Relative to an origin, the inertia tensor is , with . Its orthogonal eigenvectors are principal axes, with eigenvalues the principal moments of inertia. Different components of are multiplied by different eigenvalues, so angular momentum need not point along angular velocity.
For a uniform thin disc, radial integration gives . The perpendicular axis theorem and symmetry then give , hence
For the elastic eccentric impact of a thin disc, neglect gravity's impulse and take the pole's impulse as vertical. Let point up and let the contact position relative to the centre be . An upward impulse gives
For an elastic impact with a fixed pole, the contact point reverses its normal velocity, so . Therefore
Here denotes upward motion; the centre can still move down if . Substitution gives
This also follows by combining conserved angular momentum about the fixed contact with conserved kinetic energy and selecting the separating, nonzero-impulse solution. At the disc simply rebounds without rotation; at its centre is instantaneously stationary. The rebound result concerns the contacting material point immediately after this ideal impact, not its subsequent contact history.
For an elastic eccentric impact of a thin disc, take upward velocity as positive and let be the short upward impulse. The disc's moment of inertia about a horizontal diameter is . With impact offset ,
The contact point has upward velocity . Unit restitution against the fixed post requires that velocity to be , so
Since , and the centre's instantaneous speed is zero. The disc rotates with ; its rotational kinetic energy equals the entire initial energy. Gravity's impulse during the impact is neglected. Using the polar-axis moment of inertia would describe the wrong rotation axis.