We already have F=ma, what more do we need to
understand how force produces motion?

People who can ask that question are the ones who most need to
understand the real messages of kinetics and kinematics.

Position is not the same as motion (or velocity).

Acceleration is not the same as motion (or velocity).

Far
too many laymen -- golfers, instructors, and TV personalities -- don't
understand this. And they especially don't understand the second
bullet, that acceleration and velocity can have different, even
opposed, values. I have even met a fellow electrical engineer with whom
I had to break off discussion because he refused to recognize that
difference. (And he took enough physics course that this should not
have been a problem.)

So please bear with me while I preach. At
times, it may seem like I'm beating a dead horse, but without getting
this distinction you will never
understand some crucial aspects of the swing.

Introduction and examples

Because of F=ma
we correctly equate force with acceleration. The force graph over time
is the same shape as the acceleration graph over time. The mass is just
a scaling factor; it tells us how tall that shape is, but it is still
the same shape. So, in the discussion below, we will use "force" and
"acceleration" almost interchangeably, except when there is a real need
to distinguish between them.

Let's start with a very simple example of position, velocity, and
acceleration. Here's
an example
I used for a more elementary tutorial look at the basics of physics. It
shows position, velocity, and acceleration for both positive and
negative acceleration. (Negative acceleration has its own name,
"deceleration".)

Now we'll look at a live example
of how acceleration and velocity are
different. Here's a video of someone pushing a car for exercise.

He is turning force into velocity. But not directly! He is turning
force into acceleration, which gradually turns into velocity. Since
force is the same shape as
acceleration (because of F=ma),
we can sort of "see" the acceleration; it is pretty constant at the
maximum this guy can exert on the car. So we have a constant
acceleration. But, if we look at the wheels spinning, it is clear that
they keep spinning faster and faster for at least the first 7 seconds.
So we have a more-or-less constant acceleration and an increasing
velocity.

Once you are comfortable with the tutorial and the live example,
let's dig into what we can glean from the graphs we get from motion
capture systems and biomechanics tech reports.

Velocity to
position - slope

Let's
look at how velocity can tell us about position. In order to do so,
we'll take a quick journey back to grade school. Remember the problems
you had to solve "DRT" problems from the formula

Distance
= Rate * Time

That
tells us how to go from a velocity (a rate) to a position (a distance
from where we started). The way we do it is to add some distance for
every unit of time the "system is running". How much distance per unit
of time? The rate!
That's what a rate is; it's something per unit of time. In this case,
the something is distance. For instance, the rate may be 15 miles per
hour. That means for every hour, we add 15 miles to the distance.

What
does this look like on a graph? Here's a graph of a simplet DRT
(Distance = Rate * Time) problem.

For elementary DRT problems, the rate (velocity) was always a
constant. On the graph, the velocity is the green curve, and it is a
constant; it does not change with time.

The distance is a function of time. In fact, we know the exact funtion
of time. It is our favorite formula, D=RT.
As time increases, distance increases as a straight line. The slope of
the line is the velocity. In fact, that is the definition of slope --
the rate of change with respect to time.

This tells me we could find the distance curve by laying together
infinitesimal line segments, where the slope of each segment is equal
to the velocity. Note that, in this case, the velocity is a constant.
Therefore the slope is a constant. So we don't need to resort to a lot
of tiny segments. If the slope is constant, then the curve on the graph
is a straight line. And that is what we see here.

Velocity to position - area

There
is another way of looking at velocity-to-position calculations. Looking
at it from the slope point of view is fine if the velocity is constant,
but we need something a little more powerful if the velocity varies.
And, in most golf physics problems, the velocity does vary.

The diagram shows a velocity (the red curve on the graph) that varies
with time. Suppose we want to find out how far we went during the
interval from time=0 to time=100. All we know about converting velocity
to distance is D=RT,
but R (the
rate, or velocity) keeps changing. How can we do D=RT if R is varying?

