Just as we saw that torque was an angular analog of the linear
quantity
force, there is an angular energy that is an analog of the linear
energy. Both work and kinetic energy in the angular domain are
calculated as rather obvious analogs of their linear counterparts:

Work =
Angular distance * Torque in the direction of that
angle

Kinetic
energy = ½ I ω^{2}

That
isn't hard to understand in the diagram on the left, where the rotation
is about the center of mass. No linear work is done, because the CoM
never moves. Any moment of inertia involved is I_{o}, the inertia
about the CoM. So the calculations are very straightforward.

Not so for the diagram on the right. This time, the axis of rotation is
not through the CoM, so the CoM is going to move (its mothion is the
blue arrow). There had to be an overall force to move the CoM. Remember
the equivalent force
couple systems?
If you understand and remember what happened there, then you realize
there has to be a net force in this diagram, because it is providing
some linear acceleration to the CoM of the object.

Let's solve a problem like this that has a golf flavor. It might
provide you some intuition about questions that come up in golf club
fitting, though that process is much more complicated than this simple
example

Here's our favorite seven-iron that we analyzed in previous chapters,
and a few things we learned about it back then. Oh, that's right! The
last time we were using the usual American clubmakers' units, inches
and grams. Let's get used to solving problems in the MKS system that
biomechanists will generally use. Here, we have simply converted the
units:

Grams to kilograms.

Gram-inches-squared to kilogram-meters-squared.

The first conversion, g to Kg, is straightforward. If you had trouble
with this one, you need to review the early section on measurement units. The other
conversion is one you are not likely to find; even calc98 does not have the
clubmakers' unit for moment of inertia. The way you convert
g-in-squared to Kg-m-squared is:

Multiply the moment of inertia by the conversion from grams
to Kg (1/1000).

Multiply that by the conversion from inches to meters
(1/39.37).

Repeat step #2, because it has to be squared.

OK, let's move on to the problem: How much work is done by a
torque of 0.03 Newton-meters applied at the CoM for 2 seconds? What is the kinetic energy of
the club afterwards?

The
solution
First, let's write down the basic stuff we know or can easily compute.

I
= I_{o} = .0455 Kg-M^{2}

T = αI
so α
= T/I = .03/.0455 = .66 rad/sec^{2}

Now we know the angular acceleration, so we should be able to easily
compute angular velocityω and the angle
traveled Θ.
(You were told early on that you
were expected to just know this, but I'll give you the link back to it
if you forgot. Last time, though.)

ω = αt = .66 *
2 = 1.32 rad/sec

Θ = ½ α t^{2}
= ½
* .66 * 2^{2} = 1.32 rad

... which happens to be an angle of 75.6 degrees.
Now we have everything we need in order to compute work and kinetic
energy.

Work = ΘT = 1.32 *
.03 = .0396 N-m

KE
= ½
I ω^{2}
= ½
* .0455 * 1.32^{2}
= .0396 N-m

We have answer to the problem, but let's make an important observation.
The work applied to the club is
exactly equal to the kinetic energy it picked up.
That is not a coincidence! One way even the most experience
biomechanists check their work is to make sure (if the information is
available) that the work done is equal to the useful energy created.
(And, if they are not equal, where the rest of that energy went.)

Let's move on to a more interesting version
of the same problem.
Instead of applying to torque to the center of mass, we apply it where
we normally would with our hands -- to the grip.The axis of rotation is
now at the mid-hands point. But note, this is not a full swing, just a
waggle as shown in the video. For purposes of this example, the major
difference between this waggle and a full swing is that the hands,
arms, and body never move; the only thing whose energy we care about
here is the club itself.

Go ahead and try it yourself. Vigorously waggle a proper golf club back
and forth, holding the mid-hands point still. No, not a shaft or dowel,
a full club with plenty of mass and a CoM near the clubhead. You will
find that, in addition to the torque (the hand couple) you must apply,
you also have to apply a force to keep the hands in one place. The
club wants to rotate around its CoM. If the CoM is far away from the
hands, that means the handle end of the club wants to move opposite
from the clubhead -- and it takes a force from your hands to prevent
that motion. We will see this in more detail below.

