Lesson 1 · 25 min
Why Mass Moments of Inertia?
Push a door near its hinges and it barely moves; push at the handle and it swings open. How hard a body is to spin up depends on its mass and on how far that mass sits from the axis. The mass moment of inertia puts a number on that.
Learning objectives
- Explain the role of the mass moment of inertia \(I\) in \(M = I\alpha\) and \(T = \tfrac12 I\omega^2\), as the rotational counterpart of mass.
- Compute \(I = \sum m r^2\) for a set of particles, measuring each \(r\) from the axis.
- Compute and interpret the radius of gyration \(k = \sqrt{I/m}\).
- Distinguish mass moments of inertia (kg·m²) from the area moments of inertia of statics (m⁴).
The rotational counterpart of mass
In particle dynamics, mass measures resistance to a change in velocity: \(\sum F = ma\). A rigid body that rotates about a fixed axis has a matching law. The resultant moment about the axis sets the angular acceleration \(\alpha\), and the constant of proportionality is the body's mass moment of inertia about that axis:
Rotation about a fixed axis through O (or about G)
\[ \sum M_O = I_O\,\alpha, \qquad T = \tfrac12 I_O\,\omega^2, \qquad H_O = I_O\,\omega \]Compare \(\sum F = ma\), \(T = \tfrac12 m v^2\) and \(L = mv\) for a particle: \(I\) plays the part of \(m\), \(\alpha\) of \(a\) and \(\omega\) of \(v\).
Mass is a single number for a body. The moment of inertia is not: it depends on the axis you choose. A slender rod is easy to spin about its own long axis, harder about a perpendicular axis through its middle, and harder still about a perpendicular axis through one end. Choosing the right axis, and computing \(I\) about it, is the skill this module builds.
Particles: \(I = \sum m r^2\)
Think of a body as many small particles. A particle of mass \(m\) at perpendicular distance \(r\) from the axis moves on a circle of radius \(r\), with speed \(v = r\omega\) and tangential acceleration \(a_t = r\alpha\). Its kinetic energy is \(\tfrac12 m v^2 = \tfrac12 (m r^2)\,\omega^2\). Adding up all the particles gives
Moment of inertia of a set of particles
\[ I = \sum_i m_i\,r_i^2 \]\(r_i\) is the perpendicular distance from particle \(i\) to the axis, not its distance from the origin. Units: kg·m².
Because \(r\) is squared, position matters more than mass: doubling a particle's mass doubles its contribution, but doubling its distance from the axis multiplies it by four. That is why a flywheel puts most of its metal in a heavy rim, and why figure skaters spin faster when they pull their arms in.
For a particle in 3D, the distance to each coordinate axis comes from the two other coordinates. The distance from \((x, y, z)\) to the \(z\)-axis is \(\sqrt{x^2 + y^2}\), so
\[ I_{xx} = \sum m\,(y^2 + z^2), \qquad I_{yy} = \sum m\,(z^2 + x^2), \qquad I_{zz} = \sum m\,(x^2 + y^2). \]The double subscript is the notation of Lessons 5–8, where products such as \(I_{xy}\) also appear. Many books write \(I_x\), \(I_y\), \(I_z\) for the same three moments.
Example 1.1 — Three masses on a light frame
Small masses are fixed to a light frame in the \(xy\)-plane: \(2\ \text{kg}\) at \((0.3,\ 0.4)\ \text{m}\), \(1.5\ \text{kg}\) at \((-0.5,\ 0.2)\ \text{m}\) and \(3\ \text{kg}\) at \((0.1,\ -0.6)\ \text{m}\). Find the moments of inertia about the \(z\)-axis and about the \(x\)-axis.
Show solution
About \(z\). Every mass has \(z = 0\), so \(r^2 = x^2 + y^2\):
\[ \begin{aligned} I_{zz} &= 2(0.3^2 + 0.4^2) + 1.5(0.5^2 + 0.2^2) + 3(0.1^2 + 0.6^2) \\ &= 0.500 + 0.435 + 1.110 = 2.045\ \text{kg·m}^2 \end{aligned} \]About \(x\). Now \(r^2 = y^2 + z^2 = y^2\):
\[ I_{xx} = 2(0.4^2) + 1.5(0.2^2) + 3(0.6^2) = 0.32 + 0.06 + 1.08 = 1.460\ \text{kg·m}^2 \]The \(3\ \text{kg}\) mass contributes most to both: it is the heaviest and the farthest from each axis. (Check: \(I_{yy} = \sum m x^2 = 0.585\ \text{kg·m}^2\), and \(I_{xx} + I_{yy} = I_{zz}\). That is no coincidence for a flat body; Lesson 2 explains it.)