We
can solve the problem by slicing time into small pieces, so small
that the velocity is pretty close to constant over the small piece. The
diagram shows the slices as rectangles, whose width is the time slice
and whose height is the velocity at that particular time. Let's look at
a few things about the
rectangles:

The velocity (the red curve) does not change a lot over a
single rectangle.

We have chosen a rectangle height that tries to intersect
the
velocity in the center of the top line of the rectangle -- sort of
equals out the velocity above and below the rectangle.

Now we will say, "OK, let's ignore the small change in
velocity over a
single rectangle, we'll consider it to be a constant velocity. The
magnitude of the velocity is the height of the rectangle."

Let's look at the distance traveled over an individual time
slice. Suppose the width of the slice (as a time) is t_{0} and the height
(which is a velocity) is v_{0}. Then v_{0}t_{0} is the distance
traveled during t_{0}, because D=RT.

Let's look again! v_{0}t_{0} is not just the distance. It is also the area of that
rectangle. Remember your geometry; the area of a rectangle is
base times height.

Bottom line: if we add together
the v_{0}t_{0} for all the
individual rectangles, we get:

A very good approximation of the distance traveled, and also...

A very good approximation of the area under the velocity
curve.

What that means is that the distance
is equal to the area under the curve.
That should be true no matter how you find the area. Of course, you
have to be careful about units of velocity and time, and keep the area
in those units -- but if you do, then the area is the distance traveled.

Also worth noting: as you reduce t_{0}
(and therefore increase the number of slices you'll have to add up),
the more accurate your approximation of both distance and area. That's
what a lot of computer usage is in biomechanics research -- adding up
little increments of D=RT to get a
total distance.

Examples of position, velocity, and acceleration: clubhead [speed] in
downswing.

Acceleration to
velocity

Acceleration has the same relationship to velocity that velocity does
to position. Acceleration is the
rate of change of velocity. If you have a graph of velocity,
acceleration is the slope of the graph -- just as velocity is the slope
of the position graph.

Please please please remember that acceleration
is not velocity. Too often I see golf instructors confuse
acceleration and velocity, and wind up with ridiculous conclusions. One
of the more common confusions sounds something like, "You say there is
a backwards force on the club? But then it would go backwards, and we
both know that doesn't happen." Let's deconstruct this and see what is
really being said.

A backwards force will produce an acceleration in the direction
of the force. - Perfectly true.

A backwards acceleration means the club would start going
backwards. - Ummm, no!
That is a statement about velocity,
not acceleration. A backwards acceleration means the club will
begin to slow down. If it is applied for a long enough time, then it will start to go backwards.

Yes, there are situations where a reverse acceleration will look like an instantaneous reversal
of direction. If the force is large enough and the mass is small
enough, then the acceleration will be so big that our eyes will miss
the slowing-down phase altogether. Let's look at a couple of examples.

An elite tennis player returns a stroke. The ball arrives at
50mph and the player hits it back at 75mph. (Yes, those are pretty
typical speeds.) The ball is on the racquet face for only 5msec. When
you do the f=ma
math, that takes a force of 143lb. Given that the tennis ball weighs
only 57g (that's about 2oz), 143 pounds of force changes its direction
so fast it looks instantaneous.
It isn't instantaneous; the first 2msec of the 5msec impact it is
slowing down.

A freight train hits the brakes to stop. The brakes cause a
backward acceleration, but the velocity keeps going forward for a while
before the train finally stops. From a typical speed of 55mph, it takes
almost three minutes to come to a halt -- with a backward acceleration
the whole time. The problem is that the train has a huge mass and the
brakes can apply only so much force, so the backward acceleration is
pretty small.

You probably knew this already, but backward acceleration, or negative
acceleration, is called "deceleration". I have been calling it
"backward acceleration" to make a point.

If you want to better understand force, acceleration, velocity, and
position, you should review the calculations for examples #1 and #2
above. I have the detailed solutions to both at the end of this section.

You can use the same methods to go from acceleration to velocity that
we did to go from velocity to position. For instance:

Use V=aT
the way we used D=RT
for position (distance) problems.