Here is information we will need to solve the waggle
problem, which we will state similarly to the problem we just
solved. How much work is done by a
torque of 0.03 Newton-meters applied at the mid-hands point for 2 seconds? (Note that a force may be
needed at the mid-hands point to assure that is the axis of rotation.) What is the kinetic energy of
the club afterwards?

This
works very much like the previous problem, where the axis of
rotation was the CoM. The numbers will be different, but the formulas
are the same, as is the order in which we use them.

Again, let's start with the easy stuff. We need the parallel axis theorem to go from I_{o} (at the CoM) to I (at the mid-hands point).

I
= I_{o} + mr^{2} = .0455 + .41*.64^{2} = .213

T = αI
so α
= T/I = .03/.213 = .141 rad/sec^{2}

Now we know the angular acceleration, so we should be able to easily
compute angular velocityω and the angle
traveled Θ.

ω = αt = .141 *
2 = .282 rad/sec

Θ = ½ α t^{2}
= ½
* .141 * 2^{2} = .282 rad

... which happens to be an angle of 16.2 degrees.
Now we have everything we need in order to compute work and kinetic
energy.

Work = ΘT = .282 *
.03 = .0085 N-m

KE
= ½
I ω^{2}
= ½
* .213 * .282^{2}
= .0085 N-m

Again, the work done by the hands on the club equals the kinetic energy
of the club. Or does it?!?
We have a couple of linear things to worry about here.

There is a force at the hands. Is it doing work? Well, no,
it isn't doing work. It is applied at the axis, which does not move.
Without motion in the direction of the force, the force does no work?
So we're OK here.

The CoM of the club finishes with some linear motion. Isn't
that kinetic energy? Actually, it is. We've miscomputed the kinetic
energy by (a) ignoring this linear KE, and (b) using a moment of
inertia other than I_{o} for the energy
computation. Let's go back and do that KE calculation strictly rather
than the loosey-goosey way we just did it.

Angular KE
= ½
I_{o} ω^{2}
= ½
* .0455 * .282^{2}
= .0018 N-m Linear KE = ½ mv^{2} ^{
}And we know that, for a spot on a rotating body a
distance r
from the axis

v
=
rω = .64 * .282 = .18 m/s so Linear KE = ½ m v^{2}
= ½
* .41 * .18^{2}
= .0066 N-m

Finally

Total KE = Angular KE
+ Linear KE = .0018 + .0066 = .0084 N-m

Yes,
it comes out the same. (There is a round-off
problem in the last digit, but that is all that is wrong.) It took me
less than 20 minutes to prove that they will always come out the same,
no matter where on the club you place the axis. But I won't bore you
with the proof.^{
}

Other
forms of energy

So far, we have talked about work, potential energy, and kinetic
energy. Let's look at the other forms of energy we are taught in
Physics 101. We'll review them quickly, because they are mostly
irrelevant to the study of the golf swing. We'll note where there might
be some relevance, and explain it.

Heat

When I think of heat energy, the first thing that comes to my
engineering-trained mind is the sort of heat engine that has been the
world's most important source of energy for the last century or two.
I'm talking here about steam engines, internal combustion engines, and
turbines. They power our transportation, and indirectly almost
everything else we use. The vast majority of our electric power comes
from generators turned by heat engines. All of them have heat as input,
and some useful energy as output.

But the importance of heat energy in golf is quite the opposite. The
heat is output, not input. It is the form of wasted energy, the unusable
energy in the golf swing -- and almost all other enterprises that use
energy. Let's stop for a moment and define efficiency, because
inefficient use of energy is exactly where that heat comes from.
Efficiency is:

Efficiency
η =

Energy_{out} Energy_{in}

For
example, it you only get out half the energy you put in, the efficiency
is 50%. (In golf and most technologies, there are other metrics that
they call "efficiency". For instance, I have seen clubhead speed
divided by the maximum possible clubhead speed called "speed
efficiency" of a swing. But, strictly speaking, efficiency really
refers to the useful energy you get divided by the total energy you
supply.)

