Lesson 3 · 30 min
The Parallel-Axis Theorem
Tables give moments of inertia about axes through the center of mass. Real parts turn about hinges, pins and shafts that are somewhere else. One short theorem connects the two, as long as you remember where it starts.
Learning objectives
- State and prove the parallel-axis theorem \(I = I_G + m d^2\).
- Move a centroidal moment of inertia from the table of Lesson 2 to any parallel axis.
- Transfer between two axes that do not pass through \(G\), by going through \(G\).
- Use the radius-of-gyration form \(k^2 = k_G^2 + d^2\), and measure \(I\) with a compound pendulum.
The theorem
Let \(G\) be the center of mass of a body of mass \(m\). Take an axis through \(G\), and a second axis parallel to it at perpendicular distance \(d\). Then
Parallel-axis theorem
\[ I = I_G + m d^2 \]\(I_G\): moment of inertia about the axis through \(G\). \(d\): perpendicular distance between the two parallel axes. \(I\): moment of inertia about the other axis.
Why it works. Put the \(z\)-axis through \(G\) and the parallel axis through the point \(A = (x_A, y_A)\) of the \(xy\)-plane, so \(d^2 = x_A^2 + y_A^2\). A mass element at \((x, y, z)\) is at squared distance \((x - x_A)^2 + (y - y_A)^2\) from the second axis:
\[ \begin{aligned} I_A &= \int \left[(x - x_A)^2 + (y - y_A)^2\right] dm \\ &= \int (x^2 + y^2)\,dm - 2x_A \int x\,dm - 2y_A \int y\,dm + (x_A^2 + y_A^2)\int dm \\ &= I_G - 0 - 0 + m d^2 \end{aligned} \]The two middle integrals are first moments of mass about axes through \(G\), and those are zero: that is what "center of mass" means. The whole proof hinges on it, so the theorem only works when one of the two axes passes through \(G\).
Since \(m d^2 \ge 0\), the axis through \(G\) has the smallest moment of inertia of all the axes parallel to it. A body is easiest to spin about an axis through its center of mass.
Example 3.1 — The rod again
Use the parallel-axis theorem to find \(I\) of a slender rod about a perpendicular axis through one end.
Show solution
\(I_G = \tfrac1{12} m l^2\) and the end is \(d = l/2\) from \(G\):
\[ I_\text{end} = \tfrac1{12} m l^2 + m\left(\tfrac l2\right)^2 = \tfrac1{12} m l^2 + \tfrac14 m l^2 = \tfrac13 m l^2 \]the same as the direct integration in Lesson 2.
Example 3.2 — A disk pivoted at its rim
A thin \(4\ \text{kg}\) disk of radius \(0.15\ \text{m}\) swings in its own plane about a pin through a point on its rim. Find its moment of inertia about the pin.
Show solution
The pin is perpendicular to the disk, so start from \(I_G = \tfrac12 m r^2\), with \(d = r\):
\[ I_O = \tfrac12 m r^2 + m r^2 = \tfrac32 m r^2 = 1.5(4)(0.15)^2 = 0.1350\ \text{kg·m}^2 \]Example 3.3 — A cylinder about an axis through an end face
A solid steel cylinder has \(m = 8\ \text{kg}\), \(r = 50\ \text{mm}\) and length \(h = 500\ \text{mm}\). Find its moment of inertia about a diameter of one end face.
Show solution
A diameter of the end face is a transverse axis, parallel to the centroidal transverse axis and \(h/2\) from it:
\[ I = \tfrac1{12} m (3r^2 + h^2) + m\left(\tfrac h2\right)^2 = \tfrac1{12} m (3r^2 + 4h^2) \] \[ I = \tfrac{8}{12}\left(3(0.05)^2 + 4(0.5)^2\right) = \tfrac{8}{12}(1.0075) = 0.6717\ \text{kg·m}^2 \]For a long cylinder the \(3r^2\) term hardly matters: a slender rod would give \(\tfrac13 m h^2 = 0.6667\ \text{kg·m}^2\).
Every transfer goes through \(G\)
Suppose you know \(I_A\) about an axis through \(A\) and need \(I_B\) about a parallel axis through \(B\), where neither passes through \(G\). The theorem cannot jump straight from \(A\) to \(B\). Step back to \(G\) first, then out to \(B\):
Between two axes that miss G
\[ I_G = I_A - m d_A^2, \qquad I_B = I_G + m d_B^2 \]\(d_A\) and \(d_B\) are the distances from \(G\) to each axis, not the distance between \(A\) and \(B\).
Measuring \(I\) with a pendulum
For a part too complex to calculate, such as a forged connecting rod, you can measure \(I\). Hang it from a pin through \(O\) and let it swing with small amplitude, as a compound pendulum. With \(G\) a distance \(d\) below \(O\), the period is
\[ \tau = 2\pi\sqrt{\frac{I_O}{m g d}} \qquad \Longrightarrow \qquad I_O = \frac{m g d\,\tau^2}{4\pi^2}, \]and the parallel-axis theorem then gives \(I_G = I_O - m d^2\).
Example 3.4 — A connecting rod
A \(1.2\ \text{kg}\) connecting rod hangs from a knife edge through its small-end bore \(A\). Its center of mass is \(0.12\ \text{m}\) below \(A\), and it swings with a period of \(1.121\ \text{s}\). (a) Find \(I_A\) and \(I_G\). (b) The big-end bore \(B\) is \(0.20\ \text{m}\) from \(A\), on the far side of \(G\). Find \(I_B\).
Show solution
(a) From the period:
\[ I_A = \frac{(1.2)(9.81)(0.12)(1.121)^2}{4\pi^2} = 0.04497\ \text{kg·m}^2, \qquad I_G = 0.04497 - 1.2(0.12)^2 = 0.02769\ \text{kg·m}^2 \](b) \(B\) is \(0.20 - 0.12 = 0.08\ \text{m}\) from \(G\):
\[ I_B = I_G + m d_B^2 = 0.02769 + 1.2(0.08)^2 = 0.03537\ \text{kg·m}^2 \]The tempting shortcut \(I_A + m(0.20)^2 = 0.09297\ \text{kg·m}^2\) is off by more than a factor of two.
The radius-of-gyration form
Divide \(I = I_G + m d^2\) by \(m\), with \(I = m k^2\) and \(I_G = m k_G^2\):
\[ k^2 = k_G^2 + d^2 \]So radii of gyration add like the sides of a right triangle, with \(k\) as the hypotenuse. Radii of gyration do not add directly: \(k \ne k_G + d\).
Check your understanding
Key takeaways
- \(I = I_G + m d^2\), for two parallel axes, one of them through \(G\).
- The centroidal axis has the smallest \(I\) of all axes parallel to it.
- To go between two axes that miss \(G\), go through \(G\): \(I_B = I_A - m d_A^2 + m d_B^2\).
- \(k^2 = k_G^2 + d^2\). A compound pendulum measures \(I_O = m g d\,\tau^2/4\pi^2\).
- Next: Lesson 4 uses the theorem on every part of a composite body.