Fluid Drag and Terminal Velocity: MDCAT Physics notes
Fluid Drag and Terminal Velocity notes for MDCAT: viscous drag, Stokes’ law F = 6πηrv, net force during fall, terminal velocity formula and traps.
Viscosity and drag
A body moving through a fluid (liquid or gas) experiences a retarding force called fluid drag or fluid friction. It arises from the viscosity of the fluid, its internal friction between layers. Viscosity is measured by the coefficient $\eta$ (SI unit $\text{N s m}^{-2}$ or $\text{Pa s}$). Honey and glycerine are highly viscous; air and water much less so.
The drag force always acts opposite to the velocity of the body relative to the fluid, and it increases as the speed increases.
Stokes' law
For a small sphere of radius $r$ moving slowly (laminar flow) with speed $v$ through a fluid of viscosity $\eta$:
$$F_D = 6\pi\eta r v$$
So drag is proportional to the speed, the radius and the viscosity.
Falling through a fluid: net force
Take a ball of weight $F_g$ falling vertically. Drag $F_D$ acts upward. Ignoring buoyancy, the net downward force is
$$F_{net} = F_g - F_D$$
- At release: $v = 0$, so $F_D = 0$, and acceleration $= g$.
- Speeding up: $F_D$ grows, $F_{net}$ shrinks, and acceleration decreases. The weight stays the same; only the drag changes.
- Terminal velocity: $F_D = F_g$, net force is zero, acceleration is zero, and the body falls at a constant maximum speed $v_t$.
A heavy coin dropped a short distance does not reach terminal velocity because its speed never becomes large enough for drag to equal its weight: weight remains greater than air resistance throughout the short fall.
A 10 kg body that has reached terminal velocity in a viscous medium has zero net force on it, not 98 N; its weight is balanced by the resistive forces.
Terminal velocity of a sphere
Setting $mg = 6\pi\eta r v_t$ with $m = \tfrac43\pi r^3\rho$:
$$v_t = \frac{2g r^2\rho}{9\eta}$$
(for a fluid whose density is small compared with $\rho$). Hence $v_t \propto r^2$: larger droplets fall faster. This is why fog droplets (tiny) float, while raindrops fall steadily.
| Stage | Drag vs weight | Acceleration | Velocity |
|---|---|---|---|
| Start | $F_D = 0$ | $g$ | Zero |
| During fall | $F_D < F_g$ | Decreasing | Increasing |
| Terminal | $F_D = F_g$ | Zero | Constant $v_t$ |
Key formulas
- $F_D = 6\pi\eta r v$
- $F_{net} = F_g - F_D$
- $v_t = \dfrac{2gr^2\rho}{9\eta}$
Common MDCAT traps
- At terminal velocity the net force is zero, not $mg$.
- Acceleration at terminal velocity is zero, not negative or variable.
- Weight does not increase as air resistance increases; drag grows, weight is constant.
- Net force while falling is $F_g - F_D$, not $F_g + F_D$.
- Terminal velocity depends on $r^2$, not $r$.
Quick revision
- Drag opposes motion and grows with speed.
- Stokes' law: $F = 6\pi\eta rv$.
- Terminal velocity: weight = drag.
- Short falls: weight > drag, so no terminal velocity.
- Bigger droplets have higher terminal velocity.