The Parallel Axis Theorem

Why a Shifted Axis Changes the Moment of Inertia

A lamina’s moment of inertia depends not only on how its mass is distributed, but also on the axis about which the distribution is measured. Mass farther from an axis contributes more because its contribution is proportional to the square of its perpendicular distance. The parallel axis theorem gives a precise way to adjust a known moment of inertia when the axis is shifted.

The most useful reference axis is often a line parallel to the axis of interest that passes through the lamina’s centroid. The theorem says exactly how much the moment of inertia increases when we move from that centroidal axis to another parallel axis.

Definitions and the Theorem

Let a lamina occupy a planar region $R$, with nonnegative mass density $\rho(x,y)$. Its mass is $m=\iint_R \rho(x,y)\,dA$. For a line $L$ in the plane, the mass moment of inertia about $L$ is the integral of the square of each mass element’s perpendicular distance from $L$.

Definition: If $r$ is the perpendicular distance from a mass element $dm$ to the line $L$, then the lamina’s moment of inertia about $L$ is $I_L=\iint_R r^2\,dm$. In particular, for the vertical line $x=a$, $I_L=\iint_R (x-a)^2\rho(x,y)\,dA$.

The centroidal axis for a given direction is the line through the lamina’s centroid that is parallel to the axis being considered. For example, the centroidal axis parallel to the vertical line $x=a$ is the vertical line $x=x_c$, where $x_c$ is the centroid’s $x$-coordinate. The distance $d$ between the two axes is their shortest, perpendicular separation—not a distance measured along either line.

Theorem (Parallel Axis Theorem): Let $I_c$ be the moment of inertia of a lamina about an axis through its centroid, and let $I$ be its moment of inertia about a parallel axis a perpendicular distance $d$ away. If the lamina has finite mass $m$ and finite moments of inertia about these axes, then $$ I=I_c+md^2. $$

Here $m$ is the lamina’s total mass. For mass in kilograms and distance in metres, both terms on the right have units $\text{kg}\,\text{m}^2$, as a mass moment of inertia should.

Why the Theorem Works

Use a coordinate $u$ measured perpendicular to the parallel axes. Let the centroidal axis have coordinate $u_c$, and let the shifted axis have coordinate $u_a$. A mass element $dm$ at coordinate $u$ is at perpendicular distance $|u-u_a|$ from the shifted axis. Its squared distance is $(u-u_a)^2$.

By the definition of the centroid, the average perpendicular coordinate of the mass is $u_c$. Equivalently, the first moment of mass about the centroidal coordinate is zero:

\int (u-u_c)\,dm = \int u\,dm-u_c\int dm = mu_c-mu_c = 0.

Write the distance from a mass element to the shifted axis as the sum of its displacement from the centroidal axis and the displacement between the axes. Expanding the square then gives the result:

Express the shifted distance.
$u-u_a=(u-u_c)+(u_c-u_a)$.

Square and integrate over the mass.
The squared distance contains a centroidal term, a cross term, and a constant term.

Use the centroid property.
The cross term integrates to zero because $\int(u-u_c)\,dm=0$.

In full, with $d=|u_c-u_a|$, the calculation is

\begin{aligned} I &=\int (u-u_a)^2\,dm\\ &=\int\big((u-u_c)+(u_c-u_a)\big)^2\,dm\\ &=\int (u-u_c)^2\,dm +2(u_c-u_a)\int(u-u_c)\,dm +(u_c-u_a)^2\int dm\\ &=I_c+0+md^2\\ &=I_c+md^2. \end{aligned}

This argument explains why the centroid matters: it makes the cross term vanish. If the reference axis does not pass through the centroid, that cross term is not generally zero, so the theorem cannot be applied by treating an arbitrary axis as though it were centroidal.

Using the Theorem in Practice

To transfer a known moment of inertia, first identify the centroidal axis parallel to the desired axis. Then find the perpendicular separation $d$, square it, multiply by the total mass, and add the result to $I_c$. The distance is squared, so its sign does not matter. A shifted axis cannot have a smaller moment of inertia than the parallel centroidal axis: for $m>0$, $md^2\geq 0$.

Worked Example: Shift the Axis of a Triangular Lamina

A triangular lamina occupies $0\leq x\leq 1$, $0\leq y\leq x$, with all lengths measured in metres. Its constant density is $6\,\text{kg}/\text{m}^2$. Find its moment of inertia about the vertical axis $x=0$, using the parallel axis theorem.

First find the mass. At each $x$, the $y$-values range from $0$ to $x$, so

m=6\int_0^1\int_0^x dy\,dx =6\int_0^1 x\,dx =6\left[\frac{x^2}{2}\right]_0^1 =3\,\text{kg}.

The centroid’s $x$-coordinate is the first moment of mass about the $y$-axis divided by the total mass:

x_c=\frac{1}{m}\int x\,dm =\frac{6}{3}\int_0^1\int_0^x x\,dy\,dx =2\int_0^1 x^2\,dx =2\left[\frac{x^3}{3}\right]_0^1 =\frac{2}{3}\,\text{m}.

