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 $I$.

Thin spherical shell: $I = \frac{2}{3}MR^2$

$a_{shell} = \frac{g\sin\theta}{1 + 2/3} = \frac{g\sin\theta}{5/3} = \frac{3g\sin\theta}{5}$

$a_{shell} = \frac{3(9.8)\sin(30°)}{5} = \frac{3(9.8)(0.5)}{5}$

$\boxed{a_{shell} = 2.94 \text{ m/s}^2}$

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

$a_{sphere} = \frac{g\sin\theta}{1 + 2/5} = \frac{g\sin\theta}{7/5} = \frac{5g\sin\theta}{7}$

$a_{sphere} = \frac{5(9.8)(0.5)}{7}$

$\boxed{a_{sphere} = 3.50 \text{ m/s}^2}$

(b) Which reaches bottom first?

Since $a_{sphere} > a_{shell}$:

$\boxed{\text{Solid sphere reaches bottom first}}$

(Less rotational inertia = faster)

(c) Ratio of speeds:

For same distance L down incline: $v^2 = 2aL$

$\frac{v_{sphere}}{v_{shell}} = \sqrt{\frac{a_{sphere}}{a_{shell}}} = \sqrt{\frac{3.50}{2.94}}$

$\boxed{\frac{v_{sphere}}{v_{shell}} = 1.09}$

Sphere is 9% faster!

Mass of shell: $dm = \rho dV = \rho \cdot 2\pi r h \, dr$

where density $\rho = \frac{M}{\pi R^2 h}$

Moment of inertia contribution: $dI = r^2 \, dm = r^2 \cdot \rho \cdot 2\pi r h \, dr = 2\pi\rho h r^3 \, dr$

Total: $I = \int_0^R 2\pi\rho h r^3 \, dr = 2\pi\rho h \left[\frac{r^4}{4}\right]_0^R$

$I = 2\pi\rho h \cdot \frac{R^4}{4} = \frac{\pi\rho h R^4}{2}$

Substitute $\rho = \frac{M}{\pi R^2 h}$:

$I = \frac{\pi R^4 h}{2} \cdot \frac{M}{\pi R^2 h} = \frac{MR^2}{2}$

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

Numerical calculation:

$I = \frac{1}{2}(4.0)(0.2)^2 = \frac{1}{2}(4.0)(0.04)$

$\boxed{I = 0.08 \text{ kg·m}^2}$