1) Terminal velocity can be defined as, the maximum constant velocity of a body which is falling (i.e., parachute) in atmosphere with which the body will be travelling.

2) When the body is at rest and is allowed to fall in the atmosphere, the velocity of body increases due to acceleration of gravity.

3) When the velocity increases, the drag force opposing the motion also increases.

4) A stage occurs or is reached where the upward drag force is equal to the weight of the body on which it is acting.

5) Thus, the net force acting on the body will be zero, and the body will be travelling at constant speed.

6) This constant speed of falling body is called terminal velocity of falling body.

7) The instant when the body acquire terminal velocity, if the body drops in a fluid, then the net force acting on the body will be zero.

Numericals

Q1) A spherical Steel ball of diameter 40 mm and of density $8500 kg/m^3$ is dropped in a large mass of water. The coefficient of drag of the ball in water is given as 0.45. Find the terminal velocity of the ball in water. If the ball is dropped in air, find the increase in terminal velocity of ball. Take density of air = $1.25kg/m^3, C_D=0.1$

Solution:- Given:-

Diameter of steel ball, D=40mm=0.04m

Density of ball, $\rho_S=8500kg/m^3=0.45$

Let the terminal velocity in water = $U_1$

The forces acting on the spherical balls are

(i) Weight, $W=\rho_S\times g×\left[\dfrac\pi6\times(D^3)\right]$

$W=8500\times 9.81×\left[\dfrac\pi6\times(0.04)^3\right]=2.794N$

(ii) Buoyant Force

$F_B=1000\times9.81\times\dfrac\pi6(0.04)^3$

$F_B=0.3286N$

(iii) Drag Force: -

$F_D=C_D\times A\times\dfrac{\rho U^2}2$

And $\rho=1000$ for water

$F_D=C_D\times \dfrac\pi4(D^2)\times\dfrac{\rho U_1^2}2$

$F_D=0.45 \times \dfrac\pi4((0.04)^2)\times\dfrac{1000 U_1^2}2$

$F_D=0.2825 U_1^2…………..(1)$

Using the relation

$W=F_D+F_B$

$2.794=0.2825U_1^2+0.3286$

$U_1^2=\dfrac{2.794-0.3286}{0.2825}=8.725$

$∴ U_1=\sqrt{8.725}=2.953m/s$

When ball is dropped in air terminal velocity = $U_2$

(i) Weight, W=2.794

(ii) Buoyant force

$F_B=1.25\times9.81\times\dfrac\pi6(0.04)^3$

$F_B=0.000411N$

(iii) Drag Force: -

$F_D=C_D\times A\times\dfrac{\rho U^2}2$

And $\rho_{air}=1.25$

$F_D=C_D\times \dfrac\pi4(D^2)\times\dfrac{\rho U^2}2$

$F_D=0.1 \times \dfrac\pi4((0.04)^2)\times\dfrac{1.25 U^2}2$

$F_D=0.0000785 U_2^2…………..(1)$

The buoyant force in Air is (0.00041) while weight of ball is (2.794N). Hence buoyant force is negligible.

Therefore for equilibrium of ball in air,

$F_D =$ weight of the ball

$\therefore 0.0000785U_2^2=2.794$

$U_2=\sqrt{\dfrac{2.794}{0.0000786}}$

$U_2=188.67m/s$

Therefore Increase in terminal velocity in air

$=U_2-U_1$

$=188.67-2.9533$

$=185.717 \ m/s$