Louisiana Governor Ricky Bobby Jindal compared the poor graduation rates of the state universities to a football coach without a winning season hinting that you would get a new coach. He hints that the problem with low graduation rates lies mostly with the institutions. Sure, we have a lot to do with this. We can provide more services and help for students, but all of this can't be blamed on higher education. According to the 2008 ACT College Readiness Report only 14% of Louisiana students meet the benchmarks for college readiness in all areas. Should it be so surprising that our graduation rate is only something like 25%? Oh - also, I would like to point out another problem with Ricky's analogy. The average winning rate for coaches is 50% (for every win, there is a loss).

So, what should we do? What can we do? I think step one is to think about the purpose of college (and high school and education). This is really what people get all warped about. Education should not be job training, it should not be an entrance requirement for certain jobs. As I have said before, education is about being human. But, there is stuff we can do. Help students. Help them help themselves. Think more about education of our future teachers.

Look how far those limbs extend horizontally. That branch is about 12 meters long. Why is this amazing? Have you ever tried to hold an 8 foot 2 x 4 board horizontally by holding one end? Pretty tough. How about I calculate the forces needed to hold that branch in place? I will do a simple model and then maybe later I can make it more complicated. Suppose I replace that limb with one straight uniform limb that looks like this:

In this replacement limb, I am going to say it is a cylinder that is 9 meters long and 30 cm in diameter. Let me assume that this limb is connected at two points to the tree (the white dots). So, for this limb to stay there, the following must be true:

The first two equations say that the total force must be zero. The last one says that the torque about any point must be zero (since it is in rotational equilibrium about any point). First for the forces. There is the gravitational force. This pulls on all parts of the limb, but I can represent this as one force pulling on the limb at the center of mass (long ago, I said I would explicitly show this - but I haven't yet). Then there are two other forces. Let me pretend like there are two pins that hold the limb to the tree. Each of these pins can exert a force in the vertical and horizontal direction I will call these F_1-y F_2-y etc...where the top pin will be 1. That is 5 forces.

For these forces and the first two equations, I get:

So, already I have some constraints. The horizontal components of the forces from the two pins must be equal and opposite. The vertical components of the forces from the pins have to add up to the weight of the log. Now for the torque, I am going to add up the torques about the lower pin. Let me draw a distorted view of the log so that the important distances can be seen.

What is torque? Torque is like a rotational force. Here is an example, what if you try to open a door by pushing near the hinges? It is much harder than pushing near the handle, right? When rotating about some axis, the torque is:

Here F is the applied force, r is the distance from the point where the force is applied to the axis. Theta is the angle between F and r. I will call torques that would make a rotation counterclockwise positive (really, torque is a vector). So, what is the some of torques about axis O (that passes through point O)? First, there are some forces that have zero torque. Both of the vertical pin forces have either theta = 0 or r = 0 so that the torque is zero. The same is true for the horizontal force on the bottom pin. This just leave two forces that have non-zero torques:

Now that I have the horizontal force on the top pin, the bottom pin has the same value (but in the opposite direction). I don't have an expression for the two vertical pin forces. Let me just say that each has a force equal to have the weight of the limb.

How about some values? First, I need the mass of the limb. If this is a cylinder of wood (with a density rho) then the mass is:

So the magnitude of the two horizontal forces on the pins would be:

If I use my values from above and an estimation of the density of wood, I get:

Wow. Oh, I know I made some estimations but even 50,000 Newtons would be huge. Impressive, most impressive. I salute you mighty tree.

In this post, ZapperZ makes a very Feynman-like distinction between "teaching physics" and "teaching about physics". This is a really good point - that to learn physics you have to do physics. I completely agree. It is just like riding a bike - you have to ride a bike to learn to ride a bike.

So here is the question. What do I do here on this blog? I don't know. Do I talk about physics? Yes? I do not teach physics - because I don't believe that I ever teach. In class, I like to consider myself a learning facilitator rather than a teacher. Still, what about this blog. First, I write stuff because I enjoy analyzing situations that I see in everyday life and on the interwebs. Could this help people learn? I am sure for some people it can (at least for some people). Could my posts inspire others to learn physics? Again, I hope so.

