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)`.

Explanation

For stable equilibrium, centre of mass should be lowest possible, therefore mass `m` should be attached to lower end.
Let `y` be the immersed length of stick
`F_B = sigma pi R^2 yg, M = rho pi R^2 L`
For vertical equilibrium
`Mg + mg = F_B`
`rho pi R^2 L + m = sigma pi R^2 y`
`y = (m + rho pi R^2 L)/(sigma pi R^2 y)`
Centre of mass of system from `O`
`y_(c.m.) = (M xx (L)/(2) + m xx 0)/(M + m) = (ML)/(2(M + m)) = (rho pi R^2 L^2)/(2(rho pi R^2 L + m))`
For rotational equilibrium, `c.m.` should be below centre of buoyancy
`y_(c.m.) le (y)/(2)`
`(rho pi R^2 L^2)/(2(rho pi R^2 L + m)) le (m + rho pi R^2 L)/(2 sigma pi R^2)`
`(pi R^2)^2 L^2 rho sigma le (m + rho pi R^2 L)^2`
`pi R^2 L sqrt(rho sigma) le (m + rho pi R^2 L)`
`m ge (pi R^2 L sqrt(rho sigma) - pi R^2 L rho)`
`m_(min) = pi R^2 L rho (sqrt((sigma)/(rho)) -1)`.
.