Q.State Stokes' law for the viscous drag force on a small sphere moving through a viscous fluid. Explain, in terms of the forces acting on the sphere, why a small ball or a raindrop falling through a viscous fluid speeds up at first but eventually falls with a constant (terminal) velocity rather than continuing to accelerate indefinitely.

Stokes' law states that a small sphere of radius $r$ moving at speed $v$ through a viscous fluid of coefficient of viscosity $\eta$ experiences a retarding (drag) force:

When the sphere is first released, it is at rest, so $v = 0$ and the viscous drag is zero -- the sphere accelerates downward under the net of its own weight and the (constant) buoyant force. As the sphere's speed $v$ increases, the viscous drag $F = 6\pi\eta r v$ increases along with it (since $F \propto v$), which means the net downward force on the sphere (weight minus buoyancy minus this growing drag) steadily decreases as the sphere speeds up.

Eventually, the sphere reaches a speed at which the (now larger) viscous drag, together with the buoyant force, exactly balances its weight. From that instant onward the net force on the sphere is zero, so by Newton's second law its acceleration is also zero -- the sphere can no longer speed up (or slow down) any further, and it continues falling at this fixed, constant speed for the rest of its fall through the fluid. This constant final speed is called the terminal velocity.

State Stokes' law, then explain the terminal-velocity mechanism as a consequence of the drag force growing with speed until it, together with buoyancy, exactly balances the weight, making the net force -- and hence the acceleration -- zero.

• Saying the sphere "stops" at terminal velocity -- it keeps moving at a constant (non-zero) speed; only its acceleration becomes zero, not its velocity.
• Forgetting to include the buoyant force in the force balance, and trying to balance the viscous drag against the weight alone.