I guess this is related to why I get so riled up about shows like Sport Science. Are they trying to teach science? (I hope not since they are totally incorrect). Are they trying to inspire? Again - not a good idea to inspire with bad physics. I guess this would be like me trying to inspire kids to be healthy by showing them how to eat junk food.

Really, there is a reason for this. I saw this little stick (sticklette?) and noticed that it was very cylindrically shaped. So, what if I just pretend it is a cylinder to calculate the volume? This way I won't have to get it wet or anything (because I might need this stick later).

First the mass

Yes, there is some uncertainty in the mass - but it is small. I put the stick on balance and I will use a value of m = 28.9 g or 0.0289 kg.

Volume

The volume of a cylinder is:

And from my estimates, I get a length and diameter of:

To determine the volume (and the uncertainty in the volume) I am going to use the "max-min" method for uncertainty. The basic idea is that I will calculate the maximum value the volume could be and the minimum and base the uncertainty on these.

And for the minimum volume:

Note: I put too many digits in these number because at this point, I don't know how many to keep. Now for the uncertainty in the volume, I will just take the average change from max to average and average to min:

Putting this all together, I get a volume of:

Density

Now to do the same thing with the density. The maximum density is:

Here I divided by the minimum volume to get the maximum density. And the min volume:

This gives a density of:

I am pretty happy with this. This density is less than water (1000 kg/m³) so that means this wood would float (along with very small rocks and gravy).

If you picked picture B - you are probably correct. That is a picture of "Apolo" being catapulted into a pool of slime at the Nickelodeon awards show (click on the link to see the video - I don't think I can embed it). Ok - time to crank out an analysis.

I think I could approach this analysis from a couple of directions. Since all I have is a crappy version of the video, I could just look at "could this be possible"? The other analysis I could do would be to measure his acceleration in free fall. Let me start with the second method.

Acceleration of Apolo

Using Tracker Video Analysis, I can look at the motion of Apolo as he is landing in the slime. Really, it is not a good video. The camera pans and zooms at the same time. Also, there isn't really anything great to scale the video with and there could be some parallax errors. So, with that I get the following for the y-motion (unscaled video).

So, it should be constant acceleration. Is it? Too difficult to tell with only 5 data points. Here is the horizontal motion:

The horizontal velocity should be constant - this looks constant enough. So, from this I can't really say for sure it is fake.

Time analysis

Suppose I assume air resistance is negligible - which really isn't a terrible assumption in this case. Then if I know how far he is going to be shot, I can determine how long this should take. I can also see how long it took in the video. First, a couple of quick points about projectile motion (more info can be found here)

• For constant acceleration, the y-motion is independent of the horizontal velocity. This means that the maximum height and time of flight do not depend on how far the guy or ball goes.
• In the vertical direction, the initial speed when the ball leaves the throwing device (in the y-direction) is the opposite of the y-velocity when it returns to the same height.
• When you throw a ball straight up (or at any angle really), the time it takes to get to the highest point is the same time it takes to come down from the highest point.

Since I don't actually know how far he went (I could estimate it) and since there can be two times for the same distance, I am just going to look at the vertical motion. From the video, Apolo is in the air for 3.6 seconds. This would give a time of 1.8 seconds to go up - how high would this be? Let me start with the kinematic equation:

You see the problem here - I don't know the initial velocity. Ah HA! But I do know another trick. What if I look at the motion of going from the highest point back down to the initial level? Then the initial y-velocity IS zero. And this is the same height as going up. This gives:

This gives a height of about 16 meters or 52 feet. High school goal posts are 10 feet tall at the cross bar. Using tracker video analysis, this puts the top of the post at about 28 feet. The video isn't quite clear (because of the creative editing), but it doesn't look like Apolo goes over 50 feet above the ground.

I have caught you Nickolodeon. You can fool some people by having incredibly terrible video frame rate, but you can't fool physics.

Oh, I know you are worried about landing in that small pool of slime - I agree that could be a problem. But don't forget Professor Splash jumped from 35 feet into 1 foot of water. The slime was deeper than 1 foot - but if he came down from 52 feet WITH some horizontal velocity, that would be a problem.

For my setup, I used a glass eye-dropper. Put it in a cup to make sure it just barely floats and then put it in a completely full water bottle. If you don't have a eye-dropper, you can use anything that floats with an air space. I have done this with part of a straw before. Fold a small section of a straw in half so that air can't get out of the top. Add a paper clip or something to the bottom so that it stays with the hole pointing down.

