Friday, May 4, 2018

Curving pitches, kicks, and shots…

So… In baseball and softball, many batters batters hate the curveball… Because it curves…
NOTE: The physics of this is true for ALL sports where a ball moves through the air… tennis, golf, ping pong, volleyball, soccer (bend it like Becker)…
Now, the curve in question is in addition to the fact that it is moving from the picture to the batter as a projectile which automatically means that it's up and down is changing at the same time it moves closer and closer to the batter.
Imagine slow-pitch softball or a lobbed pitch… In the extreme, think about soft toss… The ball moves up and down as it approaches the strike zone.
So, every pitch regardless of type is, from the moment it is released affected by these three forces:
  1. Gravity pulls the ball down toward the center of the earth.
  2. There is a tiny, tiny buoyant force that is far far less than the force of gravity.
  3. There is air resistance (friction) that opposes the motion of they ball and which is always directly directly opposite the direction of the instantaneous velocity.
Now, let's talk about rotation! Because, except for a knuckle ball, pitches rotate…
Imagine a ball spinning on an axil, not moving. When the ball spins, the air around it is "drug" in the direction of the spin. Now, if you add in the movement of the ball, something happens!
First if the ball is thrown and spin is ignored, you get an object moving through the air. The air flows around the ball equally. Nothing interesting there…
However, if you account for the spinning of the ball… and the air that is being drug around it by the spin, you start to see something.

When air moves faster, is gets stretched out. There are fewer molecules of air in the same space. From the gas laws, it is known that if you have fewer molecules in the same space, the pressure is lower (given constant temperature, which can be assumed over the diameter of a baseball or softball).
So, the spinning ball drags the air around it, AND those moving molecules of air interact with the moving air as the pitch moves through the air.  Where the drug air is going the same way as the air flow, the molecules move faster, but where the drug air is going in the opposite direction, the air flow slows down.
The result is that one side of the ball has high pressure and the other side has low pressure. The difference in pressure creates a force. The air pressure difference pushes the ball form the high pressure region to the low pressure region. 
That is to say there is a force acting in the direction of the low pressure. Given Newton's Second Law, it is known that F=ma, so where there is a net force, the ball must accelerate (change velocity) in the direction of the force.
Thus, the rate of acceleration can be found based on some function of the difference in pressure (F) divided by the mass of the ball:
a = ƒ(p1 and p2)/m
So, what can a pitcher do to make the ball curve? Well… you can't control the mass of the ball. You can control, however the spin.
Since the air moving around the ball due to it moving is the same on all sides, the spin accounts for the difference in pressure. Both p1 and p2 are affected by the spin
The net air flow is on one side pitch velocity + drug air.
The net air flow on the other side is pitch velocity - drug air.
Thus, since the difference in pressure is a result of the DIFFERENCE in the net velocities of the air on opposite sides of the pitch path, the more spin, the bigger the difference in pressure.
AND, then, the bigger the difference in pressure (back to the math)…
a = ƒ(p1 and p2)/m
…the greater the rate of acceleration.
By controlling the spin allows a pitcher to control where the high and low pressure zones are. The ball will break into the low pressure zone.
A pitch, then, with backspin, will dive. If the rotation is the other way (e.g. a good overhand throw), the ball will actually float some and the effect of gravity will be lessoned.
The LESS the axis of rotation aligns with the path of the ball, the GREATER the affect of the spin on the ball's path.
Conversely, if the axis of rotation aligns with the path of the ball's movement, then the affect of the spin is zero. Think of a spiral football pass. IF the pass/punt spirals properly, the affect of the spin is zero.
An overhand fastball rotates back toward the pitcher. Thus, there is low pressure on top of the ball.
An underhand fastball (or a topspin tennis shot) rotates forward toward the batter. Thus, there is low pressure on the bottom of the ball.
Cutters, sliders, etc have various spins that move the low pressure zone to different places on the ball.
1.) Since the rate at which a ball's velocity changes can be found by
a = ƒ(p1 and p2)/m
and since the difference in pressures is caused by the spin, then changing the direction and rate of the spin will change the direction and rate of the acceleration (of the break).
2.) Increasing the rate of spin increases the difference in pressures. The greater the difference in pressure, the more the ball with break (accelerate).

