Variable-Density Flywheel Rotational Inertia (Aerospace & Mechanical Engineering) Solution

An aerospace guidance, navigation, and control (GNC) engineer at the UNLV Howard R. Hughes College of Engineering is designing an energy-storage reaction flywheel for a low-Earth-orbit (LEO) small satellite.

The flywheel is constructed as a flat circular disk of outer radius $R = 20\text{ cm}$. To maximize rotational kinetic energy while minimizing center-of-mass balancing jitter, the disk is fabricated using a radially graded carbon nanotube matrix whose planar mass density increases quadratically toward the outer rim. In Cartesian coordinates, the area mass density distribution is modeled as:

over the disk $x^2 + y^2 \le 400$.

Step-by-Step Solution:

1. Express Disk Boundary in Polar Coordinates:
   State the Circle Boundary Equation in Polar Form:

   $$x^2 + y^2 \le R^2 \implies r^2 \le 20^2 \implies 0 \le r \le 20\text{ cm}$$
   Because the flywheel is a full circular disk, the angle sweeps through the complete revolution:
   $$0 \le \theta \le 2\pi\text{ rad}$$
   Thus, the integration domain is $D = \{(r, \theta) \mid 0 \le r \le 20, \, 0 \le \theta \le 2\pi\}$.

2. Transform Integrand and Area Differential:
   State the Polar Coordinate Substitution Relations:

   $$x^2 + y^2 = r^2, \quad \rho(r) = 3r^2\text{ g/cm}^2$$
   State the Polar Area Element Formula with Jacobian:
   $$dA = r \, dr \, d\theta$$
   Assemble the full differential moment of inertia element:
   $$dI_z = r^2 \cdot \rho(r) \, dA = (r^2)(3r^2)(r \, dr \, d\theta) = 3r^5 \, dr \, d\theta$$
   Formulate the double integral:
   $$I_z = \iint_D 3r^5 \, dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{20} 3r^5 \, dr \, d\theta$$

3. Evaluate via Fubini's Product Separability:
   Apply Fubini's Separability Theorem in Polar Coordinates (defined in Separability in Polar Integrals):

   $$I_z = \left( \int_{0}^{2\pi} d\theta \right) \left( \int_{0}^{20} 3r^5 \, dr \right)$$

   Evaluate the radial integral:

   $$\int_{0}^{20} 3r^5 \, dr = \left[ \frac{3r^6}{6} \right]_{0}^{20} = \left[ \frac{r^6}{2} \right]_{0}^{20} = \frac{(20)^6}{2} - 0 = \frac{64,000,000}{2} = 32,000,000\text{ g}\cdot\text{cm}^2$$

   Evaluate the angular integral:

   $$\int_{0}^{2\pi} d\theta = [\theta]_{0}^{2\pi} = 2\pi\text{ rad}$$

   Multiply and convert units:

   $$I_z = 2\pi \times 32,000,000 = 64,000,000\pi\text{ g}\cdot\text{cm}^2 \approx 201,061,929.8\text{ g}\cdot\text{cm}^2$$
   Convert to standard SI units ($1\text{ kg} = 1000\text{ g}$ and $1\text{ m}^2 = 10,000\text{ cm}^2 \implies 1\text{ kg}\cdot\text{m}^2 = 10^7\text{ g}\cdot\text{cm}^2$):
   $$I_z = \frac{64\pi \times 10^6}{10^7}\text{ kg}\cdot\text{m}^2 = 6.4\pi\text{ kg}\cdot\text{m}^2 \approx 20.106\text{ kg}\cdot\text{m}^2$$

Meaning of the Answer & Real-Life Application:

• Quantitative Interpretation: The polar mass moment of inertia about the spin axis is exactly $6.4\pi\text{ kg}\cdot\text{m}^2 \approx 20.11\text{ kg}\cdot\text{m}^2$.
• Engineering Application: In satellite attitude control systems, reaction wheels generate control torques by accelerating or decelerating spinning disks according to Newton's rotational law $\tau = I_z \alpha$. Storing kinetic energy ($E_k = \frac{1}{2}I_z \omega^2$) and angular momentum ($L = I_z \omega$) is vastly more mass-efficient when material is concentrated radially outward ($r^5$ scaling in the integral) rather than uniformly distributed. This enables high slew rates for telescope pointing and earth observation while minimizing overall launch payload weight.