Case Study 1 — Balancing and Spinning a Machined Flywheel

Field: Mechanical engineering (rotational dynamics, structural design)

A second-year mechanical engineer named Priya is handed a deceptively simple task: a small electric go-kart's regenerative braking system needs a flywheel, a spinning disk that stores energy when the kart slows and releases it on acceleration. Her supervisor gives her two numbers to deliver. First, the flywheel must balance perfectly on its central shaft, or it will vibrate itself — and the bearings — to destruction at 8,000 rpm. Second, it must have a specified moment of inertia, because that single number determines how much rotational energy the wheel banks for a given spin rate. Both numbers, Priya realizes as she opens her calculus notes, are multiple integrals over the wheel's cross-section. This case study walks through her reasoning from blank sheet to delivered specification.

The part and its density

The flywheel is a flat steel disk of radius $R = 0.10$ m, but it is not uniform. To save weight near the rim — where centripetal stress is worst — the manufacturing process leaves the steel slightly less dense toward the edge, while the hub region is forged solid. Priya models the area density (mass per unit area of the flat face, after accounting for the disk's thickness) as

$$\sigma(r) = \sigma_0\left(1 - \tfrac{1}{2}\cdot\frac{r^2}{R^2}\right),$$

where $r = \sqrt{x^2+y^2}$ is the distance from the center and $\sigma_0$ is the density at the hub. At the rim ($r = R$) the density has fallen to half of $\sigma_0$. The disk is circularly symmetric, so polar coordinates (§32.5) are the obvious language; everything will reduce to integrals in $r$ and $\theta$ with the area element $dA = r\,dr\,d\theta$.

Step 1 — Total mass

Mass is the integral of density (§32.10), $M = \iint_R \sigma\,dA$. In polar coordinates, with the disk described by $0 \le r \le R$, $0 \le \theta \le 2\pi$:

$$M = \int_0^{2\pi}\!\!\int_0^R \sigma_0\left(1 - \frac{r^2}{2R^2}\right) r\,dr\,d\theta.$$

The $\theta$ integral is trivial — the integrand has no $\theta$ dependence — contributing a factor of $2\pi$. The radial integral is the one that matters:

$$\int_0^R \left(r - \frac{r^3}{2R^2}\right)dr = \left[\frac{r^2}{2} - \frac{r^4}{8R^2}\right]_0^R = \frac{R^2}{2} - \frac{R^4}{8R^2} = \frac{R^2}{2} - \frac{R^2}{8} = \frac{3R^2}{8}.$$

So $M = 2\pi\sigma_0 \cdot \tfrac{3R^2}{8} = \tfrac{3\pi}{4}\sigma_0 R^2$. As a sanity check, a uniform disk of density $\sigma_0$ would have mass $\sigma_0\pi R^2$; Priya's lightened disk comes in at $\tfrac34$ of that, exactly as expected from shaving density off the outer region. The factor $r$ in the area element is doing real work here: because patches far from the center are larger, the density reduction at large $r$ removes proportionally more mass than a naive average over $r$ would suggest.

Step 2 — The balance condition

For the wheel to balance on its central shaft, its center of mass must sit exactly at the geometric center. Priya could grind through $\bar x = \frac{1}{M}\iint x\,\sigma\,dA$, but she pauses to think about symmetry first — the habit the textbook drilled into her. The density $\sigma(r)$ depends only on the distance $r$, not on direction, so the mass distribution is unchanged by any rotation about the center. A rotationally symmetric distribution can only balance at its center; any off-center balance point would single out a direction, which the symmetry forbids.

To confirm, she sets up $\bar x$ explicitly. In polar coordinates $x = r\cos\theta$, so

$$M\bar x = \int_0^{2\pi}\!\!\int_0^R (r\cos\theta)\,\sigma(r)\,r\,dr\,d\theta = \left(\int_0^{2\pi}\cos\theta\,d\theta\right)\left(\int_0^R r^2\,\sigma(r)\,dr\right).$$

The integrand factored cleanly into an angular part and a radial part. And the angular part is $\int_0^{2\pi}\cos\theta\,d\theta = [\sin\theta]_0^{2\pi} = 0$. The whole moment vanishes, so $\bar x = 0$, and by the identical argument with $\sin\theta$, $\bar y = 0$. The wheel balances at its center — guaranteed by symmetry, confirmed by the integral. This is the kind of result where doing the computation is less about getting a number and more about certifying that the number is what physics demanded. Priya notes for her report that any future design with an asymmetric lightening pattern — say, drilled holes on one side — would break this and require a balancing counterweight, sized by exactly the nonzero moment integral.