But we know that energy is conserved; it can not be created nor
destroyed. So the missing energy has to go somewhere. And it almost
always manifests itself as heat energy; something in the system gets a
little warmer, heated by this "lost" energy.

Friction

In golf analysis, the lost energy is always heat, and usually due to
friction. We know that friction can supply heat; think about starting a
fire by rubbing sticks together. Maybe it doesn't supply a lot of heat;
it takes a lot of effort to start a fire that way. But have you ever
gotten a rope burn, when a rope you are holding "runs away" in your
grasp. There's a pretty dramatic example of friction work turning into
heat energy.

So how does friction work? At a "hand waving" level, something sliding
across another surface is acted on by a force that resists its sliding.
But let's be a little more precise.

The diagram shows a block resting on
the floor.

A force (green)
is applied to the block, trying to make it slide on the floor.

Does it
slide right away? No. The force has to get to some level before it
overcomes the friction force (red)
that resists sliding.

And how big is that friction force? It depends on
the normal (which means
perpendicular) force (blue),
which in this example is just the weight of the block. Specifically,
the maximum value the friction force can take is the normal force times
a number called the "coefficient of friction" (usually designated "μ")

We all have enough experience to know that μ
depends on the material. Ice is more slippery (lower friction) than
sandpaper. In fact, the coefficient of friction depends on both
materials that are touching. And it can have a big range from one
material combination to another. For instance, steel on ice has μ=0.03; it doesn't take much applied force to get ice skates sliding across the rink. At the other end of the scale, μ
can be greater than 1. That means if the surface is rough enough or
"grippy" enough, the friction force might be more than the normal
force. Here's a link to a table of coefficients of friction for a pretty wide variety of materials.

Just in case it isn't clear yet, here's how it works. In the following, A is the applied force, F is the friction force, and N is the normal force.

If A<μN, then the friction force F is equal to A and the block does not move.

If A>μN, then the friction force F=μN and the block slides. Any excess force A-F (which is a net force) accelerates the block as it slides.

One thing that should strike you is that work is being done in the second case! (In the first case, there is no motion so no work.) If the block slides a distance x across the floor:

The total work done to the block is xA, the force applied to the block times the distance the block moved. Simple so far.

The work done accelerating the block, x(A-F), turns into kinetic energy. So we have accounted for where that energy went.

But what about the work done against friction, xF?
It doesn't turn into kinetic energy; it was just enough force to
overcome friction, not to accelerate the block -- that is, add to the
block's velocity. It doesn't turn into potential energy either; the
floor is flat, so no height is gained. The only place that work can go
is to turn into heat. Unless you have some sort of heat engine working
from the temperature of the block or the floor, that work is lost. Not
destroyed; it is still energy. But we can't do anything useful with it.
That is the nature of friction forces; their work is usually just loss of useful energy.

Let's look at some examples of friction in golf:

Non-sliding friction force: This is the case where A<μN,
so the surfaces stay locked together -- no sliding. There is a very
important application of this in golf, the interface between the grip
of the club and the hand or glove of the golfer. Designers of grips and
gloves try to maximize the coefficient of friction, within the rules,
so that the golfer can control the club slip-free with a minimum of grip pressure (normal force).

Sliding friction:
There are not many places in the golf swing to point to for examples of
sliding friction. And that stands to reason; we want the swing to be as
efficient as possible. Any work we do in the swing we would like to
turn into kinetic energy -- clubhead speed. But let's look at something
we don't want to happen, but sometimes does. I'm talking about the fat
hit. When the clubhead scuffs the ground on the way to the ball, the
result is friction. The harder the club touches the ground, the more
force. The farther the club drags on the ground (i.e.- the earlier it
hits the ground), the more distance that force is acting on it. So we
have a force decelerating the club for a distance, the definition of work.
That work can be equated to a difference in kinetic energy of the
clubhead at impact, a quantifiable loss of clubhead speed.

