where $r_i$ is perpendicular distance from rotation axis.

For continuous mass distribution: $I = \int r^2 \, dm$

where $r$ is perpendicular distance from axis.

Common Moments of Inertia

Thin rod about center: $I = \frac{1}{12}ML^2$

Thin rod about end: $I = \frac{1}{3}ML^2$

Solid cylinder/disk about axis: $I = \frac{1}{2}MR^2$

Hollow cylinder about axis: $I = MR^2$

Solid sphere about diameter: $I = \frac{2}{5}MR^2$

Hollow sphere about diameter: $I = \frac{2}{3}MR^2$

Rectangular plate about center: $I = \frac{1}{12}M(a^2 + b^2)$

Calculating I by Integration

Example 1: Thin Rod About End

Rod of length $L$, mass $M$, uniform density.

Linear mass density: $\lambda = M/L$

$dm = \lambda \, dx = \frac{M}{L} \, dx$

$I = \int_0^L x^2 \, dm = \int_0^L x^2 \frac{M}{L} \, dx$

$I = \frac{M}{L}\frac{x^3}{3}\Big|_0^L = \frac{ML^2}{3}$

Example 2: Solid Cylinder About Axis

Radius $R$, mass $M$, uniform density $\rho$.

Use cylindrical shells: $dm = \rho \cdot 2\pi r \, dr \cdot h$

$I = \int_0^R r^2 \, dm = \int_0^R r^2 \cdot \rho \cdot 2\pi rh \, dr$

$I = 2\pi\rho h\int_0^R r^3 \, dr = 2\pi\rho h \frac{R^4}{4}$

Using $M = \rho\pi R^2h$:

$I = \frac{1}{2}MR^2$

Example 3: Solid Sphere About Diameter

Sphere of radius $R$, mass $M$.

Use disk method. At distance $z$ from center, disk has radius $r = \sqrt{R^2 - z^2}$.

$dI = \frac{1}{2}(dm)r^2 = \frac{1}{2}\rho\pi r^2 dz \cdot r^2 = \frac{1}{2}\rho\pi(R^2 - z^2)^2 dz$

$I = 2\int_0^R \frac{1}{2}\rho\pi(R^2 - z^2)^2 dz$

After integration: $I = \frac{2}{5}MR^2$

Parallel Axis Theorem

$I = I_{cm} + Md^2$

where:

• $I$ = moment about new axis
• $I_{cm}$ = moment about parallel axis through center of mass
• $d$ = distance between the two parallel axes

Example: Rod About End

$I_{cm} = \frac{1}{12}ML^2$ (about center)

$d = L/2$ (distance from center to end)

$I = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{3}ML^2$

Perpendicular Axis Theorem

For planar object in xy-plane: $I_z = I_x + I_y$

where axes pass through same point.

Example: Thin Disk

About axis perpendicular to disk through center: $I_z = \frac{1}{2}MR^2$

By symmetry: $I_x = I_y$

$I_x = I_y = \frac{1}{2}I_z = \frac{1}{4}MR^2$

(moment about diameter)

Composite Objects

For object composed of multiple parts: $I_{total} = \sum_i I_i$

Calculate moment of each part (using parallel axis if needed), then sum.

Example: T-Shape

Two identical rods (length $L$, mass $M$ each) forming T-shape.

Vertical rod rotating about its end (where horizontal rod attaches): $I_1 = \frac{1}{3}ML^2$

Horizontal rod about its center (perpendicular to length): $I_2 = \frac{1}{12}ML^2$

$I_{total} = \frac{1}{3}ML^2 + \frac{1}{12}ML^2 = \frac{5}{12}ML^2$

Radius of Gyration

$I = Mk^2$

where $k$ is radius of gyration.

$k$ represents the distance from axis where all mass could be concentrated to give same $I$.

$I_{center} = \frac{12}{12}$

$\boxed{I_{center} = 1.0 \text{ kg·m}^2}$

(b) Through one end:

$I_{end} = \frac{1}{3}ML^2 = \frac{1}{3}(3.0)(2.0)^2$

$\boxed{I_{end} = 4.0 \text{ kg·m}^2}$

(c) Parallel axis theorem verification:

Parallel axis theorem: $I = I_{cm} + Md^2$

Distance from center to end: $d = L/2 = 1.0$ m

$I_{end} = I_{center} + M\left(\frac{L}{2}\right)^2$

$I_{end} = 1.0 + (3.0)(1.0)^2 = 1.0 + 3.0$

$I_{end} = 4.0 \text{ kg·m}^2$ ✓

Verified!