Irodov 1.240: Two Flat Axes Make the Third – Amit & Deepali Sinha Foundation

Problem Irodov 1.240

A thin plate of arbitrary shape lies flat. Three mutually perpendicular axes, labeled 1, 2 and 3, all pass through one and the same point in the plate’s plane, with axes 1 and 2 lying in the plane of the plate and axis 3 therefore perpendicular to it. Let $I_1$ , $I_2$ and $I_3$ be the plate’s moments of inertia about those three axes. Show that $I_1 + I_2 = I_3$ . Then use that relation to find the moment of inertia of a thin uniform circular disc of radius $R$ and mass $m$ about an axis lying along one of its diameters.

A thin plate of arbitrary shape, with axes 1 and 2 in the plate through O and axis 3 leaving the page there. — A thin plate of arbitrary shape, with axes 1 and 2 in the plate through $O$ and axis 3 leaving the page there.

Given

a thin plate of arbitrary shape
axes $1$ , $2$ in its plane and axis $3$ normal to it, all through one point
a uniform disc of mass $m$ and radius $R$

Find

a demonstration that

I_1+I_2=I_3

, then

I

of the disc about a diameter

Solution

Step 1The three integrals a single element contributes to

Put the plate in the plane of axes 1 and 2 and measure a mass element by its coordinate $x$ along axis 1 and $y$ along axis 2. Because the plate is thin, every element lies in that plane, so its distance from each axis is the plain leg or hypotenuse of the right triangle drawn in the figure, and each moment of inertia is the integral of that distance squared over the mass, with the distance read straight off the triangle.

A mass element dm of the plate. The dashed lines give its distances: y from axis 1, x from axis 2, r from axis 3. — A mass element $dm$ of the plate. The dashed lines give its distances: $y$ from axis 1, $x$ from axis 2, $r$ from axis 3.

\begin{align*}I_1 &= \int y^{2}\,dm \\ I_2 &= \int x^{2}\,dm \\ I_3 &= \int \bigl(x^{2}+y^{2}\bigr)\,dm\end{align*}

Step 2Adding the two in-plane moments

The integrand of $I_3$ is the sum of the other two integrands, and integration is linear, so the two in-plane moments add to the perpendicular one with nothing left over.

\begin{align*}I_1 + I_2 &= \int y^{2}\,dm + \int x^{2}\,dm \\ &= \int\bigl(x^{2}+y^{2}\bigr)\,dm\end{align*}

I_1 + I_2 = I_3

(1)

The uniform disc seen face on. The shaded ring between radius r and radius r+dr has every one of its points at the same distance from the central axis O. — The uniform disc seen face on. The shaded ring between radius $r$ and radius $r+dr$ has every one of its points at the same distance from the central axis $O$ .

Step 3The disc about its own perpendicular axis

Before the relation can be used, the disc’s moment about axis 3 is needed. Take the ring element of the figure, with the disc’s surface density $m/\pi R^{2}$ , and integrate from the center to the rim.

\begin{align*}dm &= \frac{m}{\pi R^{2}}\,2\pi r\,dr = \frac{2m}{R^{2}}\,r\,dr \\ I_3 &= \int_{0}^{R} r^{2}\,\frac{2m}{R^{2}}\,r\,dr \\ &= \frac{2m}{R^{2}}\int_{0}^{R} r^{3}\,dr \\ &= \frac{2m}{R^{2}}\cdot\frac{R^{4}}{4}\end{align*}

I_3 = \frac{1}{2}mR^{2}

(2)

The same disc with axes 1 and 2 along two perpendicular diameters and axis 3 out of the page at the center. A quarter turn about axis 3 carries axis 1 onto axis 2 and leaves the disc unchanged.

Step 4Symmetry splits the perpendicular moment in half

Take axes 1 and 2 along two perpendicular diameters. Rotating the disc a quarter turn about axis 3 maps one of them onto the other and maps the disc onto itself, so the two moments are equal; call each of them $I$ . Then (1) has only one unknown in it, and (2) supplies the right-hand side.

\begin{align*}I_1 + I_2 &= I_3 \\ 2I &= \frac{1}{2}mR^{2}\end{align*}

I = \frac{1}{4}mR^{2}

(3)

Check

Test the relation on a body whose three moments are known separately: a thin uniform square plate of side $a$ and mass $m$ has $ma^{2}/12$ about each in-plane central axis parallel to a side, and $ma^{2}/6$ about the perpendicular central axis, and indeed $ma^{2}/12+ma^{2}/12 = ma^{2}/6$ . For the disc itself, every element is nearer to a diameter than to the center, since $|y| \le \sqrt{x^{2}+y^{2}}$ , so $I$ about a diameter must come out below $I_3$ : it does, $\frac{1}{4}mR^{2}$ against $\frac{1}{2}mR^{2}$ .

Remark

The thinness hypothesis is not decoration, and dropping it turns the equality into an inequality that holds for every rigid body. With a third coordinate $z$ measured along axis 3, the distances squared from the three axes are $y^{2}+z^{2}$ , $x^{2}+z^{2}$ and $x^{2}+y^{2}$ , so $I_1+I_2-I_3 = 2\int z^{2}\,dm \ge 0$ . No moment of inertia about any axis can exceed the sum of the moments about two axes perpendicular to it and to each other, and equality holds precisely when all the mass sits in the plane of those two axes, which is the plate of this problem. It is the moment-of-inertia analogue of a triangle inequality, and it rules out triples such as $(1,1,3)$ as principal moments of any real body.