Does a cone in $mathbb{R}^3$ admit a warped geometry?

Here is an answer to the title "does the cone admit a warped product metric" and the answer is yes:

One natural wraped product on $C$ will be induced by a spherical section. In $mathbb{R}^3$, the polar decomposition gives a natural metric on $mathbb{R}^3 setminus {0}$ which is $$ mathrm{d}r^2 + r^2 overset{circ}{g} $$ where $overset{circ}{g}$ is the round metric of the unit sphere. It gives an isometry between $(0,infty)times mathbb{S}^2$ and $mathbb{R}^3setminus{0}$.

As $C$ is a natural riemannian submanifold of $mathbb{R}^3 setminus {0}$, in polar coordinates, it is isometric to $(0,infty) times ({Ccapmathbb{S}^2})$. Note that $Ccap mathbb{S}^2$ is a "round" disk. Thus, one can find a diffeomorphism $f:D to Ccap mathbb{S}^2$ and pull-back the round metric on $D$.

One then has an isometry $(0,infty) times D to C$, in which metric is the warped metric $$ mathrm{d}r^2 + r^2 f^*left(overset{circ}{g}|_{Ccap mathbb{S}^2} right) $$ So one has a wraped metric on $(0,infty)times D$. But indeed, the metric on $D$ here is not flat (it has sectionnal curvature $1$).

Now, here is a remark, more than an answer, to the question "can we give a product warped metric that involves the flat metric on the cone":

Note that if $mathrm{d}r^2 + f(r)g_D$ is a warped product of the cone involving the flat metric, in these coordinates $(r,p) in (0,infty)times D$, the level surfaces ${r_0}times D$ are orthogonal to the geodesics $(0,infty)times{p_0}$. Then, the $"r"$ coordinate will not be the distance to the origin in $mathbb{R}^3$. Otherwise, these radial geodesics in $mathbb{R}^3$ will be orthogonal to a same flat disk, which is not true. This shows that there exists in the cone a flat disk orthogonal to a familly geodesics in $mathbb{R}^3$, and the cone would be foliated by a family of geodesic rays all orthogonal to the same flat disk. This gives me the intuition that the answer will be "no", but I do not have a formal proof for this right now, and I may be wrong.