The radius of gyration
Suppose the whole mass \(m\) of a body were squeezed into a thin ring around the axis. The ring radius that gives the same moment of inertia is the radius of gyration \(k\):
Radius of gyration about an axis
\[ I = m k^2 \qquad \Longleftrightarrow \qquad k = \sqrt{\frac{I}{m}} \]\(k\) is a length (m). It says how far from the axis the mass is, "on average" in the root-mean-square sense.
Manufacturers often quote \(k\) instead of \(I\) because it does not change when the same design is made in a different material: a steel and an aluminium version of a wheel have different masses but the same \(k\). Like \(I\), \(k\) belongs to an axis.
Example 1.2 — Spinning up a flywheel
A \(180\ \text{kg}\) flywheel has a radius of gyration of \(0.42\ \text{m}\) about its axis. A motor applies a constant \(60\ \text{N·m}\). Ignoring friction, how long does it take to reach \(300\ \text{rpm}\) from rest?
Show solution
\(300\ \text{rpm} = 300 \cdot 2\pi/60 = 31.42\ \text{rad/s}\), and with constant \(\alpha\), \(\omega = \alpha t\):
\[ t = \frac{\omega}{\alpha} = \frac{31.42}{1.890} = 16.62\ \text{s} \]Mass moments are not area moments
In statics and mechanics of materials you met the area moment of inertia (second moment of area), \(I_x = \int y^2\,dA\), which sets the bending stiffness of a beam. It has units of m⁴. The mass moment of inertia of this module is \(\int r^2\,dm\), in kg·m². The integrals look alike, but they answer different questions:
Area moment (statics)
\[ I_x = \int y^2\,dA \quad [\text{m}^4] \]A property of a cross-section's shape. Used for beam bending and buckling.
Mass moment (dynamics)
\[ I = \int r^2\,dm \quad [\text{kg·m}^2] \]A property of a body's mass distribution about an axis. Used for rotation: \(M = I\alpha\), \(T = \tfrac12 I\omega^2\).
They are connected for a thin plate of uniform thickness \(t\) and density \(\rho\): each area element has mass \(dm = \rho t\,dA\), so for axes in the plane of the plate
\[ I_\text{mass} = \rho t \int y^2\,dA = \frac{m}{A}\,I_\text{area}. \]Example 1.3 — A steel plate, both ways
A steel plate (\(\rho = 7850\ \text{kg/m}^3\)) measures \(300 \times 200\ \text{mm}\) and is \(5\ \text{mm}\) thick. Find its area moment and its mass moment of inertia about the centroidal axis parallel to the \(300\ \text{mm}\) side.
Show solution
Area moment of the \(b \times h = 0.3 \times 0.2\ \text{m}\) rectangle: \(I_\text{area} = bh^3/12 = 0.3(0.2)^3/12 = 2.000 \times 10^{-4}\ \text{m}^4\).
Mass: \(m = \rho V = 7850(0.3)(0.2)(0.005) = 2.355\ \text{kg}\), so
\[ I_\text{mass} = \frac{m}{A} I_\text{area} = \frac{2.355}{0.06}(2.000 \times 10^{-4}) = 7.850 \times 10^{-3}\ \text{kg·m}^2 \]This equals \(\tfrac{1}{12} m h^2\), the thin-plate formula of Lesson 2.
Check your understanding
Key takeaways
- The mass moment of inertia is the rotational counterpart of mass: \(\sum M_O = I_O\alpha\), \(T = \tfrac12 I_O\omega^2\).
- \(I\) always belongs to an axis. For particles, \(I = \sum m r^2\), with \(r\) the perpendicular distance to that axis.
- \(I_{xx} = \sum m(y^2 + z^2)\), \(I_{yy} = \sum m(z^2 + x^2)\), \(I_{zz} = \sum m(x^2 + y^2)\).
- The radius of gyration \(k = \sqrt{I/m}\) is the ring radius with the same \(I\).
- Mass moments (kg·m²) differ from area moments (m⁴); for a thin plate, \(I_\text{mass} = (m/A)\,I_\text{area}\) for in-plane axes.
- Next: Lesson 2 replaces the sum by an integral and builds the table of standard bodies.