Physical principles for the golf swing
Torque
Dave Tutelman
 Sept 4, 2022
Examples and more analysis tools
We have seen some ways to analyze
forces and torques. Most, like adding vectors or resolving them into
their X, Y, and Z components, are fairly straightforward conceptually,
but sometimes
they require a lot of arithmetic. If we deal with a lot of forces and
try to compute the motions they cause (like the human body doing
something complex like a golf swing, or even just walking), the
concepts still work. But the amount of computation and the collection
of numbers you need to keep around can be staggering.
Let's look at some examples to give a better idea of how to approach
problems associated with the golf swing. Even if you are not interested
in analyzing freebody diagrams in general, there are some important
insights into the golf swing in the example
where the free body is a golf club; please make it a point to
understand this one.
Moment of inertia of a golf club
A
set of golf clubs have to be matched to one another in a variety of
ways. Unless you've been living under a rock, you have heard of
frequency matching and swingweight matching. Well, an effective (and
starting to become popular) alternative to swingweight matching is MOI
matching. An MOImatched set of clubs all have the same moment of
inertia, measured around an axis that passes through the butt of the
golf club, and is perpendicular to the shaft axis.
The diagram shows a golf club swung as a pendulum from its butt; that
is the classical instrument for MOImatching a set of golf clubs. There
are better instruments today, but this shows very well the club and the
axis about which moment of inertia is being measured.

But instead of measuring the
moment of inertia, we are calculating it
here. The way we calculate it is to consider the club as an assembly of
pieces that are easy to find the I,
either easily calculated or in a reference table. Then we add all the Is together for
the I of
the assembled club.
Here are the three major pieces that make up a golf club, the head,
the shaft,
and the grip.
Our job here is to find the I of each piece around the red axis, then
add them together. Start with nomenclature:
 Mass of the head is H.
 Mass of the shaft is S.
 Mass of the grip is G.
 The length of the shaft L
can also be used to approximate (a) the length of the club itself
and (b) the distance from the axis to the middle of the clubhead.
Let's compute the moment of inertia of each of the components around
the red axis.
Shaft:
We can consider it a long rod with the axis at one end. There are lots
of tables around with the moment of inertia for that shape; it is:
I = (1/3)ML^{2}
So, in terms of our golf shaft, it is:
I_{shaft} = (1/3) S L^{2}
Head:
The dimensions of the clubhead are a lot smaller than L, the distance
to the axis. So the mr^{2} chunks are all
going to have very similar values of r, specifically ~L. So we can,
to a pretty good approximation, say:
I_{head} = H L^{2}
Grip:
The grip is a hollow tube. It is tapered, but our first approximation
will treat it as constantdimeter. We get only very small errors from
treating it like a rod of length g. (We will
reexamine the approximation below, when we introduce the parallelaxis
theorem. We will compute a more exact I and see what
error this simplification produces.)
I_{grip} = (1/3) G g^{2}
Whole
club:
We can add up all the pieces and see the moment of inertia for the
entire club. When we do that and combine common factors, we get:
I_{club} = L^{2} (H + S/3) + g^{2} G/3
Let's plug in some numbers typical of a 7iron, and
see what the moment of inertia is, both for the components separately
and for the whole club. We will be using units familiar to an American
clubmaker: length in inches and mass in grams. Yes, this is kind of a
mixedunit system, but I in graminches^{2}
is still a valid unit. Here is a table of the calculations:

Height

Mass

Moment
of inertia

Shaft

L
= 37"

80g

36,507

Head

r
= 37"

270g

359,630

Grip

g
= 10.5"

60g

2,205

Whole club

410g

408,342

Things to note:
 88% of the moment of inertia is due to the clubhead.
 Less than on half of one percent of the moment of inertia
is due to the grip. I know I tossed away a casual excuse for treating
it like a rod, but just about any approximation for grip MOI will have
no effect on the club's MOI  because the grip itself has almost no
effect on the club's MOI.
 We are able to do computations of moment of inertia, which
was the point of this section.

