Moment of Inertia of a Disc

Result

$I = \frac{1}{2}MR^2 \quad \text{(about central axis, perpendicular to disc)}$

Derivation

Divide the disc into thin concentric rings. A ring at radius $r$ with thickness $dr$:

• Area: $2\pi r \, dr$
• Mass: $dm = \sigma \cdot 2\pi r \, dr$ where $\sigma = \frac{M}{\pi R^2}$

$I = \int_0^R r^2 \, dm = \int_0^R r^2 \cdot \frac{2M}{R^2} r \, dr = \frac{2M}{R^2}\int_0^R r^3 \, dr = \frac{2M}{R^2} \cdot \frac{R^4}{4} = \frac{MR^2}{2}$

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

Other axes

About a diameter (perpendicular axis theorem):

$I_{diameter} = \frac{I_z}{2} = \frac{MR^2}{4}$

About tangent perpendicular to disc (parallel axis, $d = R$):

$I = \frac{MR^2}{2} + MR^2 = \frac{3MR^2}{2}$

About tangent in plane of disc (parallel axis applied to diameter, $d = R$):

$I = \frac{MR^2}{4} + MR^2 = \frac{5MR^2}{4}$

Summary for disc

Disc vs ring

The disc's MI ($\frac{1}{2}MR^2$) is half the ring's ($MR^2$) for the same mass and radius. This is because the disc's mass is distributed throughout the area — much of it closer to the axis than $R$ — reducing the average $r^2$.