Buoyancy (DP IB Physics: SL): Revision Note

Buoyancy

• Buoyancy is experienced by a body which is partially or totally immersed in a fluid

  • The buoyancy force is exerted on a body due to the displacement of the fluid it is immersed in

• Buoyancy keeps boats afloat and allows balloons to rise through the air

• When a body travels through a fluid, it also experiences a buoyancy force (upthrust) due to the displacement of the fluid

• Buoyancy is calculated using:

$F_b=\rho Vg$

• Where:

  • F_b = buoyancy force (N)

  • ρ = density of the fluid (kg m^–3)

  • V = volume of the fluid displaced (m³)

  • g = acceleration of free fall (m s^–2)

• If you were to take a hollow ball and submerge it into a bucket of water, you would feel some resistance

• Some water will flow out of the bucket as it is displaced by the ball

• The buoyancy force, F_b of the water will push upward on the ball

• When you let go of the ball, the buoyancy force of the water on the ball will cause the ball to accelerate to the surface

• The ball will remain stationary floating on the surface of the water

• A this point, the weight of the ball acting downward, F_g, is equal to the buoyancy force acting upwards, F_b

The ball floats when the buoyancy force and its weight are balanced

• Notice that

$F_g=\rho Vg=\frac{m}{V}Vg=mg$

• Where:

  • m = mass of the ball (kg)

  • ρ = density of the ball (kg m^–3)

  • V = volume of the ball (m³)

• The buoyancy force and the weight force are equal

Drag Force at Terminal Speed

• Terminal velocity, or terminal speed, is useful when working with Stoke’s Law

• This is because, at terminal velocity, the forces in each direction are balanced

$W_s=F_d+F_b$ (Equation 1)

• Where:

  • W_s = weight of the sphere (N)

  • F_d = the drag force (N)

  • F_b = the buoyancy force / upthrust (N)

At terminal velocity, the forces on the sphere are balanced

• The weight of the sphere is found using volume, density and gravitational field strength

$W_s={\rho }_s{V}_{s}g$ 

$W_s=\frac{4}{3}\pi r^3{\rho }_{s}g$ (Equation 2)

• Where

  • V_s = volume of the sphere (m³)

  • ρ_s = density of the sphere (kg m^–3)

  • r = radius of the sphere (m)

  • g =  acceleration of free fall (m s⁻²)

• Recall Stoke’s Law

$F_d=6\pi \eta rv$ (Equation 3)

• Where

  • F_d = viscous drag force (N)

  • η = fluid viscosity (N s m⁻² or Pa s)

  • r = radius of the sphere (m)

  • v = velocity of the sphere through the fluid (ms⁻¹)

    • In this case, v is the terminal velocity

• The buoyancy force equals the weight of the displaced fluid

  • The sphere is fully submerged in the fluid, so the volume of displaced fluid is the same as the volume of the sphere

  • The weight of the fluid is found using volume, density and acceleration of free fall

$F_b=\frac{4}{3}\pi r^3{\rho }_{f}g$ (Equation 4)

• Where

  • ${\rho }_f$ = density of the fluid (kg m^–3)

• Substitute equations 2, 3 and 4 into equation 1

$\frac{4}{3}\pi r^3{\rho }_{s}g\mathit{=}\mathit{6}\pi \eta rv\mathit{+}\frac{\mathit{4}}{\mathit{3}}\pi r^{\mathit{3}}{\rho }_{f}g$

• Rearrange to make terminal velocity the subject of the equation

$v=\frac{\frac{\mathit{4}}{\mathit{3}}\pi r^{\mathit{3}}g\mathit{(}{\rho }_s\mathit{-}{\rho }_f\mathit{)}}{\mathit{6}\pi \eta r}=\frac{4\pi r^{\mathit{3}}g\mathit{(}{\rho }_s\mathit{-}{\rho }_f\mathit{)}}{\mathit{18}\pi \eta r}$

• Finally, cancel out r from the top and bottom to find an expression for terminal velocity in terms of the radius of the sphere and the coefficient of viscosity

$v=\frac{2\pi r^{2}g\mathit{(}{\rho }_s\mathit{-}{\rho }_f\mathit{)}}{\mathit{9}\pi \eta }$

• This final equation shows that terminal velocity is:

  • directly proportional to the square of the radius of the sphere

  • inversely proportional to the viscosity of the fluid