Center of mass (center of gravity)
What
is the center of mass? I have seen it described (by people who actually
had no idea) as the point an a body where there is as much mass on one
side as on the other. Be advised
that this is wrong! It is actually the point that
the object can balance on.
If you think carefully about it, that should tell you that it is the
point where there is as much clockwise gravitational moment as
counterclockwise gravitational moment. The system balances  that is,
it doesn't move, it stays in equilibrium  because there is a zero net
torque due just to the weight of the object itself.
Let's point out here that Center
of Mass has essentially the same meaning as Center of Gravity or Balance Point.
There are subtle differences, but they don't involve golf clubs or the
golf swing. So which term you use will depend mostly on your
background. Center of Mass is the predominant term in biomechanics
papers, and I'll try to be disciplined about using it here. Center of
Gravity is the term used in my physics and engineering courses, which
is why I might slip up now and then; it's the term I've been using
almost all my life. And Balance Point is used by clubfitters and
clubmakers more often than not.
Center of Mass is often abbreviated COM or CoM.
Next
question: how do you find or calculate the center of mass for an
arbitrary object? Well, for symmetrical objects of regular shape, the
CoM is the geometric center. If the shape is regular, you can tell it
by eye or ruler, no problem.
For more complex shapes and distribution of masses, there is a
procedure for computing where the CoM is. The procedure is:
 Compute the moment of the weight of the object object
around some axis. You can choose any axis to make the computation
easier.
 Also compute the total mass of the object.
 Divide the moment (part1) by the mass (part 2), which gives
a number with a unit of distance  e.g., 2.6 meters. That is the
distance from the axis you chose to the CoM, so you now know where the
CoM is.
Let's go through the procedure, using the 7iron whose moment of
inertia we just computed. Let's find the CoM (the balance point) of the
club. We will take as the axis the butt of the club, mostly because it
is easy to do the calculations from there. We could have taken any
perpendicular axis, and we would get the same position of the CoM.

Here are the three subassemblies of the golf club  the grip, the
shaft, and the head. We will find the torque that their weight applies
to an axis at the butt of the club. Remember, torque is a force times a
distance to the axis; all of them are labeled on the diagram.
 The grip's weight is 60g, and it can be modeled as being
applied in the middle of the grip, 5.25" from the axis.
So the torque is 5.25*60=315inchgrams.
 The shaft's weight is 80g, applied in the middle, 18.5"
from the axis.
So the torque is 18.5*80=1480inchgrams.
 The head's weight is 270g, and we can approximate the
torque by considering all of that weight 37" from the axis.
So the torque is 37*270=9990inchgrams.
We add the torques together to get 11,785 inchgrams.
If we add the weights together, the total weight of the club is 410 grams.
We divide the torque by the weight and get 28.7 inches.
So
the CoM, or balance point, of the club is 28.7 inches from the axis.
As a sanity check on this number, always good to do if you can, I
measured the balance point on my own 7iron, straight from the bag. The
components are the weights we assume here, but it is cut almost an inch
longer; that's what I play. The balance point was 29.2 inches from the
butt, which matches very will with our calculations; the difference can
be explained entirely by the difference in club length.
If you are comfortable with calculus, you can
find the center of mass
of an object that can't be broken into discrete pieces of known
position
and weight. The principle is still the same. Instead of
the sum of torques divided by the sum of masses, you would use the
integral.
Note that I have added a CoM symbol to the diagram of the golf club.
Horizontally (the x
direction), it is the 28.7 inches from the axis that we calculated. But it is not on the centerline of the
shaft shaft.
That is because not all of the mass is of the club is symmetrical
around the shaft axis. Specifically, the clubhead's CoM is decidedly
off the shaft axis. If we were to repeat the calculations for the y direction, the
answer would not be zero  hence the CoM would not be on the shaft
axis.

