\Hvec_O\) lines up with \(\wvec\). Arrows show directions only; the readouts give the values.

Example 6.1 — A tilted rod on a shaft

A uniform \(3\ \text{kg}\) slender rod, \(0.8\ \text{m}\) long, is welded at its midpoint \(O\) to a vertical shaft, at \(\beta = 30^\circ\) to the shaft. The shaft spins at a constant \(\omega = 20\ \text{rad/s}\). At the instant the rod lies in the \(xz\)-plane, find \(\Imat_O\), \(\Hvec_O\), the angle between \(\Hvec_O\) and the shaft, and the moment the bearings must supply.

Show solution

Points of the rod are at \(s(\sin\beta,\ 0,\ \cos\beta)\), \(-\tfrac l2 \le s \le \tfrac l2\), with \(dm = (m/l)\,ds\). Every integral is \(\int s^2\,dm = \tfrac1{12} m l^2 = \tfrac1{12}(3)(0.8)^2 = 0.16\ \text{kg·m}^2\) times a trigonometric factor:

\[ I_{xx} = 0.16\cos^2\beta, \quad I_{yy} = 0.16, \quad I_{zz} = 0.16\sin^2\beta, \quad I_{xz} = 0.16\sin\beta\cos\beta, \quad I_{xy} = I_{yz} = 0 \] \[ \Imat_O = \begin{bmatrix} 0.12 & 0 & -0.06928 \\ 0 & 0.16 & 0 \\ -0.06928 & 0 & 0.04 \end{bmatrix}\ \text{kg·m}^2 \]

With \(\wvec = 20\,\khat\), \(\Hvec_O\) is \(20\) times the third column:

\[ \Hvec_O = -1.386\,\ihat + 0.800\,\khat\ \ \text{kg·m}^2/\text{s} \]

It makes \(\arctan(1.386/0.800) = 60^\circ\) with the shaft: it is perpendicular to the rod, as it must be for a slender rod.

Bearing moment. \(\Hvec_O\) is fixed in the rod, which turns at \(\wvec\), so \(\dot{\Hvec}_O = \wvec \times \Hvec_O\):

\[ \sum\Mvec_O = 20\,\khat \times (-1.386\,\ihat + 0.800\,\khat) = -27.71\,\jhat\ \ \text{N·m} \]

A \(27.7\ \text{N·m}\) moment, perpendicular to the plane of the rod and turning with it, even though the rod is perfectly balanced about the shaft (\(G\) is on the axis).

Example 6.2 — The bent rod, spinning

The bent rod of Example 5.2 (\(I_{xx} = 0.05933\), \(I_{yy} = 0.2080\), \(I_{zz} = 0.2567\), \(I_{xy} = 0.0840\), \(I_{yz} = 0.0120\), \(I_{zx} = 0.0160\ \text{kg·m}^2\) about \(O\)) spins about the \(z\)-axis at \(10\ \text{rad/s}\). Find \(\Hvec_O\) and its angle to the \(z\)-axis.

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\[ \Hvec_O = \begin{bmatrix} 0.05933 & -0.0840 & -0.0160 \\ -0.0840 & 0.2080 & -0.0120 \\ -0.0160 & -0.0120 & 0.2567 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 10 \end{Bmatrix} = \begin{Bmatrix} -0.160 \\ -0.120 \\ 2.567 \end{Bmatrix}\ \text{kg·m}^2/\text{s} \]

\(|\Hvec_O| = 2.574\ \text{kg·m}^2/\text{s}\), and the angle to \(z\) is \(\arccos(2.567/2.574) = 4.46^\circ\). The large product \(I_{xy}\) plays no part here: only products that involve the spin axis (\(z\)) tilt \(\Hvec\).

Kinetic energy

The kinetic energy of a body rotating about a fixed point \(O\) (or the rotational part about \(G\)) is \(\tfrac12\int|\wvec \times \rvec|^2\,dm\), which becomes

Rotational kinetic energy

\[ T = \tfrac12\,\wvec^\mathsf{T}\Imat\,\wvec = \tfrac12\,\wvec\cdot\Hvec \] \[ T = \tfrac12\left(I_{xx}\omega_x^2 + I_{yy}\omega_y^2 + I_{zz}\omega_z^2\right) - I_{xy}\,\omega_x\omega_y - I_{yz}\,\omega_y\omega_z - I_{zx}\,\omega_z\omega_x \]

For the tilted rod of Example 6.1, \(T = \tfrac12(20)(0.800) = 8.00\ \text{J}\): only the component of \(\Hvec\) along \(\wvec\) stores energy.

Moving the tensor: the parallel-axis theorem in matrix form

Lessons 3 and 5 moved one moment or one product at a time. Doing all six at once, with \(G\) at \((\bar x, \bar y, \bar z)\) from \(O\):

Parallel-axis theorem for the inertia tensor

\[ \Imat_O = \Imat_G + m \begin{bmatrix} \bar y^2 + \bar z^2 & -\bar x\bar y & -\bar x\bar z \\ -\bar x\bar y & \bar z^2 + \bar x^2 & -\bar y\bar z \\ -\bar x\bar z & -\bar y\bar z & \bar x^2 + \bar y^2 \end{bmatrix} \]

The added matrix is the inertia tensor of a particle of mass \(m\) at \(G\). The axes at \(O\) must be parallel to those at \(G\).

Example 6.3 — An offset sphere

A \(2\ \text{kg}\) solid sphere of radius \(0.1\ \text{m}\) has its center at \((0.3,\ 0.2,\ 0)\ \text{m}\). Find its inertia tensor about the origin.

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About its center, \(\Imat_G = \tfrac25(2)(0.1)^2\,\mathbf{1} = 0.008\,\mathbf{1}\) (a sphere has no products). The transfer matrix with \((0.3, 0.2, 0)\):

\[ \Imat_O = \begin{bmatrix} 0.008 + 2(0.04) & -2(0.06) & 0 \\ -2(0.06) & 0.008 + 2(0.09) & 0 \\ 0 & 0 & 0.008 + 2(0.13) \end{bmatrix} = \begin{bmatrix} 0.088 & -0.120 & 0 \\ -0.120 & 0.188 & 0 \\ 0 & 0 & 0.268 \end{bmatrix}\ \text{kg·m}^2 \]

So \(I_{xy} = +0.120\ \text{kg·m}^2\) (the sphere is in the first quadrant), entered as \(-0.120\) in the tensor.

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Key takeaways