Additional Notes:

1.) The diameter of the ball affects the velocity of the spinning surface. Thus, two balls of the same mass spinning at the same rate, but having different diameters will break differently. For example, a huge beachball with the same mass as a volleyball will curve more with the same spin.

Tuesday, May 1, 2018

Friction: A Force That Opposes Motion

Newton's Second Law leads to the formula that connects force, mass, and acceleration:

F = ma

Earlier, it was explained that there are only four fundamental forces, but that every other force that is easily observed is derived from them. Thus, friction can be understood as a force which arises due to the interaction of the atoms between two objects when they are in contact. The strong, weak, and electromagnetic forces work such that there is a resistive force when one object slides across another.

A resistive force (also know as oppositional force) works in the opposite direction of the active forces, such as pushing or pulling something.

Before working with friction, there are two more concepts that need to be understood:


Weight is the force created by gravity and its magnitude can be found by multiplying the mass of an object by the rate at which gravity accelerates things. On earth, for introductory physics students,. the value for acceleration due to gravity of 9.8 (m/s)/s or 9.81 (m/s)/s is commonly used.

In the metric system, the Newton is the unit for weight. The pound is also a unit of weight*.

The direction for weight is always "down." That means that (when on earth) weight always acts in direction that is perpendicular to the horizon and toward the center of the earth.

Normal Force

The concept of Normal Force is easy to understand. Calculating the normal force (which is usually indicated as Fn) requires using trigonometry EXCEPT when the surface on which an object slides is level (has an angle to the horizon of zero).

The formal force (Fn) is defined as the component of the weight (mg) of an object that is perpendicular to a surface. When the surface is level ALL of the normal force is perpendicular to the surface. That makes things easy. Thus, the normal force IS THE WEIGHT, but this is ONLY true with the surface is level (has and angle to the horizon of zero).

When the surface is NOT level, then, math…

Fn is labeled, and it is clear that it pushing the box in a direction that is perpendicular to the surface of the inclined plane.

Note that F3 is perpendicular to the horizon. That would mean that F3 is the weight or is found as m•g where m is the mass and g is acceleration due to gravity.

Force of Friction

There are two types of friction. One is when an object is already sliding. One is when the object is at rest. They are cleverly named sliding friction and static friction.

Sliding friction is sometimes indicated as fk where the k stands for kinetic (which means moving) and static friction is indicated as fs.

NOTE: in this nomenclature, the fi is lowercase. the k and the s may or may not be subscripted, depending on the type-setting process. 

Intuition will tell you that pushing something over ice is easier than pushing it over asphalt. There therefore must be some way to indicate the degree of slipperiness an object has (or stickiness).

There is. It is called the coefficient of friction. And there are two sets of coefficients: one for each type of friction. Sliding and static coefficients of friction are abbreviated as µk and µs respectively.

 Here is a link that will allow you to look at coefficients of friction for various materials:

Calculating the Force of Friction

So, using all those concepts, calculating the force of friction becomes pretty simple. Friction can be found by multiplying the normal force by the coefficient of friction:

fk = (Fn)(µk)


Clean dry steel sliding on steel has a coefficient of friction of μ = 0.78. If a block with a mass of 4 kg slides across, what is the force of friction.

First, calculate the normal force:

Fn = mg
Fn = (4)(9.8)
Fn = 39.2

Now, use Fn to find the force of friction.

fk = (Fn)(µk)
fk = 39.2 • 0.78

fk = 30.576 N

*The English system unit for mass is called the Slug and is approximated by taking the weight of something and dividing by 32. So a 110 pound person would have a mass of 3.4 slugs.

