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Physical principles for the golf swing

Energy and Momentum

Dave Tutelman -- November 26, 2022

Momentum

Just another way of saying F=ma

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

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 v²
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.

	Lossless collision COR = 1.0 Balls bounce perfectly		Lossy collision COR = 0 Balls stick together
	Before Collision	After Collision	Before Collision	After Collision
Picture
Momentum	MV	MV	MV	2M×½V = MV
Kinetic Energy	½MV²	½MV²	½MV²	½×2M×V²/4 = ¼MV²

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 mr². 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:

I₁ ω₁ = I₂ ω₂

Since I₂ is so much smaller than I₁, the spin has to speed up (ω₂ has to increase to a lot more than ω₁) in order to conserve angular momentum.

Last modified -- Dec 4, 2022