Let's get something straight
right off the bat. Energy and momentum are related but fundamentally different. The most
fundamental of the differences is that energy is an essential physical
quantity, while momentum is just a computational convenience. Momentum
is another way of expressing F=ma, and a
shortcut for solving problems that are fundamentally F=ma problems.
That doesn't make it any less useful in understanding golf physics, but
you need to be aware that F=ma is the only
thing about physics you need in order to derive everything about
momentum.

So let's start with F=ma. The first
thing we are going to do is remember that acceleration
is the rate of change of velocity. If you don't have a clear
idea of what this means, please review the sections on kinematics and Newton's laws, where
we explored the relationship between acceleration, velocity, and
position. Seriously, this concept is crucial; if you're confused about
it, you will not understand important things about the golf swing when
we get to that.

So, with
constant acceleration for a time interval t, we will
achieve a velocity V.
That means our favorite Newtonian equation becomes:

F
= ma = m

V t

With a bit of simple algebraic manipulation:

F t = m V

The quantity on the right, mass
times velocity, is called "momentum"; simple as that! The
quantity on the left, force times
time, is what creates momentum, and has the name "impulse".

Many momentum
transfer problems involve a change
in momentum, rather than the absolute momentum itself. So let's
repeat the last example with acceleration not occurring at time=0 and
with a positive initial velocity. Again, we assume constant
acceleration for a time interval Δt, we will see
a change in velocity ΔV. (Note that
the use of Δ
is just a way of saying "difference" or "change". It is notation,
nothing more.) The equation becomes:

F
= ma = m

ΔV Δt

With a bit of simple algebraic manipulation:

F Δt = m ΔV

The left side is again the "impulse", simply over a time interval that
does not start at zero. And the right side is the change
in momentum produced by that impulse.

In the perfectly general case,
we must also allow the acceleration graph to have much more freedom of
shape. It must be able to be something other than the red rectangle,
indicating a constant acceleration (implying a constant force). Forces
tend to build up and settle down in reality, unlike the diagrams in a
physics textbook. So let us look at the graph again, this time with a
more free-form acceleration graph.

This
graph is closer to the reality of, say, a clulbhead striking a ball, or
a quick muscle action during the golf swing. The acceleration is a
curve that rises to a rounded peak, then settles
back to zero. The velocity curve reflects that acceleration pattern,
also curving so the steepest rate of change of velocity occurs at the
same time as the maximum acceleration. (When you think about it, that
absolutely must happen just by the definition of acceleration as the
rate of change of velocity.)

But, regardless
of the shape of the curve, the change in velocity (and thus the
change in momentum, which is velocity times mass) depends only on the area under the
acceleration curve. The pink area must be proportional to ΔV.

How can we compute this pink area?

We can break the time into tiny segments of Δt, each of
which can be approximated by a constant acceleration. We can find out
the area of each segment, because it is well-approximated by aΔt, then add up all those
segments. This is the numerical
solution.

If the acceleration function is known well enough to yield
to calculus, then we can rewrite the entire impulse/momentum equation
above as
∫ F dt
= ∫ m dv

Again, the right side is momentum change, and the left is the impulse
that brings it about.

Collisions and conservation of
momentum

Now that we have defined momentum
(and its corresponding impulse)
is there anything useful we can use it for or learn from it?

Well, it turns out to be very
useful for analyzing bodies that collide
with one another. In fact, it is pretty easy to see where conservation of momentum
comes from if we look at a typical collision. Here are a couple of
balls colliding. While they are in contact, each ball exerts a force on
the other. Newton's third law says that the forces are numerically
equal and in exactly opposite directions. So every little bit of FΔt that the
brown ball exerts on the blue ball (the blue force) means that there is
a -FΔt
that the blue ball exerts on the brown ball (the brown force). What
this means is every change to the momentum of the blue ball is matched
by an equal and opposite change to the momentum of the brown ball. Therefore, at every
moment during and after the collision, the combined momentum of the two balls
remains exactly the same as before the collision. Presto -- conservation of momentum in this system composed of the two balls.

It is worth pointing out here that momentum and impulse are both
vectors. Impulse is based on force (a vector) and momentum is based on
velocity (a vector). So when we say the momentum remains the same, we
are saying the change in momentum is zero. Remember, impulse is exactly
equal to the vector change in momentum; we saw that come straight from F=ma. This means
the vector sums remain the same because the impulses are vector
opposites.

Here is the power of computing using instead of F=ma directly. In
a collision between two objects with mass and velocity, you do not need to know the details
of how force builds up at the point of contact between the bodies. You
can be assured that the combined momentum of the objects is the same
after the collision that it was before. That will tell you a lot --
often all you need to know -- about the velocities afterward.

Let's look more at collisions, and see what sort of things momentum can
tell us about them. Here are three animations of 2-ball collisions I
borrowed from Rod White's article on golf swing physics.
The thing that changes from example to example is the relative masses
of the colliding balls. All the collisions are lossless; that means no
internal friction, and the coefficient of restitution is 1.0.

