Physical principles for the golf swing
Energy and Momentum
Dave Tutelman
 October 3, 2022
At first, I was not going to have this chapter in the tutorial. I changed my mind for a bunch of reasons:
 One of the goals of the tutorial is to allow you to read
important articles in the field, so you can keep up with the science,
not just carry away what I thought I knew when I wrote it. Some of the
papers that should be most interesting to instructors and golfers are
intimately entwined in the subject of energy or momentum.
 Everybody involved in golf uses the terms "energy", "momentum", "friction",
and probably others introduced here. It would be nice if you had a
precise understanding of what the words mean, not just a vague,
colloquial idea.
 If you wanted to know what gravity or shaft bend might contribute to clubhead speed, this is the first place to look.
So let's get started!
Among the most famous laws of physics are the "conservation laws",
commonly stated as:
 Conservation of energy =
Energy can neither be
created nor destroyed; it is a constant.
 Conservation of momentum
= In a closed system, total
momentum is a constant.
Let's put some precision and understanding into these statements.
Let's
understand what energy and momentum are, and how to think about them in
the context of golf. That context keeps things relatively simple. We
can stick to simple Newtonian physics, and not worry about relativity,
quantum physics, or even forms of energy beyond mechanical. (We will
talk a bit about other forms of energy, but not in detail and not
quantitatively.)
Energy
Work
Well, then, within the context of golf physics, what is energy anyway? Energy is the
ability to do work!
Not helpful? We need to know what "work" is? You're absolutely right 
and physics has a precise and quantifiable definition of work.
Work
is a distance that a body moves times the force in the
direction the body is moving.
Let's look at some examples of work and nonwork under this definition.
 If I push against a wall, I am not doing work. I may be
exerting a force and expending a lot of effort, but nothing moves so it
is not work
under the definition.
 If I push against a heavy door and it opens, I am doing
work. The force I exert (the push) times the distance the door moves
while I push with that force is the work I did... or, put another way,
the energy I expended.
 If I push against that same door but it is locked, am I
doing work? If you said "yes", go back to the beginning of this section
and start over.
 Here is a tug of war between the green team and the red
team. Let's look at it in terms of work.
 The teams are well
matched. They both exert a force of exactly F. Nothing
moves, they just pull. No work is being done.
 The red team puts forth a little more effort. The forces
are still both F
within a fraction of a percent, but the red force is slightly larger.
The resulting acceleration may be small, but the red team incrementally
pulls the green team over the line. The red team did work (expended
energy) equal to F
times the distance the rope moved. The green team had work done to them equal to the
same amount, F
times the distance the rope moved.

Wait! "Work done to them"? What does that
mean? It means the green team was receiving
energyfrom
the red team. In this case, the energy received was not useful,
but there are cases where the work done to an object is recoverable
energy. Sisyphus spent eternity learning this. He did work pushing the
rock up the hill. As he neared the top of the hill, the rock would get
away from him and roll back to the bottom.
But a mass moving at speed is a form of energy we will see later,
called "kinetic energy".
The Sisyphean myth can be described in terms of physics and energy.
Sisyphus did work on the rock, pushing it up the hill. The fact that it
was near the top of the hill meant that it had absorbed energy from
Sisyphus, and now had a lot of "potential
energy",
something else we'll see later. With Sisyphus no longer pushing it
upward, it rolled downward, picking up speed and converting that
potential energy to kinetic energy.
So, under the right circumstances, work done to an object may be
recoverable or partly recoverable.
When work is not recoverable, it is converted to some other form of
energy. For example, work done against the force of friction is converted into heat,
another form of energy. That is why you can start a fire by rubbing two
wooden sticks together. At the end of this section, we'll look very
briefly at nonmechanical forms
of energy. But first, let's look at the most important kinds of
mechanical energy, potential energy and kinetic energy.

