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Radius of Gyration 📂Classical Mechanics

Radius of Gyration

Definition1

For a rigid body of mass $m$ and moment of inertia $I$ about a given axis, the radius of gyration $k$ is defined as follows.

$$ I = m k^{2} $$

Rearranging this for $k$ yields:

$$ k = \sqrt{\frac{I}{m}} $$

Explanation

The radius of gyration $k$ is the distance such that, if the total mass $m$ of the rigid body were concentrated at a single point at distance $k$ from the axis of rotation, the moment of inertia would remain the same. In other words, it is an effective radius that replaces the distributed mass by a single point mass. This becomes clearer if we compare the definition of moment of inertia $I = \sum_{i} m_{i} r_{i}^{2}$ with the defining equation $I = mk^{2}$. Since the total mass is $m = \sum_{i} m_{i}$, setting the two expressions equal gives:

$$ k^{2} = \frac{\sum_{i} m_{i} r_{i}^{2}}{\sum_{i} m_{i}} $$

Thus $k^{2}$ is the mass-weighted average of the squared distances $r_{i}^{2}$ to the individual point masses. In other words, the radius of gyration $k$ can be regarded as a kind of mean distance (with mass as weight) of the point masses from the axis of rotation.

For example, the moment of inertia of a thin rod of length $a$ about an axis passing through one end is $I = \dfrac{1}{3}ma^{2}$, so its radius of gyration is:

$$ k = \sqrt{\frac{I}{m}} = \sqrt{\frac{\frac{1}{3}ma^{2}}{m}} = \frac{a}{\sqrt{3}} $$

Thus, intuitively, the rod’s rotation is equivalent to a point mass of mass $m$ rotating at a distance $\dfrac{a}{\sqrt{3}} \approx 0.577a$ from the axis. Note that this value lies outside the rod’s midpoint $a/2 = 0.5a$. Because the weighting uses the square of the distance $r^{2}$, regions farther from the axis contribute more to the moment of inertia, causing the average distance (the radius of gyration) to be biased outward relative to the simple geometric center.

Radii of Gyration for Various Bodies

Let the mass of each body be $m$. The squared radius of gyration $k^{2}$ for various shapes is given in the table below. Since $I = mk^{2}$ is the moment of inertia, $k^{2}$ equals the moment of inertia of each body divided by its mass.

BodyParametersAxis of rotation$k^{2}$
Thin rodlength $a$through rod center$\frac{1}{12}a^{2}$
Thin rodlength $a$through rod end$\frac{1}{3}a^{2}$
Thin rectangular platesides $a$, $b$axis through center parallel to side $b$$\frac{1}{12}a^{2}$
Thin rectangular platesides $a$, $b$axis through center perpendicular to plate$\frac{1}{12}(a^{2}+b^{2})$
Diskradius $a$axis through disk center and parallel to disk$\frac{1}{4}a^{2}$
Diskradius $a$axis through disk center and perpendicular to disk$\frac{1}{2}a^{2}$
Ringradius $a$axis through ring center and parallel to ring$\frac{1}{2}a^{2}$
Ringradius $a$axis through ring center and perpendicular to ring$a^{2}$
Cylindrical shellradius $a$, length $b$central longitudinal axis$a^{2}$
Solid cylinderradius $a$, length $b$central longitudinal axis$\frac{1}{2}a^{2}$
Solid cylinderradius $a$, length $b$central axis perpendicular to longitudinal axis$\frac{1}{4}a^{2}+\frac{1}{12}b^{2}$
Spherical shellradius $a$axis through sphere center$\frac{2}{3}a^{2}$
Sphereradius $a$axis through sphere center$\frac{2}{5}a^{2}$

See also


  1. Grant R. Fowles and George L. Cassiday. Analytical Mechanics (7th Edition, 2005), p335-338. ↩︎