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Terminal velocity of steel sphere in glycerine

Short answer

Terminal velocity under Stokes' flow: V = (γ_s-γ_f)d²/(18μ) = (75000-12500)×0.003²/(18×0.893) = 0.035 m/s. Drag force F_D = 3πμVd = 8.84×10^-4 N (Re=0.15).

A 3 mm diameter sphere made of steel (sp. wt. 75 KN/m³) falls in glycerine (sp. wt. 12.5 KN/m³) of viscosity 0.893 NS/m² at a terminal velocity. Determine the terminal velocity and drag force on the sphere.

At terminal velocity, the weight minus buoyancy equals the drag. For a small sphere in a viscous liquid, assume Stokes' law, and check ReRe afterwards.

Given

d=3d = 3 mm =0.003= 0.003 m, γs=75 kN/m3\gamma_s = 75\ \text{kN/m}^3, γf=12.5 kN/m3\gamma_f = 12.5\ \text{kN/m}^3, μ=0.893 N s/m2\mu = 0.893\ \text{N s/m}^2.

Terminal velocity

Net downward force: (γs−γf)πd36(\gamma_s - \gamma_f)\dfrac{\pi d^3}{6}. Stokes' drag: FD=3πμVdF_D = 3\pi\mu V d.

3πμVd=(γs−γf)πd36V=(γs−γf) d218μ=(75 000−12 500)(0.003)218(0.893)=0.0350 m/s\begin{aligned} 3\pi\mu V d &= (\gamma_s - \gamma_f)\frac{\pi d^3}{6} \\ V &= \frac{(\gamma_s - \gamma_f)\,d^2}{18\mu} = \frac{(75\,000 - 12\,500)(0.003)^2}{18(0.893)} = 0.0350\ \text{m/s} \end{aligned}

Check of Reynolds number

ρf=12 500/9.81=1274 kg/m3\rho_f = 12\,500/9.81 = 1274\ \text{kg/m}^3

Re=ρfVdμ=1274(0.0350)(0.003)0.893=0.15<1Re = \frac{\rho_f V d}{\mu} = \frac{1274(0.0350)(0.003)}{0.893} = 0.15 < 1

So Stokes' law is valid.

Drag force

FD=3πμVd=3π(0.893)(0.0350)(0.003)=8.84×10−4 NF_D = 3\pi\mu V d = 3\pi(0.893)(0.0350)(0.003) = 8.84\times10^{-4}\ \text{N}

Check: (γs−γf)πd3/6=62 500×1.414×10−8=8.84×10−4(\gamma_s - \gamma_f)\pi d^3/6 = 62\,500 \times 1.414\times10^{-8} = 8.84\times10^{-4} N.

Answer: terminal velocity ≈0.035\approx 0.035 m/s (3.5 cm/s); drag force ≈0.88\approx 0.88 mN (8.84×10−48.84\times10^{-4} N).

✓ Answer checked by Claude Opus 5.5 (high effort)

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