Monday, April 16, 2018

Finding Time Using the Distance Equation

Any calculation of time (t) is relatively easy in the cases where either vi or a was zero. In the cases where neither are zero, the math results in a second degree polynomial equation such as:
0 = x2 + 3x - 12
Where this occurs in physics of motion is in the full distance equation:
df = di + (vi)(t) + (1/2)(a)(t2)
Although this does not exactly match the expected form for a quadratic equation, it can easily be rearranged as such:

df = di + (vi)(t) + (1/2)(a)(t2)
0 = di - df + (vi)(t) + (1/2)(a)(t2)
0 = (1/2)(a)(t2) + (vi)(t) + di - df

Keep in mind that di - df will yield a number when the values are plugged in and simplified.
To solve these problems, the steps are the same for any problem in science, BUT the algebra becomes harder.
After you have plugged in the numbers, combine like terms and simplify. Suppose the following:
df = 89
di = 50
vi = 4
a = 6
and you need to find t
df = di + (vi)(t) + (1/2)(a)(t2)
89 = 50 + (4)(t) + 1/2(6)(t2)
0 = -39 + 4t + 3t2
(Put it in normal quadratic form.)
0 = 3t2 + 4t - 39
Now you can either factor or use the quadratic formula to find the values for t:

0 = ( t  - 3 )( 3t + 13)
0 = t - 3    AND   0 = 3t + 13
3 = t        AND   -13/3 = t

Since time cannot be negative within the context of classical physics, only t = 3 is a valid answer.
Using the quadratic equation will yield the same results.

While it is far easier to find t when either a or vi is zero, the math to find t when that is not the case is not beyond the skills of a student taking an introductory physics class.
For another look at this process, check out this video:

Thursday, March 22, 2018

Physics of Motion: What is What?

Physics of Motion: What is What?

The equations:

df = di + (vi)(t) + (½)(a)(t2)   This will answer almost all “how far” questions.
Use this if:
Only 1 velocity is given
You have a distance given (df is rarely 0, but sometimes it is the unknown)

vf = vi + (a)(t)    This will answer almost all “how fast” questions.
Use this if:
2 velocities are given
You don’t have distance

v(ave) = (vi + vf)/2
Use this if:
You need to find the average velocity
You know BOTH vi and vf (you might have to calculate vf!

vf = (vi + ∆v)
Use this if:
You need to find the final velocity AND
You know BOTH vi and how much v changes (e.g. speed increases by 10 m/s)

df = (di + ∆d)
Use this if:
You need to find the final distance, total distance, or final position AND
You know BOTH di and how much d changes (e.g. it moves 10 meters)

The hints and helps:

It is vi if…
“…traveling at a rate of…”
“…moving at…”
"…has a velocity of…"

If it is “at rest” then vi = 0 and di is probably 0

The words "begin" and "starts" generally go with the initial values.

If it is “moving at a constant” rate or if it “has a constant velocity” then a = 0

It is possible that something not given (but which is not the thing to be found) should have a value of 0...

Monday, October 23, 2017

Summary of Atomic Theory

The 21st Century understanding of atomic theory is very effective in helping explain and predict how substances interact. It is the result of a growing body of knowledge that dates back centuries.


In the middle of the 5th Century BCE, Democritus proposed ideas that were correct in many ways. Democritus believed that all matter consisted of extremely small particles that could not be divided. He called these particles atoms from the greek word transliterated as atomos, which means "uncut" or "indivisible." 
He thought there were different types of atoms with specific sets of properties for the different substances in creation. The atoms in liquids, for example, were round and smooth, but the atoms in solids were rough and prickly. 
Though Democritus's ideas didn’t catch on, he was surprisingly accurate in his assumptions. Scientist of the 21st Century eventually concluded that many of Democritus’s ideas were correct, and they extended the understanding of atomic theory in stages that led to the current model.

John Dalton

Dalton proposed the theory that all matter is made up of individual particles called atoms, and in it, he identified several things that have lasted into the modern atomic theory:
  • All elements are composed of atoms.
  • All atoms of the same element have the same mass, and atoms of different elements have different masses.
  • Compounds contain atoms of more than one element.
  • In a particular compound, atoms of different elements always combine in the same ratios.

J. J Thompson
Thompson's experiments provided the first evidence that atoms are made of even smaller particles that have positive and negative charges.

