Stability of conical buoy with light

Data: sea-water $\rho=1025\ \text{kg/m}^3$; cone vertex down, top diameter 1.2 m, vertex angle $60^\circ$ (half-angle $30^\circ$); buoy mass 300 kg, $G_1$ at 0.75 m from the vertex; light mass 55 kg at height $y$ above the top.

      light at y
      ____
     \    /   top dia 1.2 m
      \  /
 water \/  vertex (down)

Floating depth $d$ (vertex to waterline). Total mass $M=355$ kg:

$$V=\frac{355}{1025}=0.3463\ \text{m}^3=\frac13\pi\,(d\tan30^\circ)^{2}d$$ $$d^{3}=\frac{3V}{\pi\tan^{2}30^\circ}=0.9922\ \Rightarrow\ d=0.9974\ \text{m}$$

(The buoy height is $0.6/\tan30^\circ=1.0392$ m, so it floats with the top just above water.)

Waterline radius: $r_w=d\tan30^\circ=0.5758$ m.

Centre of buoyancy: $KB=\tfrac34d=0.7480$ m from the vertex.

$$I=\frac{\pi r_w^{4}}{4}=0.0863\ \text{m}^4,\qquad BM=\frac{I}{V}=0.2493\ \text{m}$$ $$KM=KB+BM=0.7480+0.2493=0.9974\ \text{m}$$

Limit of stability: $KG\le KM$. Combined centre of gravity from the vertex:

$$KG=\frac{300(0.75)+55(1.0392+y)}{355}\le0.9974$$ $$225+55(1.0392+y)\le354.07\ \Rightarrow\ 1.0392+y\le2.3468$$ $$y\le1.308\ \text{m}$$

Answer: the centre of gravity of the light unit may be at most about 1.31 m above the top of the buoy (2.35 m from the vertex) for the assembly not to be unstable.