Irodov 1.239: Rings and Slabs Build the Inertia – Amit & Deepali Sinha Foundation

Problem Irodov 1.239

Find the moment of inertia:

of a uniform copper disc of thickness $b = 2.0$ mm and radius $R = 100$ mm, taken about the axis through its center and perpendicular to its flat faces;
of a uniform solid cone of mass $m$ and base radius $R$ , taken about its own axis of symmetry.

Part (a). The disc seen face on, radius R, its uniform thickness b running into the page; the axis passes through its center O, out of the page. — Part (a). The disc seen face on, radius $R$ , its uniform thickness $b$ running into the page; the axis passes through its center $O$ , out of the page.

Given

$b = 2.0$ mm
$R = 100$ mm
$\rho = 8.9\times 10^{3}$ kg/m $^{3}$ (copper)
cone of mass $m$ , base radius $R$

Find

(a)

I

of the disc about its symmetry axis, (b)

I

of the cone about its symmetry axis

Solution

Step 1The ring that all sits at one distance from the axis

The moment of inertia is $I = \int r^{2}\,dm$ , and that integral is trivial only when every part of the element shares one value of $r$ . For a disc turning about its own axis the ring between $r$ and $r+dr$ , taken through the whole thickness $b$ , is exactly such an element: unroll it and it is a strip of length $2\pi r$ , width $dr$ and depth $b$ .

The element is the shaded ring between radius r and radius r+dr, every point of it the same distance from the axis O. — The element is the shaded ring between radius $r$ and radius $r+dr$ , every point of it the same distance from the axis $O$ .

dV = 2\pi r\,dr\,b

Multiplying that volume by the uniform density turns it into the ring’s mass.

dm = \rho\,dV = 2\pi\rho b\,r\,dr

(1)

Step 2Summing the rings from the center to the rim

Every element of the ring stands at the same $r$ , so its share of the integral is $r^{2}\,dm$ with $dm$ from (1). Letting $r$ run over the whole disc gives the total.

\begin{align*}I &= \int r^{2}\,dm \\ &= \int_{0}^{R} r^{2}\,\bigl(2\pi\rho b\,r\,dr\bigr) \\ &= 2\pi\rho b\int_{0}^{R} r^{3}\,dr \\ &= 2\pi\rho b\left.\frac{r^{4}}{4}\right|_{0}^{R} \\ &= 2\pi\rho b\,\frac{R^{4}}{4}\end{align*}

Copper’s tabulated density is $\rho = 8.9\times 10^{3}$ kg/m $^{3}$ , and the disc measures $b = 2.0$ mm and $R = 100$ mm.

I = \frac{1}{2}\pi\rho b R^{4} = 2.8\times 10^{-3}\ \mathrm{kg}\cdot\mathrm{m}^{2}

(2)

Step 3The same result written with the disc’s own mass

Part (b) needs this result as a formula in the disc’s mass rather than in its dimensions, so replace the geometry by the mass it carries, $\pi\rho R^{2}b$ .

\begin{align*}I &= \frac{1}{2}\pi\rho b R^{4} \\ &= \frac{1}{2}\bigl(\pi\rho R^{2} b\bigr)R^{2}\end{align*}

I_{\mathrm{disc}} = \frac{1}{2}m_{\mathrm{disc}}R^{2}

(3)

The cone in axial section, height h and base radius R. The shaded slab a distance z below the apex is a disc of radius r and thickness dz, spinning about the cone's own axis. Similar triangles fix r from z. — The cone in axial section, height $h$ and base radius $R$ . The shaded slab a distance $z$ below the apex is a disc of radius $r$ and thickness $dz$ , spinning about the cone’s own axis. Similar triangles fix $r$ from $z$ .

Step 4The cone as a stack of discs

Slice the cone perpendicular to its axis. A slab at distance $z$ from the apex is a uniform disc of thickness $dz$ whose radius is fixed by the similar triangles in the figure, and its own moment of inertia about the cone’s axis is (3) applied to that slab.

\begin{align*}r &= \frac{R}{h}\,z \\ dm &= \rho\,\pi r^{2}\,dz = \pi\rho\,\frac{R^{2}}{h^{2}}\,z^{2}\,dz \\ dI &= \frac{1}{2}r^{2}\,dm = \frac{1}{2}\,\frac{R^{2}z^{2}}{h^{2}}\cdot\pi\rho\,\frac{R^{2}}{h^{2}}\,z^{2}\,dz \\ &= \frac{\pi\rho R^{4}}{2h^{4}}\,z^{4}\,dz\end{align*}

Step 5Integrating the stack and trading density for mass

Summing the slabs from apex to base gives the cone’s moment of inertia in terms of $\rho$ and the geometry.

\begin{align*}I &= \int_{0}^{h} \frac{\pi\rho R^{4}}{2h^{4}}\,z^{4}\,dz \\ &= \frac{\pi\rho R^{4}}{2h^{4}}\int_{0}^{h} z^{4}\,dz \\ &= \frac{\pi\rho R^{4}}{2h^{4}}\cdot\frac{h^{5}}{5} \\ &= \frac{\pi\rho R^{4}h}{10}\end{align*}

The problem states the mass, not the density, so integrate the same slabs once more without the $r^{2}$ weighting to find what $\pi\rho$ is worth.

\begin{align*}m &= \int_{0}^{h}\pi\rho\,\frac{R^{2}}{h^{2}}\,z^{2}\,dz \\ &= \pi\rho\,\frac{R^{2}}{h^{2}}\cdot\frac{h^{3}}{3} \\ &= \frac{\pi\rho R^{2}h}{3}\end{align*}

Solving that for the combination $\pi\rho$ which the inertia above contains,

\pi\rho = \frac{3m}{R^{2}h}

and substituting it back,

\begin{align*}I &= \frac{\pi\rho R^{4}h}{10} \\ &= \frac{3m}{R^{2}h}\cdot\frac{R^{4}h}{10}\end{align*}

I = \frac{3}{10}\,m R^{2}

(4)

Check

Part (a) two ways: the disc’s mass is $m_{\mathrm{disc}} = \pi\rho R^{2}b = \pi(8.9\times10^{3})(0.100)^{2}(2.0\times10^{-3}) = 0.559$ kg, and $\frac{1}{2}m_{\mathrm{disc}}R^{2} = \frac{1}{2}(0.559)(0.100)^{2} = 2.80\times10^{-3}$ kg $\cdot$ m $^{2}$ , the boxed value. Part (b) against a bound: every slab of the cone has radius $r \le R$ , with equality only at the base, so the cone’s $I$ must be strictly less than that of a cylinder of the same mass and radius $R$ , which is $\frac{1}{2}mR^{2}$ . It is: $3/10 < 1/2$ .

Remark

The cone’s answer carries no $h$ at all, so at fixed mass and base radius a squat cone and a needle-sharp one have the same moment of inertia about their axis. The reason is visible in the slab relations: $dm \propto z^{2}\,dz$ and $r \propto z$ together give $dm \propto r^{2}\,dr$ , so the fraction of the mass lying inside radius $r$ is $(r/R)^{3}$ whatever $h$ is. Stretching the cone along its axis slides mass parallel to the axis without moving any of it radially, and only radial distance enters $I$ . The claim is about the symmetry axis alone: about a transverse axis through the apex the height matters a great deal.