Basketball Hoop
Consider a basketball hoop’s backboard is placed in parallel with the direction of wind such that it experiences a 0.5 N drag force. If the backboard’s Reynolds Number (Re) is 4 x 10^4 and its projected area is 0.75 square meters, what is the wind speed in m/s? Assume the density of air is 1.22 kg/m^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
For flat plates placed parallel with the flow and have
$$10^4 < Re<5\cdot10^5$$
:
$$$C_D=\frac{1.33}{\sqrt{Re}}$$$
where
$$Re$$
is the Reynolds Number.
First, let’s find the backboard’s coefficient of drag. For flat plates placed parallel with the flow and have
$$10^4 < Re<5\cdot10^5$$
:
$$$C_D=\frac{1.33}{\sqrt{Re}}$$$
where
$$Re$$
is the Reynolds Number.
$$$C_D=\frac{1.33}{\sqrt{4\cdot 10^4}}=\frac{1.33}{200}=0.00665$$$
Next, 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.
$$$0.5N=\frac{1}{2}(1.22\frac{kg}{m^3}) U^{2}(0.00665)(0.75m^2)$$$
Solving for velocity:
$$$U^{2}=\frac{1N}{0.00608\frac{kg}{m}}=\frac{1\frac{kg\cdot m}{s^2}}{0.00608\frac{kg}{m}}=164.34\:m^2/s^2$$$
Thus,
$$$U=\sqrt{164.34m^2/s^2}=12.8\:m/s$$$
12.8 m/s
Time Analysis
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021. Optimizing for Drag
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