I've been struggling with this question, which would probably be more appropriate to ask on Physics Stack Exchange which I've already done, but am hoping for some guidance here too.
A hemispherical bowl of inner radius $r$, density $ϕ$ and thickness $t$ is floating in a fluid of density $ρ$. The bowl is being filled with sand at a constant mass-rate $α$. What is the length of time $T$ before the bowl sinks below the surface?
The bowl is in equilibrium, I drew a diagram of the bowl and all the forces acting on it. Using Archimedes' principle, there must be a buoyancy force acting on the bowl equal to the weight of the volume of water the bowl displaces. With the information given in the question, the volume of water displaced if the bowl were completely solid (not hollow) would be $\frac{2}{3}π(r+t)^3ρ$. The volume displaced by the "outer ring" of the bowl, i.e. the non-hollow part, would be $\frac{2}{3}πr^2tϕ$. Therefore, the total volume of water displaced by the bowl would be the difference between these, i.e. $\frac{2}{3}π(r+t)^3ρ - \frac{2}{3}πr^2tϕ$. I think that the bowl would sink when the weight of the sand is greater than this buoyancy force, and therefore the length of time, $T$, should be the volume displaced divided by $α$, so $\frac{2}{3α}π(r+t)^3ρ - \frac{2}{3α}πr^2tϕ$, ($g$ gets cancelled). However, this is not the correct answer so I'm not sure if there is something wrong with my logic.
The external volume of the hemisphere is
$ V_e = {\frac 2 3} \pi (r + t)^3 $.
The internal volume of the hemisphere is
$ V_i = {\frac 2 3} \pi r^3 $.
The net volume of the hemispheric bowl (i.e. the volume of its solid part) is
$ V_d = V_e - V_i = {\frac 2 3} \pi \left( (r + t)^3 - r^3 \right) = {\frac 2 3} \pi (3 r^2 + 3 r t + t^2) t $.
Now, the volume of displaced fluid is equal to the external volume of the hemisphere (!!)
$ V_w = V_e = {\frac 2 3} \pi (r + t)^3 $,
whence the weight of displaced fluid is
$ W_w = V_w * \rho = {\frac 2 3} \pi (r + t)^3 \rho $,
while the weight of the solid part of the bowl is
$ W_d = V_d * \phi = {\frac 2 3} \pi (3 r^2 + 3 r t + t^2) t \phi $.
The (minimum) weight of the sand needed to sink the bowl is the weight of the displaced fluid minus the weight of the solid part of the bowl (since this is already contributing to the sinking)
$ W_s = W_w - W_d = {\frac 2 3} \pi \left( (r + t)^3 \rho - (3 r^2 + 3 r t + t^2) t \phi \right) $.
Finally, the time it takes until the bowl sinks is the weight of the sand needed divided by the rate of filling
$ T_s = W_s / \alpha = {\frac 2 3} {\frac \pi \alpha} \left( (r + t)^3 \rho - (3 r^2 + 3 r t + t^2) t \phi \right) $.