Potential energy
When you do work on an object and that work is recoverable but has not
yet been recovered, the object has acquired potential energy.
(Let me be a bit more specific  some might say anal  about that
definition. We're talking about mechanical work and mechanical
recovery. It is certainly what we mean in the golf context, and most
seem to use the term that way in general.) Let's look at two important
examples, gravity and a spring. Each example teaches us something we
can use later in golf physics.
Gravity
We
have a ball
of
weight W
sitting on the
lower shelf in the picture. We pick it up and put it on the upper
shelf, a height h
higher. How much potential energy do we add to the ball?
The potential energy is equal to the amount of work we do against
gravity to lift the ball to the upper shelf. Let's see...
 We are lifting the ball, which will take an upward force of
W
to oppose the weight.
 We apply a force of W to the ball
through a distance equal to the height h.
 Therefore, the work done is W*h.
So the added potential energy is equal to Wh.
But we glossed over the fact that the path of the ball is not just
"straight up and stop"; the ball has to move the the shelf as well. It
turns out that it doesn't
matter. Let's look at the problem again, with a specific path,
to see why we can ignore the path.

Here is the same example, but
with a path made of four segments. We'll call the height of each segment y. Let's look at each segment to see how its
work is Wy.
 This part of the path is obviously Wy, because we
are lifting straight up a distance y with a force of W.
 Now the path is off vertical by an angle of a. That means
the path is longer; it is y times 1/cos a. So
wouldn't the work be greater? Remember, work is the distance times the
force along
the path. W
is a vertical force, and the path is not vertical. So let's get the
components of W
that are parallel and perpendicular to path B. Turns out that the
component in the direction of the path is smaller than W; it is W multiplied by cos a. So for
path B, the work is still W times y, the
height of path B, not the length of path B. That's because the length
is divided by the same amount that the force component is multiplied by
 so they cancel out.
 The path is horizontal, but the force is vertical. So there
is no component of the force in the direction of the path. No work is
done while traversing path C.
 The force is upward but the path is downward. The distance the ball moves is y, and that minus sign is important It means
that negative work is being done on the ball. Whatever extra work was
done getting the ball well above the upper shelf is given back when the
ball is put down on the shelf (along path D).
These are the basic ways the path can be something different from
straight up. (What about a curved path? It is composed of small
segments of the style of path B.) And none of them matter. The work is
"adjusted" for each
path so that the final work done is just Wy on that path, and a total work of Wh.
So we can state that the potential energy of any object of weight W at a height h above our zero
reference is exactly Wh.

Spring
A spring, so matter what its size,
shape, or intended use, is a device that can exert a force against
being deflected, that force being proportional to the amount that it is
deflected. Its behavior can be described very concisely by its "spring
constant" K,
which is the force divided by the deflection. K is the same
value at any deflection, since the force and deflection are
proportional. (Of course, this breaks down once a spring is deflected
beyond its useful range, but for this example we'll assume we are
staying in the useful range.)
We
have a coil spring with spring constant K. If we
compress the spring an amount x, how much
potential energy have we given to the spring?
Because of the definition of the spring constant, we know that, for any
x,
F = Kx
So the force changes as we press the spring. All we have to do
to calculate work is add up all the tiny contributions of the force at some
value of x,
times the tiny distance to the next x we look at.
Adding together miniscule pieces like this is the job of integral
calculus, which gives us a very simple formula. (If you ever took
integral calculus 101, this one
was covered in the first week.)
Work
= ∫
F dx = ∫ Kx dx
Work
= ½ Kx^{2}
The potential energy of the compressed spring is the work done to
compress it, so the potential energy is ½Kx^{2}

