Question 10.16: Figure 10.18 illustrates that for a specific range of R, for...

(a)–(b) The frontal area is used to compute the drag force for the golf ball, and the drag force is determined by applying Equation 10.12 F_{D} = C_{D} \frac{1}{2} \rho v^{2}A . The rate of deceleration for the golf ball is determine by applying Newton’s second law of motion in the x-direction. The pressure and viscous forces in the direction of velocity (the x-direction) are represented by the drag force. And there is no gravitational force component along the x-axis because the weight of the golf ball is assumed to be negligible; thus, \sum{F_{x}} = F_{D} = ma_{x} .

(a) The drag force and the rate of deceleration of a standard (roughened/dimpled) golf ball are determined as follows:

slug: = 1 lb \frac{sec^{2}}{ft}                                         V: = 155 mph = 227.333 \frac{ft}{s}

\rho : = 0.0023768 \frac{slug}{ft^{3}}                                 \mu: = 0.37372 . 10^{-6} lb \frac{sec}{ft^{2}}

D: = 1.68 in = 0.14 ft                                 A_{front}: = \frac{\pi .D^{2}}{4} = 0.015 ft^{2}

R: = \frac{\rho .V .D}{\mu} = 2.024 \times 10^{5}                                 W: = 1.62 oz = 0.101 lb

g: = 32.174 \frac{ft}{sec^{2}}                               m : = \frac{W}{g} = 3.147 \times 10^{-3} \frac{S^{2}.lb}{ft}

Guess value:                               F_{Drough}: = 1 lb                               a_{rough}: = 1 \frac{ft}{sec^{2}}

Given

F_{Drough} = C_{Drough} \frac{1}{2} \rho.V^{2} . A_{front}                               – F_{Drough} = – m . a_{rough}
\left ( \begin{matrix} F_{Drough} \\ a_{rough} \end{matrix} \right ) : = Find (F_{Drough}, a_{rough})

F_{Drough} = 0.227 lb                               a_{rough} = 72.103 \frac{ft}{s^{2}}

(b) The drag force and the rate of deceleration of a smooth golf ball are determined as follows:

Guess value:                               F_{Dsmooth}: = 1 lb                               a_{smooth}: = 1 \frac{ft}{sec^{2}}

Given

F_{Dsmooth} = C_{Dsmooth} \frac{1}{2} \rho.V^{2} . A_{front}                               – F_{Dsmooth} = – m . a_{smooth}
\left ( \begin{matrix} F_{Dsmooth} \\ a_{smooth} \end{matrix} \right ) : = Find (F_{Dsmooth}, a_{smooth})

F_{Dsmooth} = 0.482 lb                               a_{smooth} = 153.22 \frac{ft}{s^{2}}

Therefore, because for a given velocity (and thus a given R), the drag coefficient, C_{D} for the smooth golf ball is about 2 times that for the roughened golf ball, the corresponding drag force for the smooth golf ball is also about 2 times that for the roughened golf ball, as the drag force, F_{D} is directly proportional to the drag coefficient, C_{D} (see Equation 10.12 F_{D} = C_{D} \frac{1}{2} \rho v^{2}A ). Furthermore, the rate of deceleration of the smooth golf ball is also about 2 times the rate of deceleration of the roughened golf ball, as the rate of deceleration is directly proportional to the drag force, F_{D} (or summation of forces, in general) (Newton’s second law of motion) for a given mass.