A wooden stick of length `L`, radius `R` and density `rho` has a small metal piece of mass `m` ( of negligible volume) attached to its one end. Find the minimum value for the mass `m` (in terms of given parameters) that would make the stick float vertically in equilibrium in a liquid of density `sigma(gtrho)`.

Answer

`m ge pir^(2)L(sqrt(rhosigma - rho)`

Explanation

Let `M =` Mass of stick `= piR^(2)rhoL`
`l =` immersed length of the rod
`G =` COM of rod
`B =` Centre of buoyant force `(F)`
`C =` COM of rod `+` mass `(m)`
`Y_(atm) =` Distance of `C` from bottom of the end
Mass `m` should be attached to the lower end because otherwise `B` will be below `G` and `C` will be above `G` and the forque of the couple of two equal and opposite force `F` and `(M + m) g` will be counter clockwise on diplacing the rotational equalibrium. See the figure `3` given alongwide.
For vertical equilibrium
`Mg + mg = F` (upthrust)
or `(piR^(2)Lrhog) + mg = (piR^(2)l) sigma g`
`:. l = ((piR^(2)Lrho + m)/(piR^(2)sigma)).....(1)`
Position of `COM` (of rod `+` m) from bottom
`Y_(com) = (M(L)/(2))/(M + m) = ((piR^(2)Lrho)(L)/(2))/((piR^(2)Lrho)+(m)) .....(2)`

Centre of buoyancy `(B)` is at a height of `(l)/(2)` from the bottom.
We can see from figure `(2)` that for rotational equilibrium of the rod. `B` should either lie above `C` or at the same level of `B`.
Therefore
`(l)/(2) ge Y_(com)` or `(piR^(2)Lrho + m)/(2piR^(2)sigma) ge (piR^(2)Lrho)^((L)/(2))/((piR^(2)Lrho)+m)`
or `m + piR^(2)Lrho ge piR^(2)L sqrt(rhosigma)` or `m ge piR^(2)L(sqrt(rhosigma) - rho)`
`:.` Minimum value of `m` is `piR^(2)L (sqrt(rhosigma) - rho)`