VISCOSITY: Dropping Solid Cylinder

PROBLEM STATEMENT:
A solid cylinder of diameter D, length L, density ρ_s falls due to gravity inside a tube of diameter D₀. The clearance, D₀ - D << D, is filled with a film of viscous fluid. Derive a formula for terminal fall velocity and apply to SAE 30 oil at 20° C for a steel cylinder with D=2 cm, D₀ = 2.04 cm, and L = 15 cm. Neglect the effect of any air in the tube.

SOURCE:
unknown

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PROCESS:

1) Convert all givens to SI units

D = 0.02 m
L = 0.15 m
D₀ = 0.0204 m

2) Look up material properties

ρ_steel = 7859 kg/m³ [source]
ρ_{SAE 30 oil} = 8.815×10^-4 kg/m³ [source]
μ_{SAE 30 oil} = 0.310 N⋅s/m² [source]

3) Analyze the problem & draw a picture

So it asks for us to derive a formula for the terminal fall velocity.
"Terminal velocity is the highest velocity attainable by an object as it falls through a fluid. It occurs when the sum of the drag force and the buoyancy is equal to the downward force of gravity."

4) Pick out appropriate equations & formulas

Shear Stress ≡ τ = μ⋅dv/dy = F/A
Area_circle ≡ A_ο = πr²
Circumference_circle ≡ C_ο = 2πr
Volume_cylinder ≡ V_cyl = πr²⋅L
Mass ≡ m = ρ⋅V
Force ≡ F = m⋅a

5) Plug, Chug, Solve

Cross sectional area of steel cylinder: (in j^ direction)
A_ο = πr² 
r = D/2

→ Area_{steel cylinder} ≡ A_s = πD²/4 j^

Shear Stress: τ = μ⋅dv/dy
dv/dy = V(y)/clearance → dv/dy = V(y)/(½⋅(D₀ - D))

→ dv/dy = 2⋅V(y)/(D₀ - D)
→ τ = μ⋅2⋅V(y)/(D₀ - D)

Drag: τ = μ⋅dv/dy = F/A
F_drag = ∫_A τ dA
dA refers to the surface area around the outside of the steel cylinder 
dA = circumference⋅dy = 2π(1/2 D)⋅dy

→ dA = πD⋅dy

plugging equations for dA, dv/dy, & τ into F_drag equation
→ F_drag = ∫₀^L μ⋅2⋅V(y)/(D₀ - D)⋅πD⋅dy

→ F_drag = [2π⋅μ⋅D⋅L⋅V(y)]⋅[D₀ - D]^-1​

Volume of steel cylinder: V = πr²⋅L

→ V_s = πLD²/4​

Mass of steel cylinder: m = ρ⋅V

→ m_s = ρ_s⋅πLD²/4​

Gravitational force on steel cylinder: F = ma

→ F_g = ρ_s⋅πLD²/4 ⋅ g​

Sum of forces: +ΣF_y = 0: 0 = F_drag - F_g
→ F_drag = F_g
→ [2π⋅μ_oil⋅D⋅L⋅V(y)]⋅[D₀ - D]^-1 = ρ_s⋅πLD²/4 ⋅ g

⇒ V = ρ_s⋅D⋅g⋅(D₀ - D)/(8μ)

6) Plug in actual values

V = (7859 kg/m³)⋅(0.02 m)⋅(9.81 m/s²)⋅[(0.0204 m) - (0.02 m)]/[8⋅(0.310 N⋅s/m²)]

⇒ V = 0.2487 m/s     [WolframAlpha Solution]

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SOLUTION:
 V = ρ_s⋅D⋅g⋅(D₀ - D)/(8μ)
V = 0.2487 m/s

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TOPICS COVERED: Viscosity, Solid Cylinder, Oil, Density, Terminal Velocity, Shear Stress, Force, Chapter 10