How does this work?

This could be explained on several different levels. I will go with the shortest.

When you squeeze the bottle, you increase the pressure in the liquid AND in the air in the diver. This makes the air bubble get smaller so that the diver displaces less water. The buoyancy force on the diver is equal to the weight of the water it displaces. Simple - no?

The current course is pretty traditional. Your basic physical science stuff. It has the following content.

• Forces and Motion
• Newton's Laws
• Projectile motion and gravity (these first three are like 4 chapters)
• Pressure, fluids, buoyancy
• Thermodynamics and stuff
• Circuits
• Electric interactions
• Waves and Sound
• Light and electromagnetic waves

Maybe not exactly that order, but you get the idea. Pretty traditional stuff.

But here is what I am thinking. What is the purpose of this course? Why are students taking it? What do I want them to get out of it? Does it need to prepare them for any future courses? The answer to last question is "no". The second physical science course (the chemistry one) does not have any pre-reqs. In fact, students can take these in any order. The other questions can maybe be answered by describing this as a course for non-science majors that satisfies their general education requirements.

Here is my new plan. Maybe structure the course something like this:

• Fundamental forces (basically gravity and electromagnetic forces)
• Force and motion (very simple - saying forces CHANGE motion)
• Energy and conservation of energy
• Atomic nature of matter
• Basic quantum nature of matter - you know, Bohr model type stuff
• Light and spectroscopy
• Cool stuff that can be explained with the above for the rest of the semester

I would still use the same textbook, it basically has all these ideas in there, just not in the same order.

Could all the cars in the USA be used to change the rotation of the Earth?

Well, I shouldn't have phrased the question that way. Of course 1 car technically is all you need to change the rotation of the Earth - but by how much. Suppose everyone in the USA got in their car and drove East at the same time. How much would this change the rotation rate of the Earth?

Starting points:

• What is the rotation rate of the Earth now? I will call it omega-0, but the value is about 1 rotation in 23.9345 hours (don't confuse this with 1 rotation 24 hours - that is how long it takes the Sun to be back in the same position in the sky).
• Should I assume the Earth is a sphere? Well, I am going to even though it isn't. Also, I don't know the mass distribution of the Earth. Maybe I will just assume uniform density - again this is wrong (but it is wrong in a direction favorable to no change in motion).
• There are about 300 million people in the USA. I am going to totally estimate 100 million cars.
• I will use an average car the same weight as a Crown Victoria. Why? Because this is my estimation. How about 1,800 kg?
• These cars will be moving West at a speed of 60 mph (27 m/s).

The basic physics idea is that the Earth is rotating and has angular momentum. If the cars are moving with respect to the rotating Earth, the angular momentum of the two must be constant. For a rigid object (the Earth) the angular momentum is (I will assume rotation about a fixed axis because that makes things so much easier):

If I assume the Earth is on a fixed rotation axis (which it totally is not), then the angular momentum vector and the angular velocity vector are in the same direction with the moment of inertia as a scalar value. (here is an example of a case where L and omega are not in the same direction). But, back to the Earth. If the Earth is a uniform sphere, then the moment of inertia about a fixed axis is:

What about a car? If I consider a car to be a point particle, then as it move around the Earth, it would have an angular momentum of:

But what is the angular velocity for the car? Let me draw a picture.

So first, the car is stationary with respect to the rotating Earth. (here I am just showing one typical car) Then, the car drives with a velocity v with respect to the surface of the Earth. Before the car moves, the angular momentum of the Earth plus cars is: (magnitude only since the direction does not change)

Since the cars have the same angular velocity as the Earth, I can add the moments of inertia together. Also, I have included an n for the number of cars. Now that the cars are moving, the angular momentum would be:

Angular momentum is conserved. This means that L₀ = L₂. I want to solve for the change in angular velocity. To do this, I will set the two angular momentum expressions equal to each other.

