Lesson 2 · 35 min

Moments of Inertia by Integration

A real body is not a handful of particles but a continuous distribution of mass. Replace the sum by an integral, choose a mass element cleverly, and a few lines of calculus give the formulas that fill every dynamics textbook's inside cover.

Learning objectives

From sums to integrals

Slice the body into mass elements \(dm\), each at perpendicular distance \(r\) from the axis, and let the slices shrink. The sum \(\sum m r^2\) of Lesson 1 becomes

Moment of inertia of a continuous body

\[ I = \int_m r^2\,dm, \qquad dm = \rho\,dV \ \ (\text{solid}), \quad \sigma\,dA \ \ (\text{thin plate}), \quad \lambda\,dx \ \ (\text{slender rod}) \]

For a uniform body the densities are constants: \(\rho = m/V\) (kg/m³), \(\sigma = m/A\) (kg/m²) and \(\lambda = m/L\) (kg/m).

The whole art is in the choice of element. A good element either has all of its mass at the same distance \(r\) from the axis (a thin strip parallel to the axis, a thin ring around it), or is a shape whose own moment of inertia you already know (a thin disk). Then the integral is one-dimensional.

A slender rod

Take a uniform slender rod of mass \(m\) and length \(l\) along the \(x\)-axis, centered on the origin, and find \(I\) about a perpendicular axis through its middle. A short piece \(dx\) at position \(x\) has mass \(dm = \lambda\,dx = (m/l)\,dx\), and every point of it is at distance \(|x|\) from the axis:

\[ I_G = \int_{-l/2}^{l/2} x^2\,\frac{m}{l}\,dx = \frac{m}{l}\left[\frac{x^3}{3}\right]_{-l/2}^{l/2} = \frac{1}{12}\,m l^2 \]

Through one end, the limits become \(0\) to \(l\):

\[ I_\text{end} = \int_0^{l} x^2\,\frac{m}{l}\,dx = \frac13\,m l^2 \]

Four times as large, because the far end of the rod is now twice as far away. About the rod's own long axis, every element is (nearly) on the axis, so \(I \approx 0\): that is what "slender" means.

Figure 2.1 A \(3\ \text{kg}\), \(2\ \text{m}\) rod cut into \(N\) equal pieces, each treated as a particle at its midpoint. The bars show each piece's contribution \(m_i r_i^2\): the pieces far from the axis dominate. Increase \(N\) and the sum closes in on the integral.

Rings and disks

In a thin ring (hoop) of radius \(R\), all the mass is at distance \(R\) from the axis through its center, perpendicular to its plane. No integral is needed: \(I = m R^2\).

A thin disk is a set of such rings. The ring between \(r\) and \(r + dr\) has area \(dA = 2\pi r\,dr\) and mass \(dm = \sigma\,2\pi r\,dr\), with \(\sigma = m/(\pi R^2)\):

\[ I_z = \int_0^R r^2\,\sigma\,2\pi r\,dr = 2\pi\sigma\,\frac{R^4}{4} = \frac{m}{\pi R^2}\cdot\frac{\pi R^4}{2} = \frac12\,m R^2 \]

A disk has half the moment of inertia of a ring with the same mass and radius, because much of its mass is near the center.

Example 2.1 — A thick-walled steel tube

A steel tube (\(\rho = 7850\ \text{kg/m}^3\)) is \(2\ \text{m}\) long, with outer radius \(50\ \text{mm}\) and inner radius \(40\ \text{mm}\). Derive its moment of inertia about its own axis, and evaluate it.

Show solution

Use thin cylindrical shells of radius \(r\), thickness \(dr\) and length \(L\): \(dm = \rho\,(2\pi r L)\,dr\), all at distance \(r\):

\[ I_z = \int_{r_i}^{r_o} r^2\,\rho\,2\pi r L\,dr = \frac{\pi \rho L}{2}\left(r_o^4 - r_i^4\right) = \frac{\pi\rho L}{2}\left(r_o^2 - r_i^2\right)\left(r_o^2 + r_i^2\right) \]

and since \(m = \rho\pi(r_o^2 - r_i^2)L\),

\[ I_z = \tfrac12\,m\left(r_o^2 + r_i^2\right) \]

Numbers: \(m = 7850\pi(0.05^2 - 0.04^2)(2) = 44.39\ \text{kg}\), and \(I_z = \tfrac12(44.39)(0.05^2 + 0.04^2) = 0.09100\ \text{kg·m}^2\).

Checks: with \(r_i = 0\) this is the solid cylinder, \(\tfrac12 m r_o^2\); as \(r_i \to r_o\) it becomes the thin-walled tube, \(m r^2\).

Solid cylinders and spheres: stack thin disks

A solid cylinder of radius \(R\) is a stack of thin disks, each with \(dI_z = \tfrac12 R^2\,dm\). Adding them gives \(I_z = \tfrac12 m R^2\): the length of the cylinder does not matter for its own axis.