Parallel axis theorem
This is a very handy way to compute the moment of inertia of an object,
if you know it about some axis but really need to know it about a
different axis. The way it works is:
 The smallest moment of inertia of the object (any object) is about an axis
through the CoM. We refer to this moment of inertia as I_{o}.
 The moment of inertia of the object around an axis parallel
to the axis for I_{o}
and a distance r
from that axis, is given by:
I = I_{o} + mr^{2}
Here are a couple of very simple examples before we get to more
interesting examples. For these simple examples, the object is a thin
ring of mass m
and radius r,
for instance the rim of the bicycle wheel we saw earlier. The moment of
inertia around the axle we have already seen to be mr^{2}. So...
 What is the
moment of inertia of that ring about an axis at the rim but parallel to
the axle?
The new axis is at the rim, so it is a distance r from the
original axis (the axle). So the new I is
I = I_{o} + mr^{2} = mr^{2}
+ mr^{2} = 2mr^{2}
 What is the moment of inertia of
that ring about an axis 10r from the axle?
By the same reasoning:
I = I_{o} + m(10r)^{2} = mr^{2}
+ 100mr^{2} = 101mr^{2}
So moving the axis away
from the object only 10 times the radius of the object (roughly five
times the size of the object), we can ignore any properties of the
object except of its mass, and only introduce at most a 1% error. That
justifies what we did earlier when we considered the clubhead a point
mass and didn't worry about its actual size or shape.
Example: hollow tube
Earlier, when we computed the moment of inertia of a golf club, we
approximated the grip as a hollow tube, which was pretty believable.
Then I suggested the further step of evaluating its moment of inertia
about the end cap by the same formula we would use for a slender rod.
We neglected the diameter as well as the open center. Now let's justify
that step.
Here is a reference table entry showing the moment of
inertia of a hollow tube. I have added a green axis labeled E, parallel to
the Xaxis
but at the end of the tube (where the end cap of the grip would be)
instead of the middle. The table gives I_{o} for the Xaxis as:
I_{o} = (m/12) *
(3(R_{2}^{2} + R_{1}^{2}) + h^{2})
We need to move the axis to the Eaxis, which is
a distance of h/2
from the Xaxis.
The parallel axis theorem tells us that moment of inertia is:
I_{E} =
(m/12)*(3(R_{2}^{2} + R_{1}^{2}) + h^{2}) + m*(h/2)^{2}
With a little factoring, we can come up with:
I_{E}
= m (

h^{2}
3

+

R_{2}^{2}
+ R_{1}^{2}
4

)

We know from other tables that the first term (mh^{2}/3) is the
moment of inertia of a slender rod about its endpoint  which is
exactly the approximation we made. We can see how much of an error we
introduced by comparing the first term (our approximation) to the
second term. We know from the dimensions of a typical golf grip that h=10.5", R_{1}=0.3", and R_{2}=0.5". Therefore:
 The first term (which our approximation was based on) is
36.75
 The second term (what our approximation ignored) is 0.085
So what we ignored is only 0.23% of the approximation  negligible
unless it has to be very accurate. Our approximation was warranted.

Example: I_{o} of the golf
club
In the previous exercises, we found:
 The moment of inertia of a golf club, taken about an axis
at the butt. (Often used to match clubs within a set.)
 The center of mass of that club, measured from the butt.
Now let's find I_{o} of the same
club, a conventional 7iron. We could do it as we did it for the I at the butt,
trying to compute the moment of inertia about the balance point for
each component. (That sound like a lot of work, and it is. But
conceptually at least, it would not be too hard since we have the
parallel axis theorem as a tool.) But now we have a much easier way to
do it. We can use the parallel axis theorem, applied to the totalclub
information we already have.
Let's review what we already know:
 The total mass of the club is 410 grams.
 The total moment of inertia about the butt is 408342
graminchessquared.
 The balance point is 28.7 inches from the butt.
How can we attack this with the parallel axis theorem? At first glance,
it doesn't sound right. The parallel axis theorem tells us how to find
any other moment of inertia, given that we know the moment of inertia
through the CoM. But it is still a pretty simple equation with only one
unknown, so let's set it up and see if we can solve it.
I = I_{o}
+ mr^{2} (the
theorem)
408342
= I_{o} + 410 * 28.7^{2} = I_{o}
+ 337713 (our equation)
This equation is simple to solve, even by eighth grade standards. The
base moment of inertia I_{o} is
70,629 graminchessquared.

