Radius Of Gyration

Introduction

A moment of inertia measures how a mass distribution is arranged relative to a chosen axis: mass farther from the axis contributes more strongly than mass close to it. The radius of gyration turns that information into a length. It answers this question: at what distance from the axis could the entire mass be concentrated while keeping the same moment of inertia?

The answer depends on both the object and the axis. It is not necessarily a physical radius or the distance to any particular point of the object.

Definition and Interpretation

For a planar lamina occupying a region \(D\), let \(\rho(x,y)\) be its mass density, and let \(\ell\) be the chosen axis in the plane. The distance from a point \((x,y)\) to that axis is denoted by \(d((x,y),\ell)\). Its moment of inertia and mass are

These formulas require a nonnegative density with finite, positive total mass, and a finite moment of inertia about the chosen axis. Once \(I_{\ell}\) and \(m\) are known, the radius of gyration about that same axis is

The axis subscript matters. Changing the axis changes the distances in the moment integral, so the radius of gyration can change as well. For example, about the \(x\)-axis the perpendicular distance of \((x,y)\) is \(|y|\), and \(I_x=\iint_D y^2\rho(x,y)\,dA\). About the \(y\)-axis, the corresponding formula is \(I_y=\iint_D x^2\rho(x,y)\,dA\).

Why It Represents an Equivalent Distance

A point mass \(m\) located a perpendicular distance \(r\) from an axis has moment of inertia \(mr^2\). For a distributed object, the moment of inertia adds the contributions from all its mass elements. The definition of \(k\) is chosen so that the same mass, if placed at distance \(k\), has exactly the original moment of inertia.

Here is the full justification. Each mass element \(dm\) at distance \(d\) contributes \(d^2\,dm\), so \(I_{\ell}=\int d^2\,dm\), where the integral is over the object's mass. Since \(m=\int dm>0\), the definition gives

Multiplying by \(m\) yields \(m k_{\ell}^2=I_{\ell}\). Conversely, if a single mass \(m\) at a nonnegative distance \(r\) is required to have moment \(I_{\ell}\), then \(mr^2=I_{\ell}\). Dividing by \(m>0\) gives \(r^2=I_{\ell}/m\), and the unique nonnegative solution is \(r=\sqrt{I_{\ell}/m}=k_{\ell}\). This proves both the interpretation and why the nonnegative square root is used.

Worked Example: Two Point Masses

Calculating It for a Lamina

For a lamina, the calculation has two distinct parts: determine the total mass, and determine the moment of inertia about the chosen axis. The radius of gyration is then the square root of their ratio. Keep the same axis throughout; using the moment of inertia about one axis with a mass-distance interpretation about another would not give the required equivalence.

Worked Example: Uniform Triangular Lamina

Worked Example: A Lamina with Variable Density

Radius of Gyration as a Root-Mean-Square Distance

The formula also clarifies what kind of average the radius represents. For point masses \(m_i\) at perpendicular distances \(d_i\), total mass and moment are \(m=\sum_i m_i\) and \(I=\sum_i m_i d_i^2\). Consequently,

This is a mass-weighted root-mean-square distance: square the distances, weight them by mass, divide by total mass, and take the square root. It is not generally the ordinary average distance. For a continuous lamina, the corresponding expression is \(k_{\ell}=\sqrt{(\int d^2\,dm)/m}\).

This interpretation gives a useful bound. If every part of the mass lies at distance at most \(R\) from the chosen axis, then \(d^2\leq R^2\) everywhere. Since \(dm\) is nonnegative,

A calculated radius larger than the greatest distance of any mass from the axis signals an error in the mass, moment, or units.

Common Mistake: Using the Ordinary Average Distance

Consider two equal \(1\text{ kg}\) point masses, one on the axis and one \(2\text{ m}\) away. Their distances are \(0\text{ m}\) and \(2\text{ m}\). The ordinary average distance is \(1\text{ m}\), but that is not the radius of gyration. Using it would give the incorrect moment

The actual moment is

The correct radius is

Substitution verifies it: \((2\text{ kg})(\sqrt{2}\text{ m})^2=4\text{ kg}\,\text{m}^2\). Squaring distances before averaging is essential because moment of inertia depends on the square of distance.

Why the Radius Is Useful

The radius of gyration gives a compact length-based description of how mass is spread relative to an axis. Two objects with the same total mass and the same radius of gyration about that axis have the same moment of inertia about it, even if their shapes and density distributions differ. Conversely, the same object can have different radii of gyration about different axes.

The unit check is straightforward: moment of inertia has units of mass times length squared, while mass has units of mass. Therefore \(I/m\) has units of length squared, and its square root has units of length. A radius of gyration should not be reported as an area or as a moment of inertia.