Balls of equal
mass:
Think of this as a straight shot in billiards or pool. Two balls of
equal mass collide. All the momentum of the striking ball (blue) is
transferred to the stricken ball (red). After the collision, the blue
ball is standing still and the red ball is moving at the speed the blue
ball had before the collision. Clearly, momentum is conserved:

Before the collision, the blue ball's momentum was mV, where m is the mass of
each ball and V
is the blue ball's velocity. Since the red ball was at rest, its
momentum is zero. So the whole system's momentum is mv.

After the collision, the red ball's momentum is mV, where V
is now the velocity of the red ball. Since the blue ball is now at
rest, its momentum is zero. So the whole system's momentum is still mv, and momentum
is conserved.

Striking ball
has more mass than stricken ball:
This is the same model as a clubhead striking a golf ball. The
clubhead's mass is something between 200 and 300 grams (more if we
include putters), while just about every golf ball is within a half gram of 45.5 grams.

The red ball leaves the blue ball faster than the blue ball
itself was traveling beefore impact. This is because keeping a constant mV for a smaller
m
requires a larger V.

Even so, the blue ball retains a small amount of velocity
after the collision. It hasn't given up all its momentum to the red
ball, but retains some. In the case of a golf driver striking a ball,
the clubhead typically retains about 2/3 of its initial velocity,
while imparting to the ball almost 1½ times its initial velocity.

Striking ball
has less mass than stricken ball:
The model for this is almost the same as if the blue ball were striking a red
wall. Almost, but not quite. If the red ball had infinte mass, it would
function as an immovable wall. Even if not infinite, a red ball several
orders of magnitude heavier between than the blue ball would be hard to
distinguish from a wall. In this example, the ball is several times
heavier, but not several orders of magnitude, so it is only slightly
wall-like.

The blue ball bounces back at a significant fraction of its
original speed, but in the opposite direction.

The red ball moves in the original direction of the blue
ball, but much more slowly. Still, its velocity must be large enough so
that its momentum is equal to the original momentum of the blue ball
before impact, plus the absolute
value
of the blue ball's final momentum. Why? Because the blue ball finishes
with negative momentum -- it is going in the opposite direction now --
and the red ball has to make up for it.

Example: the essential divot

I have written a whole
article about this, but here is a capsule summary.

Let's
start with the conclusion: Any good strike of a ball
on the ground should produce a divot. Yes, even if your thing is
to pick the ball. The explanation lies in momentum transfer between the
clubhead and the ball.

Here's a prime photographic example. Luke Donald's strike with a middle
iron has a slightly negative angle of attack, but really not much at
all. At impact, several things happen:

The ball takes off at a substantial launch angle. I
measured it with a protractor as about 10°. Not really high, but far
from negligible.

The clubhead slows down. We expect this because the
clubhead has transferred some of its momentum to the ball.

The clubhead angles down more than the AoA we saw as it
approached the ball. Its original AoA might not have been enough to
take a divot, but this new downward angle is.

#3 on the list is a bit of a
surprise, but we can explain it easily as
just another piece of momentum transfer. Let's look at the momentum
changes
of the ball and the clubhead due to the collision.

The green angled arrow is the momentum change of the ball;
since it had zero momentum (zero velocity) before the collision, it is
a vector of magnitude mv at an angle
equal to the launch angle.

The change of momentum of the clubhead (the red arrow) is a
vector equal and opposite to the
green arrow. That is what conservation of momentum means!

Note that each of the momentum vectors has a resolution to horizontal
and vertical components in the diagram. Since the main vectors are the
same magnitude and exactly opposite directions, the resolution into
horizontal and vertical components gives the same magnitudes for both
the green (ball) and red (delta clubhead) arrows. In fact, if the
launch angle is a and the ball velocity after impact is V, then the
vertical momentum change from this collision is:

delta momentum
= m_{ball} V sin(a)

I said the launch angle was 10°, and sin(10)=0.17.
Therefore, roughly 17% of the momentum change shows up as vertical
momentum change. Not to be sneezed at.

So the collision does impart a downward change in velocity to the
clubhead, for the simple reason that it imparts an upward change to the
ball's velocity... and momentum must
be preserved.

Comparing energy and momentum

There are a few things about energy and momentum that are so similar,
one to the other, that it might be confusing. Let's single out these
similarities and explain why they are different, so we can refer to them if we get confused:

Conservation
laws - These are real similarities. If you take a closed system and examine it at one
moment and again at a later moment in time, the energy will be the same before and
after, and also the momentum will be the same before and after.
Conservation works for both. But be aware that energy may change its
form, often to heat, which can make it appear that energy is lost. We
will discuss this in a bullet point below.

I used the term "closed system" above. What does that mean? It means we
are considering a collection of objects such that, during the time
interval we are talking about:

For conservation of energy, no energy goes into or out of
that collection of objects.

