HARD · FRICTION AND DRAG
TERMINAL VELOCITY STOKES SPHERE
A steel sphere of radius r = 3 mm and mass m = 0.15 g falls through still water (dynamic viscosity η = 0.001 Pa·s). Using Stokes' law for a sphere, find the linear drag coefficient b, the terminal velocity v_t the sphere reaches, and the time constant τ that describes how quickly it approaches v_t.
Linked equations:F_d = bvv_t = \frac{mg}{b}\text{ (linear)},\quad v_t = \sqrt{\frac{2mg}{\rho C_d A}}\text{ (quadratic)}
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Step-by-step solution
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Step 1
What is the Stokes drag coefficient b for this sphere?
Hint
Stokes' law for a sphere: F_drag = 6πηrv, so b = 6πηr. Units of b are N·s/m (or kg/s).
Step 2
At what speed does the sphere reach terminal velocity?
Step 3
What is the time constant τ for the exponential approach to v_t?
Solution walkthrough
Stokes showed in 1851 that a sphere of radius r moving slowly through a fluid of viscosity η experiences a drag force F = 6πηrv. The drag coefficient b = 6πηr captures all the geometry and fluid properties in a single constant; multiplied by velocity it gives the force.
With r = 0.003 m and η = 0.001 Pa·s: b = 6π × 0.001 × 0.003 ≈ 5.655 × 10⁻⁵ kg/s. This is a very small number, consistent with the fact that water is not particularly viscous and the sphere is tiny.
Terminal velocity is reached when gravity and drag balance: m·g = b·v_t, giving v_t = m·g/b = 1.5 × 10⁻⁴ × 9.807 / 5.655 × 10⁻⁵ ≈ 26.0 m/s. This is surprisingly large for a 3 mm sphere — a reminder that the Stokes regime (Re ≪ 1) may not actually hold at this speed in water, where the Reynolds number would be Re ≈ ρvr/η ≈ 1000 × 26 × 0.003 / 0.001 ≈ 78 000. In practice the sphere would enter the quadratic regime long before reaching this speed. The Stokes formula is exact for very slow motions; here it sets the scale.
The time constant τ = m/b = 1.5 × 10⁻⁴ / 5.655 × 10⁻⁵ ≈ 2.65 s. The approach is exponential: v(t) = v_t·(1 − e^(−t/τ)). After 5τ ≈ 13 s the sphere is within 1 % of terminal velocity.
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