Ernest Rutherford
Rutherford's model extended what was previously understood by identifying that atoms have a central, relatively dense (compared to the entire volume of an atom) nucleus around which electrons move. 

Niels Bohr

  • In Bohr's model, electrons move with constant speed in fixed orbitals around the nucleus.
  • Electrons must orbit the nucleus in one of several fixed, specific orbits, and each orbit represents a specific energy level.
  • The first orbital represents the lowest electron energy level, and the other orbitals represent progressively higher and higher energy levels.

Electron Cloud Model

  • Evidence following Bohr's work led to the understanding that the electrons do not orbit the nucleus like a planet.
  • While they do exist at specific energy levels and occupy orbitals, their position in the orbital is never 100% certain. They are somewhere in the orbital, but exactly where cannot be known specifically. Each orbital can be, therefore, conceived as an electron cloud. 

The progression of Atomic Theory in the modern era took place in various steps and stages. The resulting body of understanding results in a model of atoms that is highly effective and useful in understanding and predicting the behavior of substances.

Thursday, October 5, 2017

Solving Science Word Problems (and any other kind of word problem, too.)

Throughout science, there inevitably comes a time when it is necessary to use some "known relationship" to figure out what happened or what might happen. In physics and chemistry, especially, this is the case, but sciences like sociology use "known relationships," such as population growth models, to explain or predict certain happenings.

In many, many cases, the "known relationship" is expressed as a mathematical equation. The relationship between the rate something happens and the amount of something produced is very common, so this discussion will use a couple of rate problems as its example.

In most general terms, the amount of something produced is equal to the rate of production times the amount of time production took place. For instance, in a displacement problem, rate is how far something moves in a given time and motion of an object is produced. In a general form a rate problem might look like this: 
O = R•t 
where O is the output, R is the rate, and t is the time.

So what is this process of solving science problem that will use rate problems as its example?

There are three steps to solving ANY science problem (or word problem of any kind, for that matter).

The following two examples will be used as the steps are discussed:

EXAMPLE 1: A baker's oven will hold only 1 pan of cookies, and each pain has space for 12 cookies. The baking time on cookies (including putting the dough on the pan) is 20 minutes. Therefore, the rate cookies are baked is 12 cookies per 20 minutes. How many cookies can be baked in 80 minutes? 
EXAMPLE 2: A car travels at an average rate of 20 MPH for 3 hours. How far does it go.

The first step is to write down what is given AND what is asked.

Example 1:
R = 12 cookies/20minutes
t = 80 minutes
LOOKING FOR "How many cookies"
Example 2:
R= 20 MPH
t=3 hours
LOOKING FOR "How far does it go"
A student familiar with physics motion problems would see Example 2 as a "distance" problem and would use the distance equation variables. In such a case, R would be velocity, v, t would be time, t, and LOOKING FOR would be distance, d (or in some cases, s). 
The second step is to write the equation that relates to the question.

Example 1:
Cookies = R•t

Example 2:
d = vt

From time to time, it is necessary to rearrange an equation so that it yields the answer you are trying to find. For example, if you trying to find time in Example 2 above, you will need to isolate the t variable. MANY TIMES it is easier to do the algebra BEFORE you plug in the numbers and units!

d = vt               (To find t divide both sides by v.) 
d/v = vt/v           (The v on the right side cancels.)
d/v = t
Once you have an equation in the form you need, you are ready for the next and final step.

The third step is to plug in what was given and solve. Assuming the second step was completed, the following solutions would emerge:

Example 1:
Cookies = R•t
Cookies = 12 Cookie/20 min • 80 min
Cookies = 960 Cookie•min / 20 min
Cookies = 48 Cookie 

Example 2:
d = vt
=  20 mi/hr * 3hr
d = 60 (mi•hr)/hr
d = 60 mi

For additional examples and a video explanation, check this out:


There are three steps to solving problems.

1: Write down what is given.
2: Write down the relevant equation.
2b: Rearrange the equation so that it yields the answer you want.
3. Plug in and solve.

