Chapter 8, Problem 8.102P

To determine

(a)

To analyze:

Impact point without spin.

Answer to Problem 8.102P

Impact point without spin:

$\Delta x_{impact}\approx 1250\text{ft}$

Explanation of Solution

Given:

Golf ball diameter, $D=1.7\text{in}$

Inlet velocity, $V_0=250\text{ft}/s$

Upward angle, ${\theta }_0={20}^{\circ }$

For sea − level air,

Take $\rho =0.00238slug/{\text{ft}}^3$

If drag and spin is neglected, we will use classical particle physics to predict the distance traveled:

When,

$t=\frac{V_0\sin \theta }{g}$

$x=V_{ox}\left(2t\right)=\frac{V_0^2}{g}.2\sin {\theta }_0\cos {\theta }_{\circ }$

Substitute the given values,

$\Delta x_{impact}=\frac{{\left(250\right)}^2}{32.2}\sin {20}^{\circ }\cos {20}^{\circ }\approx 1250\text{ft}$.

To determine

(b)

To analyze:

Impact point with backspin of $\text{7500}r/\min$.

Answer to Problem 8.102P

Impact point:

$\Delta x_{impact}\approx 1570\text{ft}$

Explanation of Solution

Given:

Golf ball diameter, $D=1.7\text{in}$

Inlet velocity, $V_0=250\text{ft}/s$

Upward angle, ${\theta }_0={20}^{\circ }$

Estimate the lift of spinning ball:

$\omega =7500\left(\frac{2\pi }{60}\right)=785\frac{rad}{s}$ ;

$\frac{\omega R}{U}=\frac{785\left(\raisebox{1ex}{$1.7$}\!\left/ \!\raisebox{-1ex}{$24$}\right.\right)}{250}\approx 0.22$

$C_L=0.02$

Average lift:

$L\approx C_L\left(\raisebox{1ex}{$\rho $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)V^2\pi R^2$

$L\approx \left(0.02\right)\left(\raisebox{1ex}{$0.00238$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right){\left(250\right)}^2\pi {\left(\frac{1.7}{24}\right)}^2\approx 0.0231lbf$

Equations of motion in z and x directions and average values to be assumed:

$m\ddot{x}=-L\sin \theta$

With

${\theta }_{avg.}\approx {10}^{\circ }$

${\ddot{x}}_{avg}\approx -\frac{0.023}{\raisebox{1ex}{$0.102$}\!\left/ \!\raisebox{-1ex}{$32.2$}\right.}\sin {10}^{\circ }\approx -1.3\frac{\text{ft}}{s^2}$

Then

$x\approx V_{ox}t-\frac{1}{2}{\ddot{x}}_{avg.}{t}^2$

Where,

$t=2x$ (time to reach the peak)

$m\ddot{z}=L\cos \theta -W$

Or:

${\ddot{z}}_{avg.}\approx \frac{32.2}{\raisebox{1ex}{$0.102$}\!\left/ \!\raisebox{-1ex}{$32.2$}\right.}\cos {10}^{\circ }-{32.2}^{\circ }\approx -25\frac{\text{ft}}{s^2}$

Using these crude estimates, the travel distance is estimated:

$t_{peak}=\frac{V_{oz}}{a_z}=\frac{250\sin {20}^{\circ }}{25}\approx 3.4s$

$t_{impact}=2.t_{peak}=6.8s$

$\Delta x_{impact}=V_{ox}-\frac{1}{2}{a}_x{t}^2$

$\Delta x_{impact}=250\cos {20}^{\circ }\left(6.8\right)-\frac{1.3}{2}{\left(6.8\right)}^2\approx 1570\text{ft}$

These are 400 − yard to 500 − yard drives.

So, it would be nice, at least during teeing off, to have zero viscous drag.