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Difficulty: HardFluids at Rest, Archimedes' Principle and Viscosity

A spherical ball bearing of radius 3.0 mm3.0\text{ mm} and density 5400 kg/m35400\text{ kg/m}^3 falls vertically through a viscous oil of density 900 kg/m3900\text{ kg/m}^3 and dynamic viscosity coefficient 0.10 Pa⋅s0.10\text{ Pa}\cdot\text{s}. Assuming the motion obeys Stokes' law and taking the acceleration due to gravity g=10 m/s2g = 10\text{ m/s}^2, what is the terminal velocity of the sphere?

  1. 0.90 m/s0.90\text{ m/s}Answer
  2. B
    1.08 m/s1.08\text{ m/s}
  3. C
    0.68 m/s0.68\text{ m/s}
  4. D
    0.18 m/s0.18\text{ m/s}

Answer

The terminal velocity of the sphere is 0.90 m/s0.90\text{ m/s}.
At terminal velocity, the downward force of gravity (weight of the sphere) is balanced by the sum of two upward forces: the buoyant force (upthrust) and the viscous drag force given by Stokes' law (Fv=6πηrvTF_v = 6\pi \eta r v_T). Using the formula vT=2r2(ρs−ρf)g9ηv_T = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta} with r=3.0×10−3 mr = 3.0 \times 10^{-3}\text{ m}, ρs−ρf=4500 kg/m3\rho_s - \rho_f = 4500\text{ kg/m}^3, η=0.10 Pa⋅s\eta = 0.10\text{ Pa}\cdot\text{s}, and g=10 m/s2g = 10\text{ m/s}^2 yields 0.90 m/s0.90\text{ m/s}.

Step-by-Step Solution

1
Convert given parameters to standard SI units
Radius r=3.0 mm=3.0×10−3 mr = 3.0\text{ mm} = 3.0 \times 10^{-3}\text{ m}, density of sphere ρs=5400 kg/m3\rho_s = 5400\text{ kg/m}^3, density of liquid ρf=900 kg/m3\rho_f = 900\text{ kg/m}^3, viscosity η=0.10 Pa⋅s\eta = 0.10\text{ Pa}\cdot\text{s}, g=10 m/s2g = 10\text{ m/s}^2.
Ensures dimensional consistency across all terms in the physical equations.
2
Apply the equilibrium condition at terminal velocity
At terminal velocity vTv_T, downward weight equals upward forces: W=U+FvW = U + F_v, where W=43πr3ρsgW = \frac{4}{3}\pi r^3 \rho_s g, U=43πr3ρfgU = \frac{4}{3}\pi r^3 \rho_f g, and Fv=6πηrvTF_v = 6\pi \eta r v_T.
Terminal velocity is reached when net acceleration is zero.
3
Rearrange Stokes' law formula for terminal velocity
vT=2r2(ρs−ρf)g9ηv_T = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}
Isolates the target unknown variable vTv_T.
4
Substitute the physical values and solve
vT=2×(3.0×10−3)2×(5400−900)×109×0.10=2×(9.0×10−6)×4500×100.90=0.810.90=0.90 m/sv_T = \frac{2 \times (3.0 \times 10^{-3})^2 \times (5400 - 900) \times 10}{9 \times 0.10} = \frac{2 \times (9.0 \times 10^{-6}) \times 4500 \times 10}{0.90} = \frac{0.81}{0.90} = 0.90\text{ m/s}.
Calculates the final quantitative answer.

Key Concept

Viscosity and Stokes' Law for terminal velocity of a sphere in a viscous fluid
Estimated Time:2m 0s
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Topics

Fluids at Rest, Archimedes' Principle and Viscosity

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