Golf Ball in Flight A 1.68-in.-diameter golf ball is driven off a tee at 70 mph. Determine the drag force acting on the golf ball by (a) approximating it as a smooth sphere and (b) using the actual drag coefficient of 0.27. Approach To find the drag force in part (a), we will begin by calculating

Question

Golf Ball in Flight

A 1.68-in.-diameter golf ball is driven off a tee at 70 mph. Determine the drag force acting on the golf ball by (a) approximating it as a smooth sphere and (b) using the actual drag coefficient of 0.27.

Approach
To find the drag force in part (a), we will begin by calculating the Reynolds number [Equation (6.6)]

[latex]R_{e}=\frac{\rho vl}{\mu } [/latex] (6.6)

with the density and viscosity of air listed in Table 6.1.

Table 6.1 Density and Viscosity Values for Several Gases and Liquids at Room Temperature and Pressure

	Density,ρ		Viscosity, μ
Fluid	[latex]kg/m^3[/latex]	[latex]slug/ft^3[/latex]	kg/(m . s)	slug/(ft . s)
Air	1.20	[latex]2.33×10^{-3}[/latex]	[latex]1.8×10^{-5}[/latex]	[latex]3.8×10^{-7}[/latex]
Helium	0.182	[latex]5.53×10^{-4}[/latex]	[latex]1.9×10^{-5}[/latex]	[latex]4.1×10^{-7}[/latex]
Freshwater	1000	1.94	[latex]1.0×10^{-3}[/latex]	[latex]2.1×10^{-5}[/latex]
Seawater	1026	1.99	[latex]1.2×10^{-3}[/latex]	[latex]2.5×10^{-5}[/latex]
Gasoline	680	1.32	[latex]2.9×10^{-4}[/latex]	[latex]6.1×10^{-6}[/latex]
SAE 30 oil	917	1.78	0.26	[latex]5.4×10^{-3}[/latex]

If Re for this situation is less than one, then it will be acceptable to apply Equation (6.16).

[latex]F_{D}\approx 3\pi \mu dv [/latex] (Special case for a sphere: Re < 1) (6.16)

On the other hand, if the Reynolds number is larger, that equation can’t be used, and we will instead fi nd the drag force from Equation (6.14)

[latex]F_{D}=\frac{1}{2} \rho Av^2C_{D} [/latex] (6.14)

with [latex] C_{D}[/latex] determined from Figure 6.22. We will use this latter approach to find the drag force in part (b) also.

Accepted Answer

(a) In dimensionally consistent units, the golf ball’s speed is

[latex]v =\left(70\frac{mi}{h} ight)\left(5280\frac{ft}{mi} ight)\left(\frac{1}{3600}\frac{h}{s} ight) [/latex]

[latex]=102.7\left(\frac{\cancel{mi}}{\cancel{h}} ight) \left(\frac{ft}{\cancel{mi}} ight) \left(\frac{\cancel{h}}{s} ight) [/latex]

[latex]=102.7\frac{ft}{s}[/latex]

With the diameter being [latex] d = 1.68 in. = 0.14 ft[/latex], the Reynolds number becomes

[latex]Re=\frac{(2.33 imes 10^ {-3}slug/ft^3)(102.7ft/s)(0.14 ft)}{3.8 imes 10^{-7}slug/(ft.s)} \longleftarrow [R_{e}=\frac{ ho vl}{\mu }][/latex]

[latex]= 8.813 imes 10^4\left(\frac{\cancel{slug}}{\cancel{ft^3}} ight)\left(\frac{\cancel{ft}}{\cancel{s}} ight) (\cancel{ft}) \left(\frac{\cancel{ft}.\cancel{s}}{\cancel{slug}} ight)[/latex]

[latex]= 8.813 imes 10^4[/latex]

Since this value is much greater than one, we can’t apply Equation (6.16) or Equation (6.15)

[latex]C_{D}\approx \frac{24}{Re} [/latex] (Special case for a sphere: Re < 1) (6.15)

for the drag coefficient. Referring to Figure 6.22, this value for Re lies in the flat portion of the curve where [latex]C_D ≈ 0.5[\latex]. The frontal area of the ball is

[latex]A= \frac{\pi (0.14 ft)^2}{4} \longleftarrow \left[A=\frac{\pi d^2}{4} ight] [/latex]
[latex]= 1.539 imes 10^{-2}ft^2[/latex]

The drag force then becomes

[latex]F_D=\frac{1}{2}\left(2.33 imes 10^{-3}\frac{slug}{ft^3} ight)( 1.539 imes 10^{-2} ft^2 )\left(102.7\frac{ft}{s} ight)^2(0.5)[/latex]

[latex]= 9.452 imes 10^{-2}\left(\frac{{slug}}{\cancel{ft^3}} ight) (\cancel{ft^2})\left(\frac{\cancel{ft^2}}{s^2} ight) \longleftarrow \left[F_{D}=\frac{1}{2} ho Av^2C_{D} ight] [/latex]

[latex]= 9.452 imes 10^{-2}\frac{slug.ft}{s^2} [/latex]

[latex]= 9.452 imes 10^{-2} lb[/latex]

where the final grouping of USCS dimensions is equivalent to the pound [Equation (3.3)].

[latex]1 lb = (1 slug)\left(1\frac{ft}{s^2} ight)=1\frac{slug.ft}{s^2} [/latex] (3.3)

(b) With [latex]C_{D} = 0.27 [/latex] instead, the drag force is reduced to

[latex]F_D=\frac{1}{2}\left(2.33 imes 10^{-3}\frac{slug}{ft^3} ight)( 1.539 imes 10^{-2} ft^2 )\left(102.7\frac{ft}{s} ight)^2(0.27)[/latex]

[latex]= 5.104 imes 10^{-2}\left(\frac{{slug}}{\cancel{ft^3}} ight) (\cancel{ft^2})\left(\frac{\cancel{ft^2}}{s^2} ight) \longleftarrow \left[F_{D}=\frac{1}{2} ho Av^2C_{D} ight] [/latex]

[latex]= 5.104 imes 10^{-2}\frac{slug.ft}{s^2} [/latex]

[latex]= 5.104 imes 10^{-2} lb[/latex]

Discussion
The simplification of treating the golf ball as a smooth sphere neglects the manner in which the dimples change the airflow around the ball, lowering its coefficient of drag. By reducing [latex] C_{D}[/latex], the ball will travel farther in flight. The aerodynamic behavior of golf balls is also significantly influenced by any spin that the ball might have when it is driven from the tee. Spin can provide extra lift force and enable the ball to travel farther than would otherwise be possible.

[latex]F_{D} = 9.452 imes 10^{-2} lb [/latex] (smooth sphere)
[latex]F_{D}= 5.104 imes 10^{-2} lb [/latex] (actual)

Question 6.7: Golf Ball in Flight A 1.68-in.-diameter golf ball is driven ...

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