Step 3 — Moment of inertia

The energy-storage spec is governed by the moment of inertia about the spin axis (the $z$-axis through the center). From §32.10,

$$I_z = \iint_R (x^2 + y^2)\,\sigma\,dA = \iint_R r^2\,\sigma(r)\,dA,$$

since $x^2 + y^2 = r^2$. Each ring of material contributes its mass times the square of its distance from the axis — distant mass counts far more, which is why flywheels concentrate material at the rim when they can. In polar coordinates:

$$I_z = \int_0^{2\pi}\!\!\int_0^R r^2\,\sigma_0\left(1 - \frac{r^2}{2R^2}\right) r\,dr\,d\theta = 2\pi\sigma_0\int_0^R \left(r^3 - \frac{r^5}{2R^2}\right)dr.$$

Evaluate the radial integral:

$$\int_0^R \left(r^3 - \frac{r^5}{2R^2}\right)dr = \left[\frac{r^4}{4} - \frac{r^6}{12R^2}\right]_0^R = \frac{R^4}{4} - \frac{R^6}{12R^2} = \frac{R^4}{4} - \frac{R^4}{12} = \frac{R^4}{6}.$$

Therefore $I_z = 2\pi\sigma_0\cdot\tfrac{R^4}{6} = \tfrac{\pi}{3}\sigma_0 R^4$. Priya now expresses this in terms of the mass she found in Step 1, which is what a downstream engineer actually wants. Since $\sigma_0 = \frac{4M}{3\pi R^2}$, substitution gives

$$I_z = \frac{\pi}{3}\cdot\frac{4M}{3\pi R^2}\cdot R^4 = \frac{4}{9}M R^2 \approx 0.444\,M R^2.$$

For comparison, a uniform disk has the textbook value $I_z = \tfrac12 M R^2 = 0.5\,MR^2$. Priya's lightened wheel has a smaller coefficient ($0.444$ versus $0.5$), because she removed mass from precisely the high-leverage outer region where the $r^2$ weighting bites hardest. That is the engineering tension in one line: lightening the rim saves weight but also lowers $I_z$, so she must check that $0.444\,MR^2$ still meets the energy spec $E = \tfrac12 I_z\omega^2$ at the design spin rate.

Closing the loop

With three integrals — mass, balance moment, and moment of inertia — Priya has fully characterized the part. Plugging $R = 0.10$ m and the measured $\sigma_0$ into $M = \tfrac34\sigma_0\pi R^2$ gives a concrete mass; the balance integral certifies zero counterweight is needed; and $I_z = 0.444\,MR^2$ feeds straight into the kinetic-energy budget. Every number on her specification sheet is a double integral evaluated in polar coordinates, and every one carried a physical sanity check that caught the kind of factor-of-$r$ slip that derails real designs.

Discussion Questions

  1. In Step 2, the balance moment vanished because $\int_0^{2\pi}\cos\theta\,d\theta = 0$. Suppose the disk had a single lightening hole drilled at $(x,y) = (R/2, 0)$. Sketch how you would compute the resulting nonzero $\bar x$ using additivity over regions (§32.1) — integrate the full disk minus the hole — and explain why a counterweight would be needed.

  2. Priya's lightened wheel has $I_z = 0.444\,MR^2$ versus the uniform disk's $0.5\,MR^2$. If instead the design added density toward the rim, would the coefficient rise above or below $0.5$? Argue from the $r^2$ weighting in the integrand, without recomputing.

  3. The moment of inertia weights mass by squared distance, while the center of mass weights by first power of distance. How does this difference explain why a figure skater spins faster when pulling her arms in (a fact the chapter mentions in §32.10)?

  4. Why were polar coordinates the right choice here, and what specific feature of the region and of the density function signaled it? (See the field guide in §32.9.)

Annotated Reading

  • Stewart, Calculus: Early Transcendentals, §15.4 (Applications of Double Integrals). The mass, moment, and moment-of-inertia formulas used throughout this case study, with several worked plate examples. Read this alongside §32.10.
  • Hibbeler, Engineering Mechanics: Dynamics, chapter on "Mass Moments of Inertia." Connects the integral $I_z = \iint r^2\,\sigma\,dA$ to the rotational form of Newton's second law $\tau = I\alpha$ and to kinetic energy $\tfrac12 I\omega^2$ — the engineering payoff of the number Priya computed.
  • OpenStax, Calculus Volume 3, §5.6 (Calculating Centers of Mass and Moments of Inertia). A free, careful treatment with both Cartesian and polar setups; good for a second pass on the balance argument.