Find the area under the acceleration-vs-time curve. Sometimes you
can just calculate it, if the curve is a shape whose area we know. Much
of the time, you will need to break it down into small slices of time,
then add the areas of those slices together.

Now a couple of simple numerical examples. Let's go back and solve the
two examples above, the tennis stroke and the frieght train.

The
tennis stroke - an elite tennis player returns a ground stroke.
What is the force applied by the racquet to the ball, averaged over the
duration of impact?

We can look up typical numbers, and find that:

The typical ground stroke in a pro tennis match is struck at
75mph, and loses speed to air resistance and friction with the ground
before it reaches the other player; let's use 50mph as the remaining
speed.

The ball is on the face of the racquet for about 5msec.

A tennis ball has a mass of 57g.

So a good ground stroke turns a velocity of -50mph into +75mph
in 5msec.

That's a change in velocity of 125mph. (75mph - (-50mph)) in
5msec.

Let's get useful units. We'll use International System (SI)
units. So 125mph becomes 56m/s. (I used Calc98 to convert units. I
strongly recommend it.) 57g becomes 0.057Kg, and 5msec becomes 0.005sec.

"Average acceleration" means we can assume the acceleration is
constant over the duration of impact. It isn't constant, but if we took
an average, the number would come out as if it were that constant
value. So we will find acceleration a.

We know that V=aT. Since we
know V=56m/s
and t=0.005s,
the equation becomes 56=0.005a

Solving for a, we get a=56/.005=11,176m/s^{2}

We can plug this, along with the tennis ball's mass, into F=ma, and come
up with F=.057*11,176

Now we just have to do the computation and find that F=637N=143lb.

It should not come as a shock that applying 143 pounds of force to a
2-ounce ball would certainly look like an instantaneous change of
direction for the ball. In fact, it took 5msec, but that is too fast
for the eye to distinguish.

The
freight train - a freight train applies the brakes. How long
does it take to stop?

Again, let's start by looking up typical numbers. I was hoping
for a time to stop, but all I could find was a distance, so we'll have
to do some calculating to find the time.

A freight train going a typical 55mph takes between 1 and 1.6
miles to stop.

Get our numbers into consistent units. This time we'll use the
American system of feet and seconds. (We don't have enough information
for mass or force, so we won't have to worry about slugs and pounds.)

55mph = 81ft/s

A mile is 5280ft. We'll pick a nice round 7000ft, which is
roughly midway between 1.0 and 1.6 miles.

What equations might we have that would give us more
information?

V=at.
But we only have V;
we have neither a
nor t.

D=½at^{2}. But we only
have D; we
have neither a
nor t.

So we have 2 equations in 2 unknowns. We should be able to
solve them, but it could take some work.

There are certainly ways of solving such sets of equations. But
if we look carefully, we can see a quick solution.

The equation for V is simply at.

But there is an at in D=½at^{2}=½(at)t
So we can plug in V
for at and
get D=½Vt. We already have values for D and V, so...

7000=½*81t.
We solve for t
and get t=7000*2/81=173sec.
That is nearly 3 minutes (180sec)!

Now we can go back to basics, V=at. We know
both V and
t now.

81=173a,
which solves to 0.47ft/s2,
or 0.32mph/s.

If you didn't see the shortcut that I saw, there are
not-too-difficult ways to solve small equation sets like this. They
justrequire more work. Sometimes (often) insight saves work. But it's
usually a good idea to go back and check -- make sure your insight was
correct. In my case, I plugged my values for a and t back into the
original equations.
Indeed, they did balance as they should have, so my work was correct.

The bottom line is that our train loses about a third of a mph every
second. Starting at 55mph, it takes almost three minutes of backward
acceleration to get down to a standstill.

Optional Math Note

If you know a little bit about calculus, this note may give you a more
quantitative insight into position, velocity, and acceleration. If you
don't know calculus, just skip over it.

Differential calculus

Let's describe position as a number x. We will
restrict ourselves in this note to positions only in the x-axis. If you
think about it a bit, that means that velocity and acceleration can
only have components in the x-axis. First,
let's look at what velocity and acceleration are in terms of
differential calculus. Then we'll look a bit at the implications for
integral calculus and differential equations.

