Golf Swing Physics
2. The swing
Guest article by Rod White

December 2008
DoublePendulum
Swing
Now
we move on the model of the swing itself. Conventionally, physicists
model it as a double pendulum: one pendulum tacked onto the end of
another.
In
this animation we show a double pendulum composed of a large blue mass
representing the arms of the golfer, and a smaller red mass
representing the clubhead. There is also a thin string connecting the
clubhead to the hub.
Initially the system rotates
at a constant speed, and the string stops the clubhead from swinging
out. No energy is put into the system for this model; the system is
rotating to begin with, and continues to rotate under its own inertia.
Since it is held in
the same shape by the string, the moment of inertia stays the same so
the speed stays the same. An essential feature
of the animations is that no external forces or torques are applied to
the system,
and the connection between the two arms of the pendulum is simply a
hinge.
Now watch what happens when the string is removed. As the red mass
swings out, the blue mass slows down, and the red mass
accelerates. When the system is fully extended (near the
point
where the head would hit the golf ball) the blue mass comes to a dead
stop.
For the physicist, this
is the magic moment! The blue
mass coming to a dead stop means that all of the kinetic energy that
was in the blue mass has been transferred to the red mass (the club
head); the swinging process is 100% efficient.
Most importantly,
there were no external forces applied to make this happen – the system
rearranged itself under its own inertia. If this was a
golf swing, the
golfer would not have to make any effort to make it happen.
In
this example, I chose the value of the red mass so the blue ball would
come to a dead stop. In general this does not happen – as we will see
with the next few animations.

The second
animation shows the double pendulum again, but with a heavy clubhead
mass  heavier than before and therefore heavier than optimum.
This
time, the blue mass slows down more quickly than before, and when the
system is fully extended the blue mass actually goes backwards (this
tells us that some of the kinetic energy remains in the blue mass).
This animation is similar to the twoball collision when a small mass
collides with the big mass and recoils.

The
third
animation shows the double pendulum again, but with a light clubhead
mass. The distribution of mass is closer to a real golf
swing than in the previous examples. This time, the blue mass slows
down but not as much as before,
and when the system is fully extended, the blue mass is still moving
forwards. Again some kinetic energy remains in the blue mass.
This
animation is similar to the twoball collision when a large mass
collides with the little mass and keeps moving.
Now
although the motion of the double pendulum is complicated, we can
simplify the problem by considering the initial motion when the string
is in place, and later when it is fully extended. If we think about
these two snapshots, and ignore the complicated movements as the red
ball swings out, we can see a strong similarity between the
twoball collision and the golf swing as it unfolds. Once
again we see
the Goldilocks principle at work. The red mass representing the club
head can
be too big, too small or just right.
We
won’t show you the curve just yet, but it is very similar to the
efficiency
curve for the 2ball collision.
The
double pendulum is much more
complicated than the simple collision, so the condition for a 100%
efficient
swing is much more complicated that the condition for a 100% efficient
collision (which was M_{1 }= M_{2}).
We
give the equation for
the optimum clubhead
mass in the math
note below.
In
addition to the mass of the blue ball representing
the arms (M_{1}),
the
condition depends on the length of the two arms of the pendulum
representing
the golfer's arm and the shaft of the club (L_{1}
and L_{2}),
and the angle
between the arm and the shaft (θ,
which is
the wrist cock angle).
The equation also tells us how to reduce the optimum clubhead mass and
increase the efficiency of the swing:
 Lengthen the shaft, and/or
 Increase the wristcock angle.
In fact the whole golf stroke: the unfolding of the swing, and the
clubball collision looks a lot like a threeball collision.
The
equations of motion for the system are still complicated, but if we
restrict ourselves to thinking only about the energy and momentum at
these two points (the point the string is released, and the point of
full extension), the system can be understood relatively easily.
Remember
that the energy and momentum are the same throughout the swing and
through impact. The momentum must be the same; that's physics. The
energy is the same because we have not yet considered energy losses in
the collision (that is, the COR is 1.00 for our model).

