The drag force FD acting on a sphere moving through a fluid is given by the expression

Question 10

(a) The drag force FD acting on a sphere moving through a fluid is given by the expression

FD = Kρv²

where K is a constant,

ρ is the density of the fluid

and v is the speed of the sphere.

Determine the SI base units of K. [3]

(b) A ball of weight 1.5 N falls vertically from rest in air. The drag force F_D acting on the ball is given by the expression in (a). The ball reaches a constant (terminal) speed of 33 m s^-1.

Assume that the upthrust acting on the ball is negligible and that the density of the air is

uniform.

For the instant when the ball is travelling at a speed of 25 m s^-1, determine

(i) the drag force F_D on the ball, [2]

(ii) the acceleration of the ball. [2]

(c) Describe the acceleration of the ball in (b) as its speed changes from zero to 33 m s^-1. [3]

Reference: Past Exam Paper – November 2017 Paper 21 Q1

Solution:

(a)

units of F: kg m s^-2                                                                       

units of ρ: kg m^-3           and      units of v: m s^-1              

{F_D = Kρv²      so, K = F_D / ρv²}

units of K: kg m s^-2 / [kg m^-3 (m s^-1)²] = m²                     

(b)

(i)

{At constant speed, the resultant force is zero. That is, the drag force F_D is equal to the weight (= 1.5 N).

Kρ = F_D / v²

The product of Kρ is a constant. That is, when FD = 1.5 N and v = 33 m s^-1, the product is}

Kρ = 1.5 / 33² = 1.38 × 10^-3   

{When the force is F_D and the speed is 25 m s^-1, the product of Kρ will still have the same value.}

{F_D = (Kρ)v²                            or F_D1 / v₁² = F_D2 / v₂²}

FD = 1.38 × 10^-3 × 25²                 or {using proportion,} FD / 1.5 = 25² / 33²

FD = 0.86 N                

(ii)

{Resultant force = ma

Resultant force = Weight – Drag = 1.5 – 0.86

Weight = mg               so, m = W / g = 1.5 / 9.81

Acceleration a = Resultant force / m

or divide the equation ‘Resultant force = Weight – Drag’  by m}

a = (1.5 – 0.86) / (1.5 / 9.81)               or a = 9.81 – [0.86 / (1.5 / 9.81)]

a = 4.2 m s^-2    

(c)

The acceleration f the ball is initially equal to the acceleration of free fall. 
{This acceleration causes the speed of the ball to increase, and thus the drag force also increases.} 
The acceleration then decreases {as the resultant force on the ball decreases} until the final acceleration becomes zero {terminal velocity is reached}.