A spherical steel ball of diameter 40 mm and density $ 8500 kg/m^3$ is dropped in the large mass of water. The coefficient of Drag of the ball in water is given by 0.45.

1. Find terminal velocity of the ball in the water if the ball is dropped in air.

2. Find increase in terminal velocity of the ball.

Take density of the air =$1.25 kg/ m^3$ & Cd = 0.1

Solution :

Given :

1. Diameter of steel ball, $D=40 \mathrm{~mm}=0.04 \mathrm{~m}$

2. Density of ball, $ \rho_{b}=8500 \mathrm{~kg} / \mathrm{m}^{3}$

3. $C_{D}$ for ball in water $=0.45$

Let the terminal velocity in water $=U_{1}$

A) The forces acting on spherical ball are :

(i) Weight

$ W=$ Density of ball $\times g \times$ Volume of spherical ball

$\begin{aligned} W &=\rho_{b} \times g \times \frac{\pi}{6} D^{3}\\ &=8500 \times 9.81 \times \frac{\pi}{6}(0.04)^{3}\\ &= 2.794 \mathrm{~N} \end{aligned}$

(ii) Buoyant Force

$F_{B}=$ Density of water $× g ×$ Volume of ball

$\begin{aligned} F_B &=1000 \times 9.81 \times \frac{\pi}{6}(0.04)^{3}\\ &=0.3286 \mathrm{~N} \end{aligned}$

(iii) Drag force

$\begin {aligned} F_{D} &=C_{D} \times A \times \frac{\rho U^{2}}{2}\\ F_{D} &=0.45 \times \frac{\pi}{4} \times(0.04)^{2} \times 1000 \times \frac{U_{1}^{2}}{2}\\ &=0.2825 U_{1}^{2} \end{aligned} $

We know that,

$\begin{aligned} W &=F_{D}+F_{B} \\ 2.794 &=0.2825 U_{1}^{2}+0.3286 \\ U_{1}^{2} &=\frac{2.794-0.3286}{0.2825}=8.725 \\ U_1&= 2.953 m/s\end{aligned}$

B) When the ball is dropped in air,

Let the terminal velocity is $=U_2$

(i) Weight

$W=2.974$

(ii) Buoyant Force

$F_{B}=$ Density of air $× g ×$ Volume of ball

$\begin{aligned} F_B &=1.25 \times 9.81 \times \frac{\pi}{6}(.04)^{3}\\ &=0.000411 \mathrm{~N} \end{aligned}$

(iii) Drag force

$\begin{aligned} F_{D} &=C_{D} \times A \times \frac{\rho_{air} U^{2}}{2}\\ F_{D} &=0.1 \times \frac{\pi}{4}(.04)^{2} \times 1.25 \frac{U_{2}^{2}}{2}\\ &=0.0000785 U_{2}^{2}\end{aligned}$

The buoyant force in air is $0.000411\mathrm{~N}$, while weight of the ball is $2.794 \mathrm{~N}$.

Hence buoyant force is negligible.

$\therefore$ For equilibrium of the ball in air,

$F_{D}=$ Weight of ball

or

$\begin {aligned} 0.0000785 U_{2}^{2} &=2.794 \\ U_{2} &=\sqrt{\frac{2.794}{0.0000785}}\\ &=188.67 \mathrm{~m} / \mathrm{s} \end{aligned}$

$\therefore$ Increase in terminal velocity in air

$=U_{2}-U_{1}=188.67-2.9533=185.717 \mathrm{~m} / \mathrm{s}$