Floating cylinder

Question

A cylinder can of mass 3kg and the diameter 5 times greater than height is floating with 1/3 part of its volume inside the water. a) find the diameter of this can b) what minimum work is required to: 1) put all can into the water 2) to raise the can above water.

Density of water 1 t/m^3

Strenght of gravitational field – 9,8 N/kg

Answer 1

Part a) From the force balance, you know that ρ_sV_sg = ρ_w(1/3)V_sg , so the solid (s) density is 1/3 that of water.

From the definition of density, you know that ρ_s = m_s/V_s or that V_s = 3kg / ρ_s in m³.

From the equation of volume of a cylinder V= πr²h = (1/4)πd²h you can substitute d/5 for h and solve for d in meters which I leave to you.

Part b) I'm assuming that the course is calculus-based because the bouyancy force is not constant with depth submerged:

The change of energy going from the equilibrium position to the submerged position (The Min W required)

ΔE = Int(from 1/3 h to h) of (A(P(z)-Patm)dz) - ρVgΔz where z is the depth from the surface.

ΔE = Aρ_wg * Int(from h/3 to h) of (zdz) - m_sgz = (πd²/4)ρ_w g(z²/2) from h/3 to h - m_sg(2/3 h)

***ΔE = (πd²/4)ρ_wg(8h²/18) - m_sg(2/3 h) (BTW, use 1000 kg/m³ for density of water.)

A way to do this without the integral is to realize that the value of a linearly changing force has an average equal to the average of the endpoints. This way the energy change due to work against the buoyant force will be F_avgΔz or 1/2 (F(h) + F(h/3)) (2h/3) = (1/2)Aρ_wg(h+h/3)(2/3 h) = Aρ_wg(8h²/18) = Aρ_wg(4/9)h²

I guess easier still is multiplying F(2/3 h)(h-h/3). These all give the same answer for the integral term.

b) I will leave this to you. The work will be due to an increase in elevation - the average bouyant force times the difference in height.