Static friction: If you looked closely at the table of coefficients of friction, you noticed that there are two columns for μ.
labeled "static" and "kinematic (sliding)". That is because it usually
takes a little more applied force to break the object free and get it
started sliding than it does to keep it sliding. So there is a
coefficient of static friction and a usually smaller coefficient of
sliding friction. In many cases they are nearly the same, but some may
show a big difference. For instance, clean and dry cast iron on cast
iron has a static μ of more than 1, but once it is sliding the μ is only 0.15.

Rolling friction:
when something round rolls across a surface ,there is some friction.
True, it is usually a lot less friction than if it were sliding, but
there is almost always some measurable friction. Ball bearings are used
to reduce force, because surfaces are no longer sliding, but that force
is reduced, not completely eliminated.

How about a golf example of rolling friction? Do you know what "stimp" is? No, really,
do you know what stimp is? It is a measure of the friction between a
green and a ball rolling on it. The less the friction, the higher the
stimp number. Here is an article I wrote
about an interesting bet, where its payoff depends on rolling friction.
Section 1 is almost a textbook example for our discussion of rolling
friction, potential energy, and kinetic energy.

Internal friction:
Friction and its associated energy dissipation can show up even without
visible surface-to-surface contact. For instance, the golf ball is
compressed on the clubface during impact. It doesn't look like there is
any sliding between clubface and ball, and in fact there is very
little. But internally, within the golf ball, the particles making up
the layers of the ball are rubbing together, with resulting frictional
losses. This does not follow any of our formulas for dealing with
friction, so another name has been associated with it: coefficient of restitution, or COR. Physicists and engineers usually designate it as e in equations.

As with friction, the restitution losses are associated with two
different surfaces. Unlike friction, there are lots of interesting
cases where the form is at least as important as the material. Our
example of a golf ball compressing on the clubface is a classic in this
regard. The clubface is titanium and the ball is usually a composite of
natural and artificial rubbers. But the thickness of the face can
account for a large proportion of the difference between the actual COR
and 1.0 (which is a lossless collision). And that is true even though
almost all the losses are in the ball, not the clubface. It is the
combination that determines the COR.

Chemical

A small corner of chemical energy is at the core of biomechanics. The
body burns nutrients to power the muscles. Our entire body is
maintained by such chemical energy. However, it isn't part of the
physics of the golf swing, and this tutorial is going to ignore it.
Nutritionists and exercise scientists get deeply involved in chemical
energy and the processes related to it. If you go on to do college
degree work in biomechanics, I'm sure you will get plenty of exposure
to this branch of the science. For our purposes, we are going to view
muscles as engineers would; they are actuators with specifications and
limitations. We'll get into that in the next chapter.

Electromagnetic

Electromagnetic energy is a very important practical branch of the
study of physics and engineering. It includes:

Electric motors and actuators, to produce forces and motion.

Electromagnetic fields and waves, to allow wireless
communication to and from devices.

Light (which is really just high-frequency electromagnetic
waves), including solar power.

While all of these can play an important part in golf instrumentation
and experiments, they have little or nothing to do with how the body
swings a club. We are not going to mention it any further in these
notes.

Nuclear

We do not have to worry about nuclear energy nor quantum mechanics in
golf physics. Period! If
anyone tells you otherwise, they need to have their heads examined --
or at least get a clue about physics. I only mention this because an otherwise
knowledgeable golf expert once insisted to me that quantum effects are
important in putting. Actually they were expert in tournament golf, not golf phsyics -- and it showed.

Last
modified -- Dec 3, 2022

Copyright Dave Tutelman
2024 -- All rights reserved