= Solution
Measure the axial coordinate $z$ from the <vertex>. The slice radius is $z\tan\alpha$ and its <mass> is $dm=3mz^2H^{-3}\,dz$, since cross-sectional area is proportional to $z^2$. Its axial moment is $\tfrac12(z\tan\alpha)^2dm$. Therefore
$$
\boxed{C_O=\frac{3mH^2}{10}\tan^2\alpha.}
$$
For an axis through the <vertex> perpendicular to the <cone> axis, a thin disc contributes $[z^2+\tfrac14z^2\tan^2\alpha]dm$: the second term is its diametral moment. Integration gives the two equal transverse principal moments
$$
\boxed{A_O=B_O=\frac{3mH^2}{5}\left(1+\frac{\tan^2\alpha}{4}\right).}
$$
Rotational symmetry eliminates the mixed moments and makes these the principal axes. The <centre of mass> is at $3H/4$, so the parallel-axis theorem would give $A_G=3mH^2/80+3mH^2\tan^2\alpha/20$ and $C_G=C_O$.
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