= Solution
Relative to an origin, the <inertia tensor> is $I_{ij}=\int(r^2\delta_{ij}-r_ir_j)\,dm$, with $\mathbf J=\mathsf I\boldsymbol\omega$. Its orthogonal <eigenvectors> are <principal axes>, with <eigenvalues> the principal moments of inertia. Different components of $\boldsymbol\omega$ are multiplied by different <eigenvalues>, so <angular momentum> need not point along <angular velocity>.
For a uniform thin disc, radial integration gives $I_z=\int r^2dm=ma^2/2$. The <perpendicular axis theorem> and <symmetry> then give $I_x=I_y=I_z/2$, hence
$$
\boxed{I_d=ma^2/4\quad\text{about a diameter}.}
$$
For 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> $P$ gives
$$
v_C=-u+P/m,\qquad \boldsymbol\omega=-\frac{rP}{I_d}\hat y.
$$
For an elastic impact with a fixed pole, the contact point reverses its normal velocity, so $-u+P/m+r^2P/I_d=+u$. Therefore
$$
\boxed{P=\frac{2mu}{1+4r^2/a^2},\quad
v_C=u\frac{a^2-4r^2}{a^2+4r^2},\quad
\boldsymbol\omega=-\frac{8ur}{a^2+4r^2}\hat y.}
$$
Here $v_C>0$ denotes upward motion; the centre can still move down if $r>a/2$. Substitution gives
$$
\boxed{v_{\rm contact}=v_C+(\boldsymbol\omega\times r\hat x)_z=u.}
$$
This 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.
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