In
terms of
differential calculus

In
terms of
geometry

Position

x

The
position itself

Velocity

dx dt

or x'

The
slope of the
position curve

Acceleration

d^{2}x dt^{2}

or x"

The
slope of the
velocity curve

Integral calculus

As you would expect if you remember that integration is
anti-differentiation, you can go from acceleration to velocity to
position by integrating at each step. If velocity x'=v and
acceleration x"=a,
then:

v = ∫ a(t) dt

x = ∫ v(t) dt = ∫∫ a(t) dt dt

Differential equations

If you get into making or solving
mathematical models for biomechanics, you will spend much time solving
the differential equations for the F=ma
of your particular model. (Also the torque equivalent, but we haven't
gotten there yet.) That is what mathematical models turn out to be;
sets of differential equations.

Let's try a very simple example. The air drag on a golf ball is
proportional to the square of the velocity of the ball, and opposite in
direction to
the velocity. So for motion in the x-axis:

F = - K x' ^{2}

since x'
is the velocity. We need the minus sign because the force is opposite
the velocity in direction. Now let's remember that F=ma. So,
substituting x"
for acceleration, this becomes:

F = m x" = - K x' ^{2}

This is a differential equation that we can solve to get the motion of
the ball (assuming that air drag is the only force acting on the ball
-- not often the case, but we're keeping it simple here). We can make
it look more like our usual idea of a differential equation with just a
little algebraic manipulation:

x" = -

K m

x' ^{2}

or
perhaps

x" +

K m

x' ^{2} = 0

Curved motion

Remember
that force is a vector quantity. That means that it might not
be pointed in the same direction as velocity, nor directly opposed to
the velocity. For example, this picture shows a force at right angles
to the velocity vector. There is no component of the force that is
either accelerating nor decelerating the magnitude
of the velocity. But the force is real, and it will impart a real
acceleration to the gray ball. That acceleration will be one to change
the direction
of the velocity, directing it more downward.

Let's
look at a picture of this situation over time. This picture shows the
initial force at a right angle to the initial velocity. The force is
downward, and it continues to be downward.

Remember, velocity is the accumulation of acceleration over time. If
there were no force, then the path would continue horizontally (the
dotted line). But the force, and thus the acceleration, is downward, so
the velocity will accumulate a downward component. The original
rightward component of velocity is still the same, but there is a
growing downward component of velocity as well. The result is the
parabolic path shown.

This is not a made-up example; it is very real and even common. If you
throw or shoot a ball horizontally, it will retain its horizontal
velocity (ignoring aerodynamic drag), but will be acted on by a
downward force -- gravity. In
a vacuum, the path would be a parabola just like this one, and for just
the same reasons.

Now
let's look at a more interesting and complex example. Suppose the force
changes direction as the velocity changes direction, so that the force
is always perpendicular to the path. But we will let the force remain
the same in magnitude; only the direction changes. What does this do to
the path of the ball?

Since the acceleration is always at right angles to the
path, the magnitude of the velocity never changes, just the direction.

That means the picture is always the same, except rotated
through some number of degrees. Therefore the curvature of the path is
always the same.

What we have just described is the ball moving in a perfect circle at a
constant speed.

A real example of this behavior is
twirling a ball on the end of a string. We've all done this at one time
or another.

The string is exerting a force that accelerates the ball toward our
hand (the center of the circle). That force is called the centripetal
force.
Since we know that every force has an equal and opposite reaction, the
ball exerts an equal and opposite force on the end of the string,
called centrifugal
force. (Note that it is a real force, but most informal mentions
of "centrifugal force" are not real at all. They refer to an apparent
force that pushes the ball or other body outward. Bear in mind that
there is no need for a force to explain why the body wants to move in a
straight line rather than a curved path.)

The magnitude of the centripetal force F needed to move
mass m at
velocity v
in a circle of radius R is:

F
=

m v^{2} R

Last
modified -- Oct 2, 2024

Copyright Dave Tutelman
2024 -- All rights reserved