I know, that seems like some messy algebra - but it really isn't bad. How can I check to see if I made a mistake. First, what if the cars are going at zero m/s? If I put in v = 0 m/s, then the change in angular speed is zero rad/s. That is good. Also, if the mass or the number of cars goes to zero, the same thing happens. Also, the units check out. Now to just put in the numbers. Doing this I get:

If I know omega, I can find the time for one rotation (a day) as:

Using this for omega-0 and omega-1, I can get the two times:

If I calculate these two times and subtract them, I get a change in time of 3.7 x 10^-10 seconds. So there is your answer. No where near the effect of the Chilean quake (even if you change my assumptions by a little).

"

1. Grades should reflect a student's progress with course material. Where an A+ indicates mastery.
2. Grades should be an amalgam of student's knowledge, behavior, and anything else the teacher wants to control.

"

I was in the middle of posting a comment to this post, but it was getting a little long. Here is what I was going to say.

Should students be intrinsically motivated - you know learning because learning is a good thing, or should they be extrinsically motivated - learning because their grade depends on it. Oh - and not just grades, but behaviors.

Let me start by examining some extreme cases. Take graduate student in physics. Why does this student study to learn physics? I would hope it is not just because of the grade. So, in this case the student should be intrinsically motivated, right?

What about a 2^nd grader? Why should this student learn to spell and add and read? Maybe reading was a bad example, but spelling and add could possibly be motivated extrinsically - by grades. Oh sure, there are some 2nd grades that just want to learn to spell (I know who you are). Also, the reading was a bad idea because there are tons of 2nd graders that want to learn to read so they can finally get to How to Train your Dragon or Diary of a Wimpy Kid (you can tell I have kids). The point is: maybe second graders need some extrinsic motivation.

What about kids? Should you just let them eat whatever they want (Fat Cakes) and stay up as late as they want? Probably not. Make them eat their veggies. It is good for them. So, I make my kids do things that they don't want to do. What if they don't eat their vegetables? Then the won't get a bad grade, but they will not get desert - that is for sure.

So, maybe grad students are intrinsically motivated and maybe 2nd graders are a mix of intrinsically and extrinsically motivated. What does this have to do with me? Well, I am going to assume that at some point students need to switch to being intrinsically motivated. I don't know when this should happen, but I am going to assume it happens before college. This means that I don't have to use grades to motivate them.

I know the arguments - if you don't assign a grade, they aren't going to do it. That may be true, but I am ok with that. I am going to use the grade to assess their understanding of the material.

PS - check out Think Thank Thunk. Shawn has some really interesting calculus posts.

What lifts up the metal block? The atmosphere. Diagram time:

But this isn't a very realistic diagram. Actually, the suction cup would be pushing down on the block because the force from the atmosphere would be too large to balance with the weight. Let me put some numbers in here. Suppose this is an aluminum block - I just going to pretend it is 4cm on a side (and a cube). In this case, the weight would be (the density of aluminum is 2700 kg/m³):

How does this weight compare to the force the atmosphere pushes up on the block? (I am assuming the suction cup on the top covers the whole block). The force will be the pressure times the area of the bottom and the atmospheric pressure is about 10⁵ N/m². This gives:

What would happen if I remove the outside air? Can you do this? I can not remove it all, but I can remove enough to make a difference. Here is a video of just that. I put the dart with the block in a jar that I can pump the air out of.

First, I am going to assume that Marshawn has a mass of about 100 kg. Also, let me say that he can produce 54,000 watts no matter what his speed.

Take a short time interval. During this time, Marshawn will increase his speed from say v₁ to v₂ this would be a change in energy of:

And this would relate to the power by:

So, if I know this small time interval and the velocity he started at (at the beginning of the interval) then I can find the final velocity:

If the time interval is short, then the velocity is essentially constant (for very short time intervals) so that I can use the average velocity to write:

You see where I am going don't you? This is all set up for a numerical calculation. Here it is - I made it as simple as I could:

I changed my mind. Instead of using the average velocity to find the new position, I just used the velocity. Trust me, it is ok. Here - you can check. One good way of checking your calculations is to make the time interval (dt in this case) smaller and see if you get the same result.

So, what do I get. Here is a plot of the speed as a function of time:

There you go - 100 meter dash in under 3 seconds. Take that Usain Bolt. Note that Usain not only has a cool name (Bolt) but has the world record at 9.58 seconds. Another note - I just noticed that lists the wind speed for these records. Boom. That is another blog post.

