6.7 Terminal Velocity

Terminal velocity is the maximum constant speed that a freely falling object eventually reaches when the resistance of the medium (like air or water) through which it is moving equals the force of gravity. At this point, the net force on the object is zero, and it stops accelerating, continuing to fall at a constant velocity.

Forces in Balance

The motion of an object falling through a fluid is governed by two primary forces:

1. Gravitational Force ($F_g$): The force of weight pulling the object downward ($F_g=mg$). This force is constant.
2. Drag Force ($F_D$): The fluid friction that opposes the object's motion, acting upward. This force is not constant; it increases as the object's velocity increases.

Terminal velocity is achieved when these two forces become equal in magnitude and opposite in direction.

$F_g=F_D$

At this equilibrium point, the net force is zero, and according to Newton's Second Law ($F_{net}=ma$), the acceleration ($a$) becomes zero.

Mathematical Derivation (for a Sphere)

We can derive a formula for the terminal velocity ($v_t$) of a spherical object using Stokes' Law for the drag force:

$F_D=6\pi \eta rv$

where $\eta$ is the viscosity of the fluid, $r$ is the radius of the sphere, and $v$ is its velocity.

Derivation Steps

1. Force Balance: At terminal velocity, the gravitational force equals the drag force.

$mg=6\pi \eta r{v}_t$

2. Expressing Mass in Terms of Density: The mass ($m$) of a sphere is its density ($\rho$) multiplied by its volume ($V=\frac{4}{3}\pi r^3$).

$m=\rho \cdot \frac{4}{3}\pi r^3$

3. Substituting and Solving for $v_t$: Substitute the expression for mass into the force balance equation:

$\left(\rho \cdot \frac{4}{3}\pi r^3\right)g=6\pi \eta r{v}_t$

Rearranging and simplifying:

$v_t=\frac{2\rho g{r}^2}{9\eta }$

This formula shows that terminal velocity is:

• Directly proportional to the square of the object's radius: $v_t\propto r^2$
• Inversely proportional to the viscosity of the fluid: $v_t\propto \frac{1}{\eta }$
• Directly proportional to the density of the object: $v_t\propto \rho$

Application: A Paratrooper's Jump

The jump of a paratrooper is a classic real-world example of manipulating drag to control terminal velocity.

• Phase 1: Freefall (Before Opening Parachute)

  • The paratrooper accelerates downwards due to gravity.
  • As their speed increases, the upward drag force also increases.
  • Eventually, the drag force balances their weight, and they reach a high terminal velocity (approximately 50–60 m/s or 120 mph).

• Phase 2: Parachute Deployed

  • Opening the parachute dramatically increases the surface area, which massively increases the drag force.
  • The upward drag force is now much greater than the downward force of gravity, causing the paratrooper to rapidly decelerate.
  • As they slow down, the drag force decreases until it once again balances their weight.
  • A new, much lower terminal velocity is reached (approximately 5–7 m/s or 15 mph), allowing for a safe landing.

Conceptual Questions

• Q: Why does a crumpled piece of paper fall faster than a flat sheet?
  A: The crumpled paper has a much smaller surface area exposed to the air, which results in a significantly lower drag force. Because its weight is the same as the flat sheet, it must reach a higher speed before the small drag force can balance its weight.

• Q: Do heavier objects always fall faster?
  A: In a vacuum, all objects fall at the same rate. In a fluid like air, a heavier object of the same size and shape will have a higher terminal velocity because a greater drag force (and thus a higher speed) is needed to counteract its greater weight.

Equilibrium→