Kinetic energy
As we noted earlier, a mass at speed has
energy due solely to its velocity. This form of energy is called
kinetic energy, and it is given by the formula:
KE = ½ mv^{2}
Where does this come from? Look
at it this way. In order to move a weight upwards a distance Wh, we had to do
work equal to Wh.
That work turned into potential
energy. That suggests a potential for somthing, it could
potentially become something, maybe some other form of energy. So let's
see what would happen if we tipped the ball off the upper shelf and let
it fall to the lower shelf. The falling ball has the potential to acquire some
velocity v.
Now let's remember what "conservation of energy" means. Energy cannot
be created nor destroyed. All that potential energy has to turn into
kinetic energy, or the work done lifting the ball (potential energy)
would not be conserved. The ball's mass m moving at a
velocity v
has to have the same energy as the potential energy was before the ball
was pushed off the shelf.
When we do the math, we find that the formula for kinetic energy has to be the one given above, or
energy is not conserved. So, reiterating once again:
KE = ½ mv^{2}

For those who are interested, here's a proof
that conservation of energy leads to our formula for kinetic energy.
It's here mostly to show the beauty and consistency of physics, but you
don't need to know it to understand golf swing biomechanics.
We'll start out by finding the velocity of the ball v when it
reaches the lower shelf.
v = at
This is just the relationship of acceleration and velocity. If you're confused by this, better go back and look at the basics. It's important.
v = gt (eq 1)
where g is
the acceleration of gravity. But we don't know t. Let's find
it. We know that distance is the integral of velocity over time, so
h = ∫ g dt = ½ g t^{2}
We can solve this for t, and plug back into equation 1.
v
= gt = sqrt ( 2hg
)
(eq 2)
But we know already that
PE = Wh = mgh
because the weight is the " mg force". And we can solve equation 2 for hg,
which we will substitute back into potential energy. This will give us
the kinetic energy, because we started out with conservation of energy
 the two energies must be the same. Solving equation 2 for hg:
hg = ½ v^{2}
Plugging back into E=mgh we get
KE = ½ mv^{2}
Q.E.D.

A couple of golf examples
Before we go further, let's look at some simple questions about golf
that might be answered by what we know now. In both cases, we don't
have enough knowledge yet to get the realworld answer, but we can
certainly answer a simpler question. It also happens to be the question so often framed by people who
know just enough physics to be dangerous.
 How much clubhead speed can we get from gravity working on the clubhead?
 How much clubhead speed can we get by recovering shaft bend as a "kick velocity"?
Gravity
We are going to do a simplified analysis of how much of our clubhead
speed comes from gravity. We are simplifying in that we are only
considering gravity's action on the clubhead. There is also a small
contribution from the weight of the shaft and a significant
contribution from the weight of body parts (mostly the arms). But we
don't have the tools yet to do those estimates. This would require both
a knowledge of angular energy and the dimensions and inertial
properties of some significant body parts.
We will solve the problem by finding:
 How much potential energy does the clubhead have at the top of the backswing?
 If all that potential energy is converted to kinetic energy
by the time the club gets to impact, how fast does that mean it is
traveling?
Here
is part of a typical swing sequence page from Golf Digest magazine, one
of many they have published of Tiger Woods' swing over the years. The
clubhead's height at the top of the backswing is the yellow arrow h. Tiger is 6'2" tall, and some measurement of the picture itself says that h
is no more than 6'9" or 81". Let's do this analysis using nonmetric
units of the foot/slug/second system; we'll do the next example in
metric.
Wait! Slugs? What is that? You perhaps thought mass would be in pounds? OK, force and mass are not the same thing; if they were, then F=ma would be meaningless. In every system, there are units of force and different units of mass. In the system we use in the USA, pounds and ounces are units of force, not mass. So, working from F=ma,
the unit of mass would be the unit of force divided by the acceleration
of gravity  32 feet per second per second. That unit is called a
slug. In other words, a mass that weighed one pound would be a mass of
1/32 slugs.
OK, let's proceed. The contribution of the gravity on the clubhead is
the loss of potential energy as the clubhead moves from the top of the
backswing to impact. (Well, not exactly "loss"; it is converted to
kinetic energy, which is the point of this problem.) The potential
energy change depends only on the mass m of the clubhead and difference in height h. That was the point of our look at several paths going from the lower shelf to the upper; only the change in height matters, not the path it takes to get there.
 The mass m of the clubhead is 200 grams (the mass of most driver heads) converted to slugs, which is .0137 slugs.
 The weight of 200g of mass is 0.44 pounds. That's the force exerted by gravity on the mass.
 The height difference h
is the 81" we identified earlier, minus about 2" between the ground and
the CoM of the clubhead at impact. That's 79", which we need to express as 6.6
feet (because the consistent system of units requires feet as the
measure of distance, not inches).
So
we multiply height and force together, and find the potential energy
loss is 2.9 poundfeet. Conservation of energy says the clubhead gains
exactly
that same amount of kinetic energy. So let's set up that equation.
Kinetic Energy = ½ mv^{2} = 2.9^{
}2.9 = ½ * .0137 v^{2}
Solving for v, we get:
v = sqrt (