Thus the vertical centroidal axis is $x=2/3$, and its separation from $x=0$ is $d=2/3\,\text{m}$. Compute the centroidal moment of inertia about that vertical axis:

\begin{aligned} I_c &=6\int_0^1\int_0^x \left(x-\frac{2}{3}\right)^2dy\,dx\\ &=6\int_0^1 x\left(x^2-\frac{4}{3}x+\frac{4}{9}\right)dx\\ &=6\int_0^1\left(x^3-\frac{4}{3}x^2+\frac{4}{9}x\right)dx\\ &=6\left(\frac14-\frac49+\frac29\right) =6\left(\frac{1}{36}\right) =\frac16\,\text{kg}\,\text{m}^2. \end{aligned}

Now transfer to the axis $x=0$:

I=I_c+md^2 =\frac16+3\left(\frac23\right)^2 =\frac16+\frac43 =\frac32\,\text{kg}\,\text{m}^2.

Check by integrating directly about $x=0$:

I_{x=0} =6\int_0^1\int_0^x x^2\,dy\,dx =6\int_0^1 x^3\,dx =6\left(\frac14\right) =\frac32\,\text{kg}\,\text{m}^2.

The direct calculation agrees with the transferred value.

Worked Example: Transfer a Given Centroidal Moment

A lamina has mass $2.4\,\text{kg}$ and centroidal moment of inertia $0.18\,\text{kg}\,\text{m}^2$ about a specified axis. Find its moment of inertia about a parallel axis $0.25\,\text{m}$ away.

Square the separation and multiply by the mass:

d^2=(0.25\,\text{m})^2=0.0625\,\text{m}^2, \qquad md^2=(2.4\,\text{kg})(0.0625\,\text{m}^2) =0.15\,\text{kg}\,\text{m}^2.

Add this shift contribution to the centroidal moment:

I=I_c+md^2 =0.18+0.15 =0.33\,\text{kg}\,\text{m}^2.

Check the difference: $I-I_c=0.33-0.18=0.15\,\text{kg}\,\text{m}^2$, which equals $md^2$. The transferred value is consistent.

Worked Example: Find the Separation Between Axes

A lamina of mass $5\,\text{kg}$ has centroidal moment of inertia $0.72\,\text{kg}\,\text{m}^2$. A parallel axis has moment of inertia $2.52\,\text{kg}\,\text{m}^2$. Find the distance between the axes.

Rearrange $I=I_c+md^2$ to isolate $d^2$, then substitute:

d^2=\frac{I-I_c}{m} =\frac{2.52-0.72}{5} =\frac{1.80}{5} =0.36\,\text{m}^2.

Since a distance is nonnegative, take the positive square root:

d=\sqrt{0.36\,\text{m}^2} =0.60\,\text{m}.

Check by transferring the centroidal moment to an axis $0.60\,\text{m}$ away:

I=0.72+5(0.60)^2 =0.72+5(0.36) =0.72+1.80 =2.52\,\text{kg}\,\text{m}^2.

This reproduces the stated moment of inertia.

A Common Mistake: Forgetting to Square the Distance

In the triangular example, the separation was $2/3\,\text{m}$, the mass was $3\,\text{kg}$, and the centroidal moment was $1/6\,\text{kg}\,\text{m}^2$. A common incorrect calculation is to add mass times distance without squaring:

I_{\text{wrong}} =\frac16+3\left(\frac23\right) =\frac16+2 =\frac{13}{6}.

This is wrong for two reasons. The theorem requires $d^2$, not $d$, and the added term $md$ has units $\text{kg}\,\text{m}$, not $\text{kg}\,\text{m}^2$. The correction is

I=\frac16+3\left(\frac23\right)^2 =\frac32\,\text{kg}\,\text{m}^2.

Always square the perpendicular separation, and check that every term has the same units.

What the Theorem Does—and Does Not—Change

The theorem changes the axis, not the lamina. Its $I_c$ must be the moment of inertia about an axis through the centroid and parallel to the target axis; $d$ is the shortest distance between those two axes. It does not supply the centroidal moment itself. That value must be known or computed separately.

The same reasoning applies if the moment is defined using squared distance from an axis perpendicular to the lamina. In that case, the theorem relates the moments about two parallel perpendicular axes, one through the centroid and one displaced from it. In either setting, the essential points are the same: use the total mass, use a centroidal reference axis, and square the separation.

This theorem is useful because an inertia integral about a convenient centroidal axis may be simpler than one about a displaced axis. Once that centroidal value is available, the formula transfers it without repeating the full integral. The increase $md^2$ also makes the effect of axis placement explicit: moving the axis farther from the centroid increases the moment quadratically.

Check Your Understanding

Use the definitions and theorem above to answer these questions.

A lamina has mass $4\,\text{kg}$ and centroidal moment $0.5\,\text{kg}\,\text{m}^2$. What is its moment about a parallel axis $0.3\,\text{m}$ away?
What must be true of the reference axis for $I_c$ in the parallel axis theorem?
Why does the cross term vanish in the derivation?
A parallel-axis moment is $1.4\,\text{kg}\,\text{m}^2$, while $I_c=0.5\,\text{kg}\,\text{m}^2$ and $m=4\,\text{kg}$. What is the axis separation?
Why is using $md$ instead of $md^2$ incorrect?

What You'll Learn