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

` pi R ^ 2 L (sqrt(rho sigma ) - rho ) `

Explanation

` pi R^2 L ( sqrt(rho sigma )- rho ) `
Let M = Mass of stick = ` pi R^2 rho L `
I = immersed length of the rod
G = CM of rod B = centre of buoyant force (F)
C = CM of rod + mass (m)
` Y _ (CM) `= Distance of C from bottom of the rod
Mass m should be attached to the lower end because otherwise B will be below G and C will be above G and the torque of the couple of two equal and opposite forces F and (M + m)g will be counter clockwise on displacing the rod leftwards. Therefore, the rod cannot be in rotational equilibrium. See the figure (iii). Now, refer figure (i) and (ii)
For vertical equilibrium ` Mg + m g = F _("upthrust") or (pi R^2 L ) rho g + mg = (pi R ^2 l ) sigma g therefore l = ((pi R^2 L rho + m)/( pi R ^2 sigma )) `
Position of CM (of rod + m ) from bottom ` Y_(cm) = ( M. (L)/(2))/(M+ m ) = ((pi R ^2 L rho ) (L)/(2))/((pi R ^ 2 L rho) + m ) `
Centre of buoyancy (B) is at a height of `(l)/(2) ` from the bottom.
We can see from figure (ii) 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_(CM)`
or ` (pi R^2 L rho + m )/( 2 pi R^2 sigma ) ge ((pi R ^ 2 L rho ) (L)/(2))/((pi R ^ 2 L rho ) + m ) or m + pi R^2 L rho ge pi R^ 2 L sqrt (rho sigma ) or m ge pi R^2 L (sqrt( rho sigma ) - rho )`