Not only would 54,000 watts give you a 100 meter time under 3 seconds, you would be going over 50 m/s. Yes, that is like 120 mph.

How about another check. What if I put in a more reasonable power of 2000 watts? Here is what I get:

Seems better, doesn't it? Still a world-record time, but I did not take into account air resistance and I assumed the power would be constant. Oh, also that would give a speed of 40 mph - so that isn't quite right.

Let me approach this first from a terminal velocity view. This requires a model for air resistance. I will use the following:

Where:

• rho is the density of air
• A is the cross sectional area of the object
• C is a drag coefficient that depends on the shape of the object
• v is the speed of the object
• And this gives a force with a direction opposite of the velocity vector

Let me assume that all the pieces of a meteor have the same density and shape - for simplicity, I will assume a sphere. Here is a diagram for two different sized pieces falling (straight down) at the same speed.

Meteor A (the big one) has a greater gravitational force because it has more mass. It also has a greater air resistance because it's cross sectional area is larger. I picked a speed so that meteor B would be at terminal velocity. This is when the air resistance has the same magnitude as the gravitational force. If I assume that meteor B has a radius of r_B and a density of rho_m then:

Where v_T is the terminal velocity. If I solve for this value, I get:

Here you can see the key point. The terminal velocity depends on the size. This is because the air resistance is proportional the area (r²) and the weight is proportional to the area volume (r³). These two things do not cancel.

Modeling a debris field

I have created a python model for shooting bullets. I can simply modify this to calculate the trajectory of a dozen or so different sized (but same shape and density) meteor pieces.

The following is a plot of the trajectory of a few pieces of a meteor. I (for random reasons) started the model at 5,000 meters above the ground moving at 350 m/s aimed 30 degrees below the horizontal. Here is what I get:

So, the bigger the piece, the farther it will go. My biggest piece was 1 meter.

Here is the deal. You are in a pool. You drop a quarter in the deep end and swim down to get it. I know the first thing you are going to ask: Why do I have a quarter in the pool? Does it matter. What matters is that your ears are killing you. Boy, that hurts. Why do your ears hurt and what can you do about it?

Pressure and depth

When you put a fluid in a gravitational field (like on Earth), the pressure in that fluid (or gas) increases as you go down. Why? There are a couple of ways to explain this. One cool way is to think about the gas as a bunch of particles (which it is) and treat each particle as though it were in projectile motion. I did this with the PhET simulator before - but maybe sometime I should come back to this.

Another way is to consider some fluid floating in a fluid. Suppose I take a chunk of air in a room with no wind or anything. That chunk of air just stays there. Here is a diagram.

If this rectangular cube thing of air is not moving and in equilibrium, then the total forces on it must be zero. The air outside of this rectangle of air pushes on this in all directions. The force on each side will be the air pressure on that side times the area of that side.

For the sides of this air rectangle, the pressure is not really important. Clearly, the forces from the sides are the same. The net force in the horizontal direction is zero. The problem with the vertical forces is that there is something extra there. Other than the force from the air above and the force from the air below, there is also the gravitational force pulling down. This means that in order for the net vertical force to be zero, the air from the bottom must push up more than the air on the top. The areas of these two things are the same, so the pressure on the bottom must be larger.

Ears

Here is a model of an ear:

In normal situations, the air inside your ear is at the same pressure as the air outside your ear. This means that the force on the two sides of the ear drum are the same and you are happy.

But what if you go underwater? In this case, there is water on the outside of this ear drum, but air on the inside. If you don't do anything, the pressure inside will still be at atmospheric pressure. However, on the outside, the pressure will be greater. This means that the force from the inside air will not cancel with the pressure from the outside. Your ear drum doesn't want to accelerate, so it stretches like a spring to produces a net force of zero. This stretching of the ear drum hurts.

What can you do?

You can fix this problem. Well, obviously people scuba dive at depths much deeper than a pool - their ears don't hurt. The solution is to add air to the inside of your ear so that the pressure inside and outside are the same. This is called "equalization". You can do this because your ears are connected to throat with the eustachian tube (that little tube on the side in my drawing). This tube is not huge so that it is mostly closed. You can force air through this tube by holding your nose and gently trying to blow out your nose. So, while you are going down to the bottom of the pool, you can do this nose-thing. If you are an expert (like me), you can do this without even holding your nose.