Math note
Here
is the basis of the mathematical model I am using. The diagram
shows how I’ve defined the various angles, shaft lengths, etc.
We are only considering the instantaneous motion of the system at the
time of two snapshots:
 The moment just before the string is released. At
this moment, everything is rotating; nothing is flying out from the
centre of rotation  though it will the instant after the string is
removed.
 The moment of impact, calculated to correspond to
"complete release"  a straight line from the golfer's center of
rotation down the arms and the club to the ball.
By picking these two snapshots, and having the
system
move under its own inertia, the system is very much simpler to analyze.
Mainly,
there are no radial terms to
consider; only tangential terms involving angular velocities.
Only two things you really need to note here, both of them at the
moment the string is released:
 The downswing angle ß, which measures
how far the arms move, and
 The wrist cock angle θ, which
measures the angle between the arms and the shaft of the club.
These are the simplified equations for the swing. (We
still have not included the collision with the ball at this
point.) They might still look horrific, but they are much
simpler than the full equations, and they can be solved algebraically
to give a simple expression for the condition for 100% efficiency.
To
read the equations you need to know that terms of the form mass
x length^{2}
measure a quantity called the moment of inertia. Moment of inertia
measures the resistance to rotational forces (torques) in the same way
that mass is a measure of resistance to a translational force.
Everybody knows " F=ma".
Well the rotational analogue is " torque =
moment of inertia x angular
acceleration".
The Greek symbols with dots over them are angular velocities.
Kinetic energies
have the form: 
moment of inertia x (angular velocity)^{2} 
Momenta have the
form: 
moment of inertia x angular velocity 
The subscripts on the angles, i
and f,
indicate initial and final velocity.
First, we require the kinetic energy and angular momentum to be the
same for the two
snapshots of the swing.
Note that the angular velocity of the two masses, betadot and
alphadot, is the same in the first snapshot.
As we did with the equivalent equations for the
collision, we can solve these two equations to calculate the final
angular
velocities for the two masses. This is how the swing efficiency curve
was
calculated. However, it is more interesting to find out the condition
for a
100% efficient swing. Ideally,
we want the blue mass to
come to a dead stop at impact, and this leads to a further
simplification, as given by the second set of equations.
This pair of equations is much simpler and has
one
nontrivial solution:
m_{1}L_{1}^{2}
+ m_{2}R^{2} =
m_{2} (L_{1} + L_{2})^{2}
The ‘bottom line’ of this analysis is that the
swing is 100% efficient when the
moment of inertia of the system in the first snap shot equals the
moment of
inertia
of the club in the second snap shot.
In words, this last equation reads, the moment
of inertia of the
initial system (with the string in place) equals the moment of inertia
of the
club when fully extended. This is analogous to the condition that the M_{1} = M_{2} in
the twoball collision.
Most importantly, this observation tells us
that the energy transfer
occurs because of the changing moment of inertia of the club as it
swing out.
We can now calculate the optimum clubhead mass
for this passive model
by substituting the value for R,
which is given by the cosine law (from diagram a
above):
R^{2}
= L_{1}^{2}
+ L_{2}^{2}  2L_{1}L_{2}
cos θ
Which then gives us the required condition for optimum clubhead mass:
m_{2}
= 
m_{1}
2[1
+ cos(θ)] 
L_{1}
L_{2} 

Ice skater analogy of the golf swing
The most important picture to keep in mind is the transfer of energy
with unfolding.
Think of this as the reverse of the ice skater effect. The ice skater
initially has the arms extended and works to move them as fast as
possible. The skater then pulls the arms in close to the body. This
causes the body speed up so that the skater spins fast. It is important
to note that the hands actually slow down in this process, but we only
watch the body. The body is spinning faster, true. But, because the
hands' radius of rotation is shorter than before, the velocity
of
the hands is lower than before.
In the golf swing, the opposite occurs. The golfer initially holds the
club close to the body using wrist cock, and works very hard to build
up the kinetic
energy in the body and arms. The golfer then allows the club to swing
away from the
body, so the body and arms slow down and the club speeds up. I
emphasise the
term “allows”,
because it can be an entirely passive process; the golfer does not have
to make the club swing out – it happens naturally.
In summary...
 The ice skater:
 First builds speed in arms,
 then folds the arms close to the body,
 to speed up the body
 The golfer:
 First builds speed in body,
 then unfolds the club from the body,
 to speed up the club.

Model golf
stroke = pendulum + collision
We are now in a position to understand the basic principles
underlying the golf stroke. There are two parts in the golf stroke: the
swing and the collision. This graph is idealised – we discuss the
nonideal bits shortly.
Part 2:
the collision between the club and ball is exactly as we described
for the twoball collision. The efficiency is represented by the left
hand of the two dotted curves, and the maximum
efficiency occurs when the two masses are equal (golf ball mass is
close to 46 g).
Part
1: the swing is where energy is transferred from the
golfer's arms and torso to the clubhead. This is represented by the
right hand dotted curve. The right hand
dotted curve for the swing efficiency is not exactly the same shape as
that for
the collision but very close. The exact position of the curve depends
on
a variety of factors that we haven’t considered in detail yet  arm
mass, arm
length, shaft length, wristcock angle, and coefficient of restitution.
The maximum
efficiency in this case for
the swing alone requires the club head mass to be about 1
kg
(according
to the complex equation given in the math notes).
As with the 3ball collision, the overall efficiency
is given by the
product of the two separate efficiency curves.
We can now see why the clubhead is so much heavier than the ball. To
maximise the efficiency of the total golf stroke, it must be midway
between
the optimum for the swing (1 kg) and the optimum for the
collision
(46 g). The right hand dotted curve is not exactly the same
shape as
the
corresponding curve in the threeball collision – but very similar.
Anything we can do to move the swing curve further left improves the
efficiency of the golf stroke. The only two factors included in this
model that are within the golfer’s
control are:
 The wristcock angle (make smaller). That is why many
of the longest hitters have a very acute lag angle well into the
downswing. (We shall see this in the videos later.)
 Shaft length (make longer). But shaft length is
limited by the rules, and in many cases further limited by the
individual golfer's ability to control a long club. Long drivers
routinely use long shafts, but they too are limited by their
competition rules.
The model we have developed so far suggests that the optimum club head
mass is about 150 g. In practice, real clubhead masses are a
little
heavier. This is because, unlike our model (which was entirely
passive), in a real golf swing, the golfer is constantly working to
build up energy and momentum. Work done early in the swing is
transferred efficiently, just as our model suggests. Work done late in
the swing, when the wrists are uncocked, is transferred less
efficiently. The overall efficiency is therefore less than
our model predicts – in effect the average wrist cock angle is greater
than we have assumed. There are other
factors too:
 Energy lost due to the compression of the ball
– as described by
the coefficient of restitution – means that the swing is less
efficient, and
the clubhead mass must be heavier to give the maximum ball speed.
 Our
muscles also
produce greater torques when they are moving slowly and this favours a
slightly
heavier mass.
 Finally,the optimum mass changes with the value of m_{1} that represents
the mass
of the golfer’s arms, and our value of 1 kg on a 0.66 m pendulum arm may
not be
representative of the moment of inertia of a typical golfer.

Last modified  Apr 10, 2010