Equivalent forcecouple
system
Suppose you have a rigid object, you know all the forces and
torques on it, and you want to know what motion that will create. Put
another way, much of biomechanic analysis is converting
kinetics into kinematics. In
our study of the golf swing, the rigid object might be:
 A body part, like a forearm.
 A collection of body parts, that can be considered
approximately rigid. For instance, standard biomechanical
representations of the human body may lump the central body into just
two or three sections, e.g. just the thorax and the pelvis.
 A golf club, assuming we are prepared to ignore the flex of
the shaft. (Whether we are will depend on the problem we are solving.
And even if we need to consider shaft flex, we can often be accurate
enough if we treat the shaft as a few rigid sections  maybe 35 of
them  connected by spring joints.)
Usually we don't have a system as simple as a single torque and a single force through
the center of mass. If we did, then we could just apply F=ma and T=Iα
and go from kinetics to kinematics  from mechanics to motion. (For it
to be this simple, the force has to be through the center of mass.
Otherwise it also contributes an additional torque, the moment of the
force.)
But there is a very powerful tool to help us here. You can always
transform a collection of forces and torques on a rigid body into a single force through the center of mass
and a single couple. This is what you need to go from known
kinetics to the motion we want to understand. Let's look at a few
examples. But first, a few caveats about this transformation, which is
called an "equivalent forcecouple system".
 The techique works for any axis you choose. But our
purposes here are to investigate the motion of a body part or a tool
(e.g. golf club) that the body is using, and the interesting axis here
is through the center of mass. Some of the early examples will be
transforms to some arbitrary axis, but as we progress through these
notes, we will be increasingly focused on equivalent forcecouple
systems about the CoM.
 The original collection of forces and torques are equivalent
to the single force and single couple. But the equivalence is
restricted to external
behavior. If you want to know how the forces and torques create
stresses within the object  internal effects  this equivalence does
not work.
 We do this in order to study dynamics  how a collection
of forces and torques create motion of a body. It is much harder to go
the other way, from observed motion to the set
of forces and torques that caused it. We will touch on this later, but
not in this chapter.
On to the examples:

Textbook example
Here is a fairly simple
example, similar to one of the first halfdozen homework problems at
the back of the chapter.
We have a rectangular gray object with a number of forces and couples
imposed on it. The object is 6 feet by 4 feet, and the Center of Mass
is in the geometric center. Our job is to find a single force and a
single couple that will produce the same dynamic effect as these forces
and couples. By now, you should not need a word description of what is
in the diagram.

The
first step in the solution is something I choose to do in the problem.
It is not absolutely necessary, but it should make solving this problem
easier. (Also many other problems, so it is worth considering this
first.) We are going to resolve the forces so that was only have Xaxis
and Yaxis forces.
Of the two forces, the 7pound force is already just an Xaxis force,
so we only need to resolve the slanted 10pound force. It resolves into
an X force of 8 pounds and a Y force of 6 pounds.
We are going to do two things with this information:
 Get the moments of all the forces about the CoM.
 Get the total net force on the object.

First let's get the
moments of the forces. Let's call clockwise moments the positive
direction. So the moments of each force is:
 (CCW) 8 lb * 2 ft = 16 ftlb
 (CW) 7lb * 2 ft = +14 ftlb
 (CCS) 6 lb * 3 ft = 18 ftlb
Adding them together, we get the total moment of forces on the object as
 16 + 14  18 = 20 ftlb
Now we can find the total
net force on the object. We need to do vector
addition for all the forces. Since we have converted all the forces
into X and Y components we just do simple signed addition of the X
forces separately from the Y forces.
 X forces: 8 (left) + 7 (left) = 15 (left)
 Y forces: 6 (down)
So we have a right triangle with sides of 6 and 15.
 The Pythagorean Theorem says that the hypotenuse of the
triangle is 16.2.
 Simple trigonometry (sine, cosine, or tangent; we can use
any of them for this) tells us the angle is 21.8°.