A sphere is also a stack of disks, but their radii change with height. That makes it a good workout.

Example 2.2 — A solid sphere

Show that a uniform solid sphere of mass \(m\) and radius \(R\) has \(I = \tfrac25 m R^2\) about any diameter.

Show solution

Slice perpendicular to the \(z\)-axis. The disk at height \(z\) has radius \(a = \sqrt{R^2 - z^2}\), mass \(dm = \rho\pi a^2\,dz\), and moment of inertia \(\tfrac12 a^2\,dm\) about \(z\):

\[ I_z = \int_{-R}^{R} \frac12\,\rho\pi\left(R^2 - z^2\right)^2 dz = \frac{\rho\pi}{2}\left[R^4 z - \frac{2R^2 z^3}{3} + \frac{z^5}{5}\right]_{-R}^{R} = \frac{8}{15}\,\rho\pi R^5 \]

With \(\rho = m\big/\tfrac43\pi R^3\):

\[ I_z = \frac{8}{15}\pi R^5 \cdot \frac{3m}{4\pi R^3} = \frac25\,m R^2 \]

By symmetry the same holds for every diameter.

Thin plates: the perpendicular-axis theorem

For a thin plate lying in the \(xy\)-plane, every element has \(z \approx 0\), so

\[ I_{zz} = \int (x^2 + y^2)\,dm = \int y^2\,dm + \int x^2\,dm = I_{xx} + I_{yy}. \]

Perpendicular-axis theorem (thin plates only)

\[ I_{zz} = I_{xx} + I_{yy} \]

for a thin, flat body in the \(xy\)-plane, with \(x\), \(y\), \(z\) meeting at the same point.

It gives quick results. For a thin disk, symmetry makes \(I_{xx} = I_{yy}\), so each is \(\tfrac12 I_{zz} = \tfrac14 m R^2\). For a thin \(a \times b\) rectangular plate, \(I_{xx} = \tfrac1{12} m b^2\) and \(I_{yy} = \tfrac1{12} m a^2\) (each is a slender rod in disguise), so \(I_{zz} = \tfrac1{12} m (a^2 + b^2)\).

The table of standard bodies

These results, all about axes through the center of mass \(G\), are the building blocks for the rest of the module. Each body's own \(z\)-axis is its axis of symmetry (the long axis of a rod, the axis of a cylinder); a plate lies in its own \(xy\)-plane.

Moments of inertia about centroidal axes (uniform bodies of mass \(m\))
Body\(I_{xx}\)\(I_{yy}\)\(I_{zz}\)
Slender rod, length \(l\) along \(z\)\(\tfrac1{12} m l^2\)\(\tfrac1{12} m l^2\)\(\approx 0\)
Thin ring, radius \(r\)\(\tfrac12 m r^2\)\(\tfrac12 m r^2\)\(m r^2\)
Thin disk, radius \(r\)\(\tfrac14 m r^2\)\(\tfrac14 m r^2\)\(\tfrac12 m r^2\)
Thin plate \(a \times b\) (\(a\) along \(x\))\(\tfrac1{12} m b^2\)\(\tfrac1{12} m a^2\)\(\tfrac1{12} m (a^2 + b^2)\)
Block \(a \times b \times c\) (along \(x, y, z\))\(\tfrac1{12} m (b^2 + c^2)\)\(\tfrac1{12} m (a^2 + c^2)\)\(\tfrac1{12} m (a^2 + b^2)\)
Solid cylinder, radius \(r\), length \(h\)\(\tfrac1{12} m (3r^2 + h^2)\)\(\tfrac1{12} m (3r^2 + h^2)\)\(\tfrac12 m r^2\)
Tube, radii \(r_o, r_i\), length \(h\)\(\tfrac1{12} m [3(r_o^2 + r_i^2) + h^2]\)same\(\tfrac12 m (r_o^2 + r_i^2)\)
Thin-walled tube, radius \(r\), length \(h\)\(\tfrac1{12} m (6r^2 + h^2)\)same\(m r^2\)
Solid sphere, radius \(r\)\(\tfrac25 m r^2\)\(\tfrac25 m r^2\)\(\tfrac25 m r^2\)
Thin spherical shell, radius \(r\)\(\tfrac23 m r^2\)\(\tfrac23 m r^2\)\(\tfrac23 m r^2\)
Solid cone, base radius \(r\), height \(h\) (about \(G\), \(\tfrac h4\) above the base)\(\tfrac{3}{80} m (4r^2 + h^2)\)same\(\tfrac{3}{10} m r^2\)
Figure 2.2 A gallery of the standard bodies, each drawn with its own centroidal axes: \(\colX{x}\) (orange), \(\colY{y}\) (violet) and \(\colZ{z}\) (blue). Choose a body and set its mass and size; the three moments of inertia update with their formulas. Drag to turn the view. Watch which dimension each moment depends on: a moment about an axis never depends on the size along that axis alone.

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Key takeaways