How do you get the air out? Well, it usually comes out on it's own. When you come back to lower depths, the greater air pressure in your ear forces the air back out through the eustachian tube. Actually, if you have a lot of mucus in your head sometimes this air doesn't come out. This is known as a reverse block. Really, at this point you have no option but to keep going up.

There are other air cavities in your head that could cause a problem. In particular, your sinus cavity. For most people, air can easily flow in and out of this making equalization a non-issue. If you have a cold, mucus can block these passages and it can hurt in that area too.

These ideas don't just apply to scuba diving and swimming in pools. When you fly in a plane, the air pressure decreases. I am sure you can feel this in your ears. Sometimes they recommend chewing gum in a plane. This makes your jaw move more and possibly open up the eustachian tube so that your ears can equalize.

More on scuba

Have you ever looked at scuba masks? Here is what most people think of (image from wikipedia):

The key feature to observe is the nose pocket. This makes it very easy for a diver to hold his/her nose closed for equalization. Check out this old-style oval shaped mask:

Notice that it also has these nose pockets. Oh, I have no idea why this girl is inside with a flooded mask - but it was the only picture I found of this kind of mask. If you ever get a chance to use one of these suckers, you should. Even though they are old, they are very comfortable.

Yes, I know I make mistakes. I try to correct them when I become aware of them, but I am just one person. Also, sometimes I say things that aren't quite true - but they are good lies. You know, like saying the weight is m*g (that isn't always true). Ok - so back to my "commentary" on a particular Sport Science episode (at least they are short). Here is one about jumping in the snow ski half-pipe. Check it out (it's not too long)

First, this episode has the same problem all the other episodes I have seen have - inconsistent use of terms. I don't think I will talk about this today - it can be saved for a later post. Instead, I want to focus on two problems.

1 mph loses 3 feet

In talking about how important speed in the half pipe, Sport Science claims that for every 1 mph of speed a skier loses, he/she will lose 3 feet in height. I can't get this one to work. First, I am not really sure about what they mean by speed. Is this the average speed going down and back up the half pipe or the speed at the instant the jumper leaves the pipe? I am going to assume it is the speed at the point the skier leaves - this will make things easier to analyze.

In this image (I found it searching for Simon Dumont, but I added my own stuff to it), I assume that this is like a half pipe jump. Since I am concerned with height, it is best to use the work-energy principle. If I consider the skier plus the Earth as the system, then while in the air there is no work done so that:

The key is that if I call the bottom zero potential energy, then at the bottom the skier only has kinetic energy and at the top, there is only potential. I can write this as:

So, here is a functional relationship between speed and height. Let me plot this from an initial speed of 1 m/s to 15 m/s (24 mph is about 10 m/s):

And here is the problem - this is not a linear relationship. Let me plot this in units of feet and mph - just so I can make it more easily agree with the video.

So, going from 24 mph to 45 25 mph gives a change in height of about 1.5 feet. However, going from 10 mph to 11 mph only has a difference of about 0.8 feet. Really, I want to know if I change the initial speed, what happens to the change in height - this could be a great set up for a calculus problem. What is the rate that h changes with change in v? This can be written as:

As is true for any parabola, the rate of change (slope) is proportional to the value of the velocity. If I use a speed of 24 mph, I get that the height changes about 1.8 feet per mph. According to this model, you would have to be going 44 mph to have a speed loss of 3 feet per mph. I am still not sure this is what the show was talking about.

Power - again

The other aspect I would like to examine is the power. Note that it seems they are using the term "power" in two different ways in this episode. I will use the definition:

And power for what? I am going to guess the only thing would be the power he produces while pushing off the lip of the half pipe? Not sure this really makes much sense, but it is the best I can do. Sport Science claims that Simon Dumont produces 24,000 watts. Seems high, but at least it is not as high as the power they claim from Marshawn Lynch - 57,000 watts - really.

There is a greater chance I can get this value to seem reasonable - mostly because if it is for a jump, it would be a very short time interval. I have no idea how Sport Science came up with 24,000. All they state is that they put motion sensors on him so that they could make an animated skeleton move like him.