So here is our
equivalent forcecouple system.
 The single force on the object, acting at the CoM is 16.2
pounds, acting to the left and downwards at an angle of 21.8°.
 The single couple on the object is 30 footpounds, or 30
ftlb acting counterclockwise.
We simply added (with directional sign) the couples applied to the
object and the total moments of force. We didn't need to worry about
where any pure torque is applied to the object; a couple
can be applied anywhere and the external dynamics
will be the same. The only thing that is different is the internal
stresses within the object.
There are lots of textbook examples out there. Here are a few links
that might be enlightening, and I'm sure you can find more.
Links to lecture notes
and lessons: example
1 example 2 example 3
Links to videos:
video 1 video
2 video
3

Golf club example
This
one is important! Repeat! This example
is important to understand!
The things you will learn here are some of the things you need to know
about the golf swing, not just about physics  and some of them run
counter to oldschool, traditional advice.
We
are going to start with a diagram that is very close to "Figure 1" for
quite some biomechanics papers involving modeling a golf swing. This
particular version of the diagram comes from a tutorial that Dr
YoungHoo Kwon posted in 2015 to the "Golf Biomechanists"
group in Facebook. I chose this for simplicity; most of the others boil
down to this, but have to be boiled down from a more complex picture.
The diagram identifies the three influences that determine the motion
of the golf club during the downswing. (The same diagram probably works
for the backswing and followthrough as well, but most biomechanics
studies focus on the downswing. During the instant of impact  for
less than a millisecond  there are other influences, but we won't
worry about them at this point.) The influences are:
 The weight
of the club, a force which acts through the CoM. It is labeled as mg because it is
equal to the mass of the club m times the
acceleration of gravity g.
 The hand force,
shown as F
on the diagram. It acts through the "midhands point", which for most
golfers is about 3.5in or 9cm from the butt.
 The hand couple,
shown as T
on the diagram. It is applied by the hands, but analytically it doesn't
matter where a pure torque is applied to the body; it has the same
effect on the body's motion wherever it is applied.
And that's it! The hand force and hand couple vary during the
downswing, which of course adds complication, but the weight vector is
perfectly constant throughout. Let's redraw this as a classical
freebody diagram, assign values, and find the equivalent forcecouple
system  and see what we can learn along the way.
Figure 1
Here is our club with the weight, the hand force, and the hand couple.
A few things I chose for this specific example:
 We are looking at the club at the moment it is horizontal
during the downswing. (The instruction community tends to refer to this
as "club parallel", but that is ambiguous. I mean, parallel to what?
They mean "the ground", but it could have been "the target line" or
perhaps "the lead arm". If you mean "parallel to the ground", then
"horizontal" is a oneword way to say that unambiguously.)
 I show the arc of the hand path in the diagram. This is
because...
 I choose to resolve the forces not to X and Y
axes, but an axis perpendicular (or "normal") to the hand path and
another axis tangential to the arc of the hand path. I'm not doing this
to confuse, but to apply some physical meaning to the components
 The normal force
is socalled because it is "normal" or perpendicular to the hand path.
It provides the acceleration to produce a curved path for the hands.
Since it is
perpendicular to the direction of motion of the hands, it can
curve that motion but not make it any faster or slower. It would be
accurate to say this is a centripetal force, keeping the club on a
circular path.
 The tangential force
accelerates the hands along the hand
path.
It is the provider of hand speed, and plays a very significant role in
clubhead speed (as we shall see several chapters down the road).
It is worth noting that club horizontal occurs fairly late in the
downswing. True, it is only about 2/3 of the way through the motion of
the club. But the club accelerates, both linearly and angularly,
through the downswing. The first two thirds of the motion will occupy a much
larger portion of the time, likely about 90% of the time.
Now let's put some numerical values on the three influences, and derive
the equivalent forcecouple system.
Figure 2
We are using the same 7iron we did when we computed the moment of
inertia and the center of mass earlier on this page. So the weight of
the whole club is 410g, which is almost a pound. Let's use 1 pound here
to keep the numbers simple. We found the CoM to be 28.7" from the butt
of the club, which is 25.2" from the midhands point where the hand
forces are applied. At this point in the swing, the hand couple is
about 10 footpounds, and the hands need to exert a 20pound normal
force just to keep the club from flying off on a straight line. The
tangential force, which is creating hand speed, is actually a lot less
than the normal force; we'll make it 9 pounds here.
Now let's find the equivalent forcetorque system.
Figure 3
The
first step is to resolve each force into its X and Y components. Then
we add the X forces and Y forces, remembering to keep the sign in mind
 add or subtract as appropriate.
 Total X force = 9.4 + 7.9 = 17.3 pounds
 Total Y force = 17.7  4.2  1 = 12.5
pounds
So now we know the X and Y components of the equivalent force.
The next step is the equivalent torque. We already have a hand couple
of 10 footpounds counterclockwise (negative). Of all the redvector
forces on the diagram, only two of them contribute to a moment around
the CoM; all the rest of the forces pass through the CoM, so the moment
arm is zero and they produce no moment. The forces of interest here are
17.7 pounds in a counterclockwise direction and 4.2 pounds clockwise.
Both are exerted at the midhands point, 25.2" from the CoM. So...
Clockwise moment of the
hand force

