Golfer Meets Physicist

Classical Mechanics Level 5

This is the first problem that I am posting.

More often than not, during the study of the motion of particles on a plane, the effects of aerodynamic forces are ignored. In this situation, an attempt is made to understand the effect of aerodynamic forces. Here, the situation is that of a golfer wanting to predict how far she is capable of hitting the ball, given the knowledge of the initial speed she can impart to it, using her club, and its angle of projection to the horizontal.

Consider a particle of unit mass, projected at an angle of 11.5 degrees to the horizontal , from the origin of the X-Y plane, with an initial speed of 70 m/s . The particle moves in the presence of an ambient gravitational field, acting in the negative Y direction, where $g = 9.8 m/s^2$ . The velocity of the particle at any instant is denoted by $\vec{v}$ . The drag and lift forces acting on the particle are as such:

$\vec{v} = v_x \hat{i} + v_y \hat{j} \\$

$\vec{F_{drag}} = -C \mid{\vec{v}}\mid \vec{v} \\$

$\mid{\vec{F_{lift}}\mid} = K \mid{\vec{v}}\mid^2 \\$

The direction of $\vec{F_{lift}}$ is perpendicular to that of the drag force and is directed away from the ground (X-axis).

Where, C = 0.00382; K = 0.0025477 .

The objective of this question is to find the range of this projectile. Let this value be $R_1$ Also compute the range of the projectile in the absence of aerodynamic forces (both drag and lift). Let this value be $R_2$ . Enter your answer as $\frac{R_1}{R_2}$ . The answer might be surprising.

Bonus: What are your thoughts about ignoring or accounting for the effect of aerodynamic forces? Also, how different would your answer be if lift force was not considered?

The answer is 1.1673.

1 solution

Steven Chase
Mar 21, 2019

Simulation code is attached. With both aerodynamic forces, the particle spends 5.558 seconds in the air and has a range of 228.047 meters. Without aerodynamic forces, the particle spends 2.848 seconds in the air and has a range of 195.366 meters. The desired ratio is therefore about 1.167.

If drag is considered and lift is ignored, the particle spends 2.561 seconds in the air and has a range of 134.170 meters. This gives the worst range, as expected. So the lift force really helps to increase the range.

import math

# Constants

theta = (11.5 / 180.0) * math.pi
v0 = 70.0
g = 9.8

C = 1.0 * 0.00382   # Can run with or without drag/lift
K = 0.0 * 0.0025477

m = 1.0

############################################

# Initialize position, velocity, acceleration
# Use microsecond time step

x = 0.0
y = 0.0

xd = v0 * math.cos(theta)
yd = v0 * math.sin(theta)

xdd = 0.0
ydd = 0.0

dt = 10.0**(-6.0)
t = 0.0

############################################

while y >= 0.0:

    x = x + xd * dt      # Euler integration
    y = y + yd * dt

    xd = xd + xdd * dt
    yd = yd + ydd * dt

    v = math.hypot(xd,yd)

    udx = xd/v          # vector in direction of velocity
    udy = yd/v

    uLx = -udy          # vector normal to velocity
    uLy = udx

    Fgx = 0.0                  # Gravity
    Fgy = -m*g

    Fdx = -C*(v**2.0)*udx   # Drag
    Fdy = -C*(v**2.0)*udy

    FLx = K*(v**2.0)*uLx    # Lift
    FLy = K*(v**2.0)*uLy

    Fx = Fgx + Fdx + FLx    # Net force
    Fy = Fgy + Fdy + FLy

    xdd = Fx/m              # Acceleration
    ydd = Fy/m

    t = t + dt

print t,x

Thank you for the solution!

Karan Chatrath - 2 years, 2 months ago

Thanks for the problem. It was a nice one

Steven Chase - 2 years, 2 months ago

Golfer Meets Physicist

The answer is 1.1673.

1 solution

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