Question

A rider on a bike with the combined mass of 100 kg attains a terminal speed of 15 m/s on a 12% slope. Assuming that the only forces affecting the speed are the weight and the drag, calculate the drag coefficient. The frontal area is 0.9 m2.

Answer 1

Answer:

7.84

Explanation:

Draw free body diagram and put all forces on it. Forces are

• Gravity in direction of slope =$\frac{12}{100} *100*12$=117.6newton
• Viscous force(opposite to slope) after attaining terminal speed =k*V=k*15

As the bike+man system has attained a terminal velocity thus acceleration is zero .

Both forces are opposite then equate them

117.6=k.15

k=7.84

Here k is drag coefficient.

Answer 2

Answer:

0·95

Explanation:

Given the combined mass of the rider and the bike = 100 kg

Percent slope = 12%

∴ Slope = 0·12

Terminal speed = 15 m/s

Frontal area = 0·9 m²

Let the slope angle be β

tanβ = 0·12

As it attains the terminal speed, the forces acting on the combined rider and the bike must be balanced and therefore the rider must be moving download as the directions of one of the component of weight and drag force will be in opposite directions

The other component of weight will get balance by the normal reaction and you can see the figure which is in the file attached

From the diagram m × g × sinβ = drag force

Drag force = 0·5 × d × $C_{D}$ × v² × A

where d is the density of the fluid through which it flows

$C_{D}$ is the drag coefficient

v is the speed of the object relative to the fluid

A is the cross sectional area

As tanβ = 0·12

∴ sinβ = 0·119

Let the fluid in this case be air and density of air d = 1·21 kg/m³

m × g × sinβ = 0·5 × d × $C_{D}$ × v² × A

100 × 9·8 ×0·119 = 0·5 × 1·21 × $C_{D}$ × 15² × 0·9

∴ $C_{D}$ ≈ 0·95

∴ Drag coefficient is approximately 0·95