For conservation of momentum, no force impinges on any
object in the collection, except for force from other objects in the
collection -- nothing from outside..

The
inputs, work and impulse - Similar, but not the same

Work
= Force * distance (with the force being the
component in the direction of the distance)

Impulse
= Force * time

So energy is a scalar;
multiplying two vectors projected into the same direction results in a
scalar. Physicists call this operation a "dot product"; you don't have
to know that, but if you run across it in a paper that is what they
mean.

Impulse is a vector; it
is the force vector scaled
up or down by multiplying it by time, which is a scalar. (That is why a simple number
is called a "scalar", anything it multiplies gets scaled but retains all its other properties except absolute size.)

The
outputs, kinetic energy and momentum
- Like the inputs, similar but not the same. And for the same reasons,
including which is a scalar and which is a vector. In the case of
kinetic energy, the velocity vector is squared. That is a dot product
of two vectors -- identical velocity vectors -- already in the same
direction.

Kinetic
energy = ½ m v^{2}

Momentum
= mv

Frictional
loss of energy but not momentum - Let's take an example to
illustrate this. This table is taken from one of my other
articles. It looks at two collisions, one lossless and the other as
lossy as we can make it:

The lossless
collision is defined by a COR of 1.0; as much energy comes out
as went in. This is well approximated by a straight billiard or pool
shot; billiard balls lose very little enrgy in collisioin. The term for
that is "nearly perfectly elastic".

The lossy
collision has a COR of 0; that is as lossy as we can make it
mathematically. Think of it as two balls of soft, wet chewing gum; they
will just stick together; no bouncing involved.

In
both cases, there are two balls of identical mass M. The green
ball moves at a velocity V until it
bounces headlong into the red ball. Here is the calculation of the
momentum and the kinetic energy, before and after the collision.

The
important thing to notice is: in the lossless case, both momentum and
energy are conserved. That is, they are the same both before and after.
But in the very lossy case, momentum is still conserved but
half the kinetic energy in the system is lost.

So what happened to the energy in the lossy case. Since it is a given
in Newtonian (and Einsteinian) physics that energy can neither be
created nor destroyed, it must still be around but in some form other
than the kinetic energy of the two balls. And indeed it is! The balls
deformed as they collided and stuck together. That implies internal
friction associated with the deformation, creating heat to exactly make
up for the lost kinetic energy. It isn't a lot of heat. It would take a
very sensitive thermometer to notice the temperature difference, but
the difference is there.

Angular momentum

Yes, there is an angular version of momentum, and it is exactly what
you might imagine it would be. We start with the angular version of
Newton's second law:

T = I α

If we go through the same steps we did to derive linear momentum, we
come up with an angular impulse-momentum equation:

T Δt = I Δω

The left side, torque times the time interval, is
angular impulse. The right side, moment of inertia times the difference
in angular velocity, is the change of angular momentum.

Unlike linear momentum, you don't need a collision to invoke
conservation of angular momentum. The reason you need a collisioin for
linear momentum is that objects don't change mass dynamically within a
closed system; it is hard to imagine such a scenario. But it isn't hard
at all to imagine a change in moment of inertia. Merely distorting the
shape of a mass can change its moment of ineria, especially if the
distortion simply moved bits of mass straight outwards from the
axis. That way, we don't have to worry about a fresh torque arising
just from distorting the shape, in order to accelerate or decelerate
part of the mass.

The most well-known visualization and demonstration of conservation of
angular momentum is an ice skater doing a spin. Here is a typical video.

The skater starts a spin (an angular velocity around a vertical axis,
through the middle of her body and down to the point where the skate
meets the ice). She has her free legs and both her arms extended out as
far as possible. Specifically, they are as far as possible from the
axis. Remember that moment of inertia is computed by adding up all the
tiny particles of mr^{2}. Extending arms and legs away from the axis of rotation ensures a high I.

Then she retracts her arms until they are folded against her body, as
close to the axis as possible. She tucks her extended leg to be as
close as possible to the weighted leg -- which itself is part of the
axis. The result is a huge freduction in r for a lot of mass, and therefore a significant reduction in I.

You can tell the moment of inertia has been reduced because the spin
speeds up dramatically. It does so even though there is no new torque
added to the system. Indeed, the spin should be slowing down because,
even as low-friction as steel-on-ice is, there is some small friction
sucking energy out of the system. But no, the spin speeds up very
visibly. That can only be explained by conservation of angular
momentum. We know the moment of inertia shrunk a lot between time 1 and
time 2. What we are saying is that:

I_{1} ω_{1} = I_{2} ω_{2}

Since I_{2} is so much smaller than I_{1}, the spin has to speed up (ω_{2} has to increase to a lot more than ω_{1}) in order to conserve angular momentum.

Last
modified -- Dec 4, 2022

Copyright Dave Tutelman
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