2 * 2.9
.0137

) = 20.6 ft/sec = 14 mph

That's out of a clubhead speed over 120mph (this is
Tiger swinging the driver). It is certainly nonnegligible, but we aren't getting a lot of "oomph"
from gravity working on the clubhead.

Shaft bend
The
shaft of a golf club functions as a spring. We can bend it (for
instance, clamp the grip and deflect the clubhead) and expect that:
 It will offer a resistance, a force, against the bending.
 It will spring back when we release the bending.
Let's treat it as a spring and compute how much clubhead speed we might
possibly get from it if it bends back during the swing then perfectly
times its straightening to occur at impact. We should recognize this as
another example of potential energy being converted to kinetic energy.
This time it is a spring rather than gravity, but the principle is the
same. And, as promised, this time we will use the MKS
(meter/kilogram/second) system of units.
Here is a driver shaft being deflected by an amount D at the start of the downswing. Most swings have the maximum deflection later than this, so let's use D
to mean the maximum deflection of the shaft during the swing before
impact. To find the clubhead speed contribution from shaft kick, we
must:
 Compute the potential energy represented by the shaft bend.
 Equate the kinetic energy of the clubhead to the potential
energy we just computed, and find the velocity that energy would
require.
We know that the potential energy of a spring is given by ½ Kx^{2}. Obviously the spring deflection x is D.
Measurements of shaft bend during the downswing tell us that, for a
golfer (not a long drive competitor) with a reasonably wellfit shaft,
the deflection of the driver shaft almost never exceeds 10cm, and is
considerably less for most golfers. So let's start with x=10cm.
How about the spring contstant K?
I have a basement full of clubmaking tools, so I went down to the
basement and measured a bunch of rather stiff shafts  that is,
springs with a high K. (We're going with a highend K and x,
because we would like to see the maximum clubhead speed advantage we
might get from shaft flex.) I found that a deflection of 1cm (or 0.01m)
gave a force of about 3.5 Newtons. That gives a K (force divided by deflection) of 350N/m. So...
Potential Energy = ½ Kx^{2} = ½ 350 * .1^{2} = 1.75 Nm
We need to find the clubhead speed representing 1.75Nm of kinetic
energy possessed by a driver clubhead of 200 grams (0.2Kg). BTW, there
is a name for the energy of 1 Nm; it is called a "Joule".
Kinetic Energy = ½ mv^{2} = 1.75^{
}1.75 = ½ * .2 v^{2}
Solving for v, we get:
v = sqrt (

2 * 1.75
.2

) = 4.18 m/sec

For the benefit of us Americans and the units we have some intuition for, that is about 9.4mph.
Actually, that is not the way it works. In the early 1990s, TrueTemper (the shaft manufacturer) found out a few interesting facts using their ShaftLab instrumentation.
 The highest kick velocity they measured in years of testing was about 11mph. (That is quite consistent with our result.)
 For
some reason, the kick velocity did not seem to show up
as added clubhead speed. (We will find out why toward the end of this
golf swing biomechanics tutorial. We're not ready yet to understand the
reason. In fact, TrueTemper did not understand the reason back in 1994.
It wasn't until 2016 that the paper was published that showed what was
really going on.)

Last
modified  Dec 3, 2022
