Question

A wooden stick of length `L`, radius `R` and density `rho` has a small metal piece of mass `m` ( of negligible volume) atached 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 1

The mass `m` cannot be at the top, because for stale equilibrium, the cenre of mass should be the lowest possible. Therefore `m` is at the lower end of the stick. Let `x` be the length of the stick immersed in liquid. For the vertical equilibrium,
`Mg+mg=B`
`impliespiR^(2)Lrhog+mg=(piR^(2)xsigma)g`
`:. x=(m+piR^(2)Lrho)/(piR^(2)sigma)`..........i
The distance of the centre of mass from `m` is
`y_(CM)=(m_(1)y_(1)+m_(2)y_(2))/(m_(1)+m_(2))=(mxx0+piR^(2)Lrho(L/2))/(m+piR^(2)Lrho)`

For rotational equilibrium of the stick, the point of application of upthrust of the liquid must pass above the centre of mass of the floating system i.e. `y_(CM)lex/2`
`(piR^(2)Lrho(L/2))/(m+piR^(2)Lrho)le1/2((m+piR^(2)Lrho)/(piR^(2)sigma))`
`implies (piR^(2)L)^(2)rhosigma,=(m+piR^(2)Lrho)^(2)`
`implies piR^(2)Lsqrt(rhosigma)lem+piR^(2)Lrho`
`implies mgepiR^(2)L{sqrt(rhosigma)-rho}`
`implies m_("min")=piR^(2)L{sqrt(rhosigma)-rho}`
`=piR^(2)Lrho{(sqrt(rho))/rho-1}`