To explore if this is a reasonable value, I am going to look at the power produced from Kobe Bryant as he appears to jump over a car - (note that I am pretty sure the video in that post is not quite real, but I do think Kobe can really jump like that). This is a good test case to look at because I know how high he jumped and I can look at the time it took to jump. This will let me calculate the power.

Here is the plan. I know the energy that Kobe produces because I know how high he went. The time can be determined from the video. This is the graph of Kobe's height has a function of time - including the height of his estimated center of mass:

Looking at this graph, Kobe's center of mass starts at 1.65 m and goes up to about 2.9 meters. If Kobe has a mass of 93 kg, then his change in energy is:

Looking back at the video, Kobe is in the process of jumping for 0.231 seconds. Note to other video analysts. You can not easily get the time for these events from something like Quicktime. Quicktime rounds the time of each frame up to the second - which is useless. I use Tracker Video Analysis for this. So, the power from Kobe is:

If Kobe produces just under 5000 watts - how is Simon supposed to produce 24,000 watts. I would love to hear from Sport Science and see how they get these power values. Please don't tell me it is this:

Sport Science: "Hey look! Tiger Woods produces 40,000 watts in his golf swing!"

This is typically the first topic in the second semester of introductory physics - the interaction between objects with electric charge. There are 4 fundamental forces that physics typically looks at:

• Gravity - an interaction between objects with mass - wow, I don't have a post on the universal law of gravity?
• Electromagnetic - an interaction between objects with electric charge.
• Weak Nuclear - an interaction between (let me just say for simplicity) leptons.
• Strong Nuclear - an interaction between hadrons.

I know those last two are complicated - but I am not going to talk about the strong and weak forces.

Coulomb's Law

Coulomb's Law is a model for the forces between two charged particles. Here is the model.

Looks complicated, doesn't it? But that is the best way to represent that. Typically, books might just show the magnitude of the force, but I think you can handle the vector form. Let me draw a picture to go along with this.

Maybe you see the first problem - which force is this equation for? If the vector r is pointing from q₁ to q₂, then this gives the force that q₁ exerts on charge q₂. Note that it is important to include the unit vector r-hat in the equation or you would not have both sides of the equation as vectors. Also, the denominator is the magnitude of this vector squared. If you did not know the vector r, but instead had the location of the two charges (this happens a lot), then:

Also, in that equation, the 1/4pi-epsilon term is a constant. The charges should be in units of Coulombs, and the distances in meters - this will give a force in Newtons. Here is the value for that constant: (some texts just call this k or something)

One final note - in this form, the sign of the charge DOES matter, and this a good thing. If the two charges have opposite signs, the direction of the force will be in the opposite direction as the r-hat vector. If both charges have the same sign, then this force will be pushing the two charges away from each other.

Maybe this seems like an unnecessarily complicated form of Coulomb's Law, but I think it is the most useful in the long run.

Comparison with gravity

It is useful to make a comparison between the universal law of gravity and the Coulomb force. Here is the gravitational force:

This looks similar to the Coulomb equation. The big difference is the negative sign. This is there because gravity is always an attractive force but the mass is always positive.

Where does this come from?

This was a great question I had introductory physics one time. Actually, I think it was something like "how do you derive Coulomb's Law?" The answer: you don't. Coulomb's law is a mathematical model based on experimental data. The basic idea is to get two spheres and put some electric charge on them. Find some way to measure the force between them and see how that compares to the distance. There is a trick here - a uniformly charged sphere looks just like a point charge. Coulomb's law deals with point charges.

An Example

This is what my students would want - a worked out example. Here is my completely made up example. Suppose I have a 3 nanoCoulomb (nC) charged ball at the origin and a -7 nC charged ball at the location < 0.3, -0.4, 0 > m. What is the force on the negatively charged ball? Here is sketch:

The first thing I need is the vector r. This is pretty easy since one of the charges is at the origin. I also need the magnitude of the r vector and the unit vector r-hat.

Now, I just have to put stuff in. Here is what I get:

Not too bad, is it?

Superposition

One last very important thing. What if you have more than two charges? The superposition principle applies to the Coulomb force. Basically, this means that if you have three charges (called A, B, and C), then the force on C will be the vector sum of the force on C due to A and the force on C due to B. Here is a picture.

I will do an example with more than two charges in a later post.