 Solution

Relative to an origin, the inertia tensor is $I_{ij}=\int (r^2{\delta }_{ij}-r_i{r}_j)dm$, with $\mathbf{J}=\mathsf{I}\mathbf{\omega }$. Its orthogonal eigenvectors are principal axes, with eigenvalues the principal moments of inertia. Different components of $\mathbf{\omega }$ are multiplied by different eigenvalues, so angular momentum need not point along angular velocity.

For a uniform thin disc, radial integration gives $I_z=\int r^{2}dm=m{a}^2/2$. The perpendicular axis theorem and symmetry then give $I_x=I_y=I_z/2$, henceFor the elastic eccentric impact of a thin disc, neglect gravity's impulse and take the pole's impulse as vertical. Let $z$ point up and let the contact position relative to the centre be $r\hat{x}$. An upward impulse $P$ givesFor an elastic impact with a fixed pole, the contact point reverses its normal velocity, so $-u+P/m+r^{2}P/I_d=+u$. ThereforeHere $v_C>0$ denotes upward motion; the centre can still move down if $r>a/2$. Substitution givesThis also follows by combining conserved angular momentum about the fixed contact with conserved kinetic energy and selecting the separating, nonzero-impulse solution. At $r=0$ the disc simply rebounds without rotation; at $r=a/2$ its centre is instantaneously stationary. The rebound result concerns the contacting material point immediately after this ideal impact, not its subsequent contact history.