Problem: A uniform disc of mass M = 2 kg and radius R = 0.5 m. Find I about an axis tangent to the disc and lying in its plane.

Solution (with units at every step):

Step 1: Find $I$ about a diameter (in-plane axis through centre) using the perpendicular axis theorem.

The disc is a planar body, so the perpendicular axis theorem applies: $I_z = I_x + I_y$ where $z$ is the axis through the centre perpendicular to the disc, and $x, y$ are two mutually perpendicular diameters.

By symmetry of the disc: $I_x = I_y = I_{diameter}$

$I_z = \frac{1}{2}MR^2 = \frac{1}{2}(2\,\text{kg})(0.5\,\text{m})^2 = \frac{1}{2}(2)(0.25)\,\text{kg·m}^2 = 0.25\,\text{kg·m}^2$

$\therefore\quad I_{diameter} = \frac{I_z}{2} = \frac{0.25}{2} = 0.125\,\text{kg·m}^2$

Step 2: Apply the parallel axis theorem to shift from the diameter axis to the tangent axis.

The tangent axis is parallel to the diameter axis and at perpendicular distance $d = R = 0.5\,\text{m}$ from it.

$I_{tangent} = I_{diameter} + Md^2$ $= 0.125\,\text{kg·m}^2 + (2\,\text{kg})(0.5\,\text{m})^2$ $= 0.125\,\text{kg·m}^2 + (2)(0.25)\,\text{kg·m}^2$ $= 0.125 + 0.5 = 0.625\,\text{kg·m}^2$

Dimensional check: $[M^1L^2T^0]$ = kg·$m^{2}$ ✓

General formula: $I_{tangent,\text{in-plane}} = \frac{MR^2}{4} + MR^2 = \frac{5MR^2}{4}$

Common error: Applying the perpendicular axis theorem directly to get the tangent-in-plane value without first finding the diameter value.