Derive the expression for terminal velocity for a sphere falling through a viscous fluid.

• Gravitational force ($F_g$) acting downwards: $$F_g = mg = (\text{volume of sphere}) \times (\text{density of sphere}) \times g$$ $$F_g = \frac{4}{3}\pi r^3 \rho g$$
• Buoyant force ($F_b$) acting upwards (by Archimedes' principle): $$F_b = (\text{volume of sphere}) \times (\text{density of liquid}) \times g$$ $$F_b = \frac{4}{3}\pi r^3 \sigma g$$
• Viscous drag force ($F_v$) acting upwards (by Stokes' Law): $$F_v = 6\pi \eta r v$$ where $v$ is the instantaneous velocity of the sphere.

The net force on the sphere is $F_{net} = F_g - F_b - F_v$.

When the sphere reaches terminal velocity ($v_t$), its acceleration becomes zero ($a = 0$). At this point, the net force is zero, meaning the upward forces balance the downward force.

Rearrange the equation to isolate $v_t$:

Divide both sides by $6\pi \eta r$:

Simplify the expression:

Graphical Variation of Velocity with Time:
When an object is dropped in a viscous fluid, its velocity initially increases from zero. As the velocity increases, the viscous drag force also increases, which reduces the net accelerating force. Consequently, the acceleration decreases over time. Eventually, the sum of the upward buoyant force and viscous drag force equals the downward gravitational force. At this point, the net force becomes zero, acceleration becomes zero, and the object attains a constant maximum velocity called the terminal velocity ($v_t$).

The graph of velocity ($v$) versus time ($t$) shows an exponential increase in velocity from zero, which then levels off to a constant value (terminal velocity) as time progresses.

Velocity
  ^
  |      /
  |     /
  |    /
  |   /
  |  /
  | /
  |/_________________ Terminal Velocity (v_t)
  |
  +----------------------> Time
  0