Irodov 1.238: Where the One Third Comes From – Amit & Deepali Sinha Foundation

Problem Irodov 1.238

Find the moment of inertia of each of the following uniform bodies:

a thin rod of mass $m$ and length $l$ , about an axis perpendicular to the rod and passing through one of its ends;
a thin rectangular plate of mass $m$ with sides $a$ and $b$ , about an axis perpendicular to the plane of the plate and passing through one of its vertices.

Part (a). The rod of mass m and length l, with the axis of rotation through its end O drawn as the gray dashed line perpendicular to it. — Part (a). The rod of mass $m$ and length $l$ , with the axis of rotation through its end $O$ drawn as the gray dashed line perpendicular to it.

Given

(a) $m$ , $l$
(b) $m$ , $a$ , $b$
both bodies uniform

Find

(a)

I

about a perpendicular axis through an end, (b)

I

about a perpendicular axis through a vertex

Solution

Step 1The rod, sliced along its length

The moment of inertia is the sum of $r^2\,dm$ over the body, so the only work is to say what $dm$ is and how far it sits from the axis. A uniform rod carries mass in proportion to length, so a slice of length $dx$ holds the fraction $dx/l$ of the total, and every point of that slice is the same distance $x$ from the axis at the end.

The shaded slice is the element: a piece of length dx whose distance from the axis through O is x. — The shaded slice is the element: a piece of length $dx$ whose distance from the axis through $O$ is $x$ .

dm = \frac{m}{l}\,dx

dI = x^2\,dm = \frac{m}{l}\,x^2\,dx

(1)

Step 2Add the slices from one end to the other

Summing means integrating $x$ across the whole rod, from the end that carries the axis to the far end.

\begin{align*}I &= \int_0^l \frac{m}{l}\,x^2\,dx \\ &= \frac{m}{l}\left[\frac{x^3}{3}\right]_0^l \\ &= \frac{m}{l}\cdot\frac{l^3}{3}\end{align*}

I_{\mathrm{rod}} = \frac{1}{3}\,m l^2

(2)

Step 3The plate, cut into little rectangles

Now the mass is spread over an area, so the element is a small rectangle $dx$ by $dy$ , and it holds the fraction $dx\,dy/(ab)$ of the plate. The new feature is the distance: the axis is perpendicular to the plate at the vertex, so what enters is the distance from the vertex within the plane, and that is the hypotenuse of the two coordinates.

Part (b). The plate, with the rotation axis out of the page through the vertex O, marked by the circled dot. The shaded element has sides dx and dy, at coordinates x and y, a distance r from the axis. — Part (b). The plate, with the rotation axis out of the page through the vertex $O$ , marked by the circled dot. The shaded element has sides $dx$ and $dy$ , at coordinates $x$ and $y$ , a distance $r$ from the axis.

dm = \frac{m}{ab}\,dx\,dy , \qquad r^2 = x^2 + y^2

dI = (x^2+y^2)\,\frac{m}{ab}\,dx\,dy

(3)

Step 4Integrate over the whole rectangle

The element must sweep the whole plate, so $y$ runs across the side $b$ for each fixed $x$ , and then $x$ runs along the side $a$ . Doing the inner integral first leaves an ordinary integral in $x$ .

\begin{align*}I &= \frac{m}{ab}\int_0^a\!\!\int_0^b (x^2+y^2)\,dy\,dx \\ &= \frac{m}{ab}\int_0^a\left[x^2 y + \frac{y^3}{3}\right]_0^b dx \\ &= \frac{m}{ab}\int_0^a\left(x^2 b + \frac{b^3}{3}\right) dx \\ &= \frac{m}{ab}\left[\frac{b x^3}{3} + \frac{b^3 x}{3}\right]_0^a \\ &= \frac{m}{ab}\left(\frac{a^3 b}{3} + \frac{a b^3}{3}\right)\end{align*}

I_{\mathrm{plate}} = \frac{1}{3}\,m\left(a^2+b^2\right)

(4)

Check

The rod result can be tested against the same integral taken about the center, where the limits run from $-l/2$ to $l/2$ and give $ml^2/12$ ; adding $m(l/2)^2$ for the shift of the axis out to the end, as the parallel axis theorem requires, returns $ml^2/12 + ml^2/4 = ml^2/3$ , the boxed value. The plate result passes two tests. Squashing it, $b\to0$ , leaves $ma^2/3$ , which is the rod of part (a) as it must be. And its two terms are separately the moments of the plate about the two edges through the vertex, $mb^2/3$ and $ma^2/3$ , each of which is the part (a) integral again with the other coordinate merely along for the ride, so the answer is their sum exactly as the perpendicular axis theorem says.

Remark

The factor $1/3$ is not a property of rods or of plates but of the integral $\int_0^L x^2\,dx = L^3/3$ against a mass proportional to $L$ : uniform mass measured from one end always has mean square distance $L^2/3$ , and measured from the center it is $L^2/12$ , four times smaller, which is the whole difference between the two standard rod formulas. There is also a small surprise in the plate. Since $a^2+b^2$ is the square of the diagonal drawn from the vertex, the plate about its corner has exactly the moment of inertia of a rod of that diagonal’s length about its end, of the same mass. The two bodies have nothing else in common; the coincidence is that both integrals are governed by the same mean square distance from the axis.