Drag Coefficient
Consider during a round of golf, a player smacks the 0.5 lb ball such that it travels 200 ft/s. If the drag force acting on the 0.1 ft^2 projected area is 2.5 lbf, what is the drag coefficient? Assume the density of air is 0.08 lb/ft^3.
Expand Hint
The drag force on objects immersed in a large body of flowing fluid or objects moving through a stagnant fluid:
$$$D=\frac{1}{2}\rho U^{2}C_{D}A$$$
where
$$C_D$$
is the drag coefficient,
$$U$$
is the flowing fluid or moving object’s velocity,
$$\rho$$
is the fluid density, and
$$A$$
is the projected area of blunt objects with axes perpendicular to the flow.
Hint 2
$$$1\:lbf=1\:lb \times gravity\approx 1\:lb \times 32.2 \frac{ft}{s^2}$$$
The drag force on objects immersed in a large body of flowing fluid or objects moving through a stagnant fluid:
$$$D_f=\frac{1}{2}\rho U^{2}C_{D}A$$$
where
$$C_D$$
is the drag coefficient,
$$U$$
is the flowing fluid or moving object’s velocity,
$$\rho$$
is the fluid density, and
$$A$$
is the projected area of blunt objects with axes perpendicular to the flow.
$$$2.5lb_f=\frac{1}{2}(0.08\frac{lb}{ft^3})(200\frac{ft}{s})^2C_D(0.1ft^2)$$$
Before solving for
$$C_D$$
, recall that
$$1\:lb_f=1\:lb \times gravity\approx 1\:lb \times 32.2 \frac{ft}{s^2}$$
. Thus,
$$$2.5lb\times32.2\frac{ft}{s^2}=\frac{1}{2}(0.08\frac{lb}{ft^3})(40,000\frac{ft^2}{s^2})C_D(0.1ft^2)$$$
$$$80.5=(160)C_D$$$
$$$C_D=\frac{80.5}{160}=0.503\approx 0.5$$$
0.5
Time Analysis
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