=
4.2*25.2  17.7* 25.2


= 340
inchpounds


= 28.3 footpounds

Figure 4
Adding the X and Yforces together at right angles, we get a 21.3
pound force at an angle 36° above horizontal So that is the force
through the center of mass.
The total couple acting on the club is the hand couple (10. ftlb
counterclockwise) plus the total moment of force (28.3ftlb
counterclockwise), for a total of 38.3 footpounds counterclockwise.
I
said that this is an important example. Richard Hamming (an innovator
in numerical analysis, among other things) famously said, "The purpose of computing is insight, not
numbers." So here are some insights we can glean from the
computation we just did.
 Gravity
doesn't matter.
Well, it matters a little. Very little. The weight of the club (the
only contribution of gravity to the free body diagram) is only one
pound, much less than any of the other forces involved and very much
less than the 21.3pound force for the equivalent system. So it
contributes rather little force to the system. Perhaps even more
important is that it contributes not at all to the equivalent torque,
which turns into angular acceleration of the club. More than 80% of the
clubhead speed at impact is due to angular velocity, and gravity seems
not to be doing much for us in that regard. Bottom line: we should not
focus our swing keys on any advantage gravity might give us; it
contributes very little of the club's energy, and that very little is
likely to happen no matter what our swing looks like.
 Moment
of the hand force is the source of most of the angular acceleration of
the club, not the hand couple.
We can't dismiss hand couple as completely as we do gravity, but most
recent biomechanical studies have shown that focusing on the moment of
the hand force is more productive.
 Most
of the force exerted by the hands late in the
downswing is not tangential, not towards the target.
It is merely
keeping the club from flying away from the curved hand path. This is
not true early in the downswing, but becomes overwhelmingly so as the
clubhead approaches impact.
We can learn a few more lessons if we analyze this example using a
different decomposition of the forces.
Figure 5
Instead of decomposing the forces on the XY axes, we could just
resolve the components of the hand force into a single force of 22
pounds acting at the midhands point and directed 38° above horizontal.
The moment arm of this net hands force is the distance from the center
of mass to an extension line of the force; it is 15.5" in this instance.
Things we learn when we look at the example this way:
 As we see here, if you exert a force on an object
that does not go through
the center of mass, it will rotate the object in such a direction to
line up
the CoM with the line of action of the force.
Therefore: As
long as you can keep the extension line of the hand force in
front of the shaft, you will keep angularly accelerating the club.
 The
larger the hand force and the more you keep the line of force in front of the shaft,
the greater the angular acceleration.

Last
modified  Oct 2, 2022