Tuesday, August 1, 2017

Physical Quantities: Equivalency and Conversion

Objects in the physical world can be described according to various physical properties, such as length, volume, mass (or weight). Since the beginning of humanity, people have come up with ways to measure things in a standard way.

Standards and Units

Measurements were usually created to compare to some reference object or other agreed-upon standard. For instance, if trading sea shells, the standard would be… a seashell. One sea shell equaled one seashell.

Okay… hang in there… keep going…

So, along the way, magic occurred (not really) and people began to equate the word "one" with "unit." So, if you said, "Give me eight units of seashells," since seashells were measured in… themselves… you would get eight seashells.

But suppose a guy sold sand. Selling grains of sand would be… dumb. Suppose (sticking to the beach motif) he had a coconut hollowed out. His thing was, one scoop of sand for one seashell…

So for sand, the unit would be scoops.

A buyer would say, "Give me eight scoops of sand," and would pay eight seashells for it.

It must be time for a definition!

Unita quantity chosen as a standard in terms of which other quantities may be expressed. (Oxford dictionary.

Not very helpful, but in the right direction!

Anne Marie Helmenstine, Ph.D. ( defines it this way:

A unit is any standard used for comparison in measurements.

Going back to strictly intuition, most people already know what units are. "Comparison" is a good word to hang onto, so do that.

Now think about things you already know. Gallons of gasoline. Pounds of lunch meat. Miles to the next town.

Using a standard unit like miles allows us to make reasonable comparisons. A mile is a mile is a mile. So if it is 200 miles to Townville and 400 miles to Villaton, since a mile is a mile is a mile, it must be twice as far to Villaton.

If I have 4 gallons of paint and you have 2, I have twice as much paint, since a gallon is a gallon is a gallon.

So, think about things around you. What are the standard units for them.

Distances between towns?
Soda (pop)?

Chances are, for at least one of those, you thought of different standards. For example, you might find 20 oz bottles of soda as well as 2 liter bottles!

There are times when converting from one standard unit to the other is desirable.

Conversion and Equivalency

The process of converting from one standard unit to another boils down to finding how many of one thing is equal to one thing of the other.

Good news, #1: The math on this is easy.

Good news, #2: You can usually look it up on Google.

Bad news, #1: An explanation follows anyway.

To start with a definition of conversion…

Conversion: The process of finding out how many units from one standard are equal to how many units from another standard.

Suppose you discover that, in a fantasy novel, the people in one town sell milk by the Nallog and in another by the Ecnuo. Some guy has a barrel on which are etched lines for both Nallogs and Ecnuos, and another dude sees that the milk comes to the 2 Nallog mark and also the 256 Ecnou mark.

In the (silly) example, you can see that 2 Nallogs = 256 Ecnous. That is, the two quantities are equivalent.

Finding out how many Ecnous are in a Nallog is easy! Divide the bigger number by the smaller. Bam!

256 ÷ 2 = 128

So 1 Nallog  (the bigger unit) is equivalent to 128 Ecnou (the smaller unit).

In the (silly) example above, the conversion factor is 2.

Conversion factor: A number that, when multiplied will convert one unit to another.

Oxford Dictionary says it like this: an arithmetical multiplier for converting a quantity expressed in one set of units into an equivalent expressed in another.

To find (or derive) a conversion factor for anything, all that is needed is to know how much of something is present in the two different units. Then, to find the conversion factor, divide the larger by the smaller of the quantities. The quotient (number after you push the equal button) will be how many of the smaller thing (the bigger number) are in the bigger thing (smaller number).

No, that's not confusing, is it?


A container has 591.5 milliliters of shampoo in it. The label also says it has 10 clarkens (which are cleverly abbreviated as "Clar"). Find the conversion factor for milliliters and clarkens.

591.5 ml
_______  =   59.15 ml/Clar
10 Clar

Thus, 1 Clar is equal to 59.15 ml.

BAM! Easy math!


It is always best to just measure things in the units you need. But sometimes, that's impossible. Sometimes your most accurate measuring device provides units that you have to convert.

A solid understanding of equivalency and conversion prepares students to face the demands of science and engineering.