Dot Physics

Slate’s Explainer has an answer and question post about moonwalking. Here is one of the very good questions answered there.

Would it be easier to moonwalk on the moon?

The Explainer says “absolutely not” and attributes this to the awkwardness of walking on the moon. The article gives an example of Earthly legs being too powerful as the “astronaut’s hop”. The explainer also says it is awkward because of the pressurized space suits.

I think the problem is almost entirely the pressurized suits. I believe that the astronauts do their moon-hop because it is difficult to bend their legs in the pressurized suits rather than because the gravitational field is smaller.

Suppose you had a space base on the moon so that you could be on the moon and NOT wear a space suit. In this case, I think moonwalking would be essentially the same as on earth. As I showed on my previous post on moonwalk, the move has a lot to do with center of mass and the ratio of the normal forces on the two feet.

On the moon, the ratio of the two forces would still be the same as on Earth. I could be wrong. The only way to know for sure is to build a moon base so that we can test this out. Well, I guess someone could go up in the vomit comet while they are simulating moon gravity.

I finally saw the movie Hancock. Yes, I know it has been out for a long time but I don’t get out much. You know me, I can’t leave something like this well enough alone. It’s not my fault, I was born this way. It shouldn’t spoil the movie too much if I tell you this one scene (you have probably already seen it anyway).

Basically, Hancock gets upset with this boy and throws him in the air to scare him or something. In case you did not time it, the kid was in the air for 23 seconds. I claim that in order for Hancock to throw a person in the air for this long, the acceleration during throw would be deadly.

For the first pass, what if there were no air resistance (clearly, there is). In this case, I can determine the initial velocity of the boy and from that his acceleration during ‘the throw’. If the time the boy is in the air is t, then I can use the definition of acceleration:

If the boy is thrown up and falls with a constant acceleration (g) then his final velocity will be the opposite of what his initial velocity was. From this, I can solve for the initial velocity:

For 23 seconds, this gives an initial velocity of 113 m/s (or 250 mph). Clearly this is fast enough that air resistance will come into play. But already, you can see if this boy is accelerated from 0 m/s to 113 m/s in the distance of about 2 meters (or less) then there is going to be trouble.

I think I have already shown my point, but that is not enough. I need to take it to the next (but not the final) level. If I include air resistance, how fast would Hancock have to throw this kid so that he is in the air for 23 seconds. Assumptions:

• I am going to assume the boy has the same terminal velocity as a grown man. This will allow me to use my sky diver falling model (from ‘can an iphone tell if your parachute didn’t open’) without much modifications. I would imagine that a smaller boy would fall about the same as a grown man because he would have both smaller surface area and mass (although these don’t change the same with scaling).
• Position. In the clip, the boy seems to come down in sky diving position, but he looks like he is thrown up in “feet down” position. This could make a difference, but I am going to model this as though the boy had the same position during the whole flight.
• Assume that the air density is constant. Of course, it isn’t – but should be close enough to constant for this. Also, this can easily be changed later.
• Finally, I will assume that the gravitational field is constant.

Ok, now on for the calculation. The basic plan is to:

• Calculate the force on the boy while in the air. This will be the gravitational force plus the air resistance. In one dimension, I need to make sure that the air resistance force is in the opposite direction as the motion.
• Calculate the acceleration. (a = F_net/m)
• Update the velocity. (v = v + a*dt)
• Update the position. (y = y +v*dt)
• Update the time.
• repeat
• Plot stuff

That is the basic idea. If you want help with numerical calculations, check out my previous introduction. Anyway, here is a plot of the boy being thrown up with an initial velocity of 113 m/s (blue line). I have also plotted (for comparison) an object with no air resistance (green line).

Both lines represent an object thrown up with the same speed. You can see that the air resistance case does not go as high (because of the air resistance). And even though it is going much slower on the way down, it still is not in the air as long as the no-air resistance case.

Next question: how fast WOULD he have to be thrown to be in the air for 23 seconds? To answer this question, I am just going to put another step in to the program. I will run it at 110 m/s, then 115 m/s then 120 m/s and so on. For each “run” I will have the program record the time. Simple, no?

Here is a plot of the time of flight for a “sky diver” with an initial upward velocity of 5 m/s to 1000 m/s.

From this graph, it seems the initial velocity for a thrown skydiver would need to be around 400 m/s in order for him (or her) to be in the air for around 23 seconds. You can see also, that this curve starts to “level off” in that to increase the flight time (or hang time if you like basketball) take a greater and greater initial velocity. Let me go ahead and re run this up to an initial velocity of 5000 m/s, you know ….just because.

By increasing the initial velocity from 1000 m/s to 5000 m/s, the flight time only increases by about 10 seconds. This is because at such high speeds, there is a tremendous air resistance force that quickly slows the skydiver down. Oh, one more thing on this part. Recall the first part above where I showed the time of flight without air resistance. Without air resistance, this graph of time of flight would be a straight line (ignoring changes in gravitational field).

Now I am ready for the second part. Let me use 400 m/s as the initial velocity of the kid to be in the air for 23 seconds. What would his acceleration be during the “throw” from Hancock? Here, I am in the situation where I am just interested in acceleration and distance and not time. Usually, I would automatically think of the work-energy theorem. However, by manipulating the kinematic equations, I can get an expression without time.

For the boy, his initial velocity is 0 m/s. Solving for the acceleration, I get:

Plug in what values you think are reasonable. I am going to use a final velocity (final for the throw is initial for the in the air part) of 400 m/s and a distance of 1.5 meters (which I think is quite generous). This gives an acceleration of over 50,000 m/s². If you like this in terms of “g’s” then this is like 5000 g’s. Danger.

This table of NASA g-force tolerance data used to be on wikipedia’s page, don’t know why they took it off, but here it is:

If the boy is being thrown while facing down, that would be “eyeballs out”. Notice that nowhere on the table is there a tolerance anywhere near 5000 gs in any position for any time. The result would be a dead bully.

I have already talked about science and why everyone should take some science courses. The short answer (in case you don’t want to read the previous post) is that everyone should take science because science (along with art and other stuff) is what humans do.

So, tomorrow I will be off to attend a workshop on a new physical science curriculum. The goal is to bring more active-learning styles courses to large enrollment classes. This is essentially a similar idea to the Physics for Elementary Teachers course that I teach with one major difference – size. I would love to teach the course for elementary ed majors to everyone, I think it would give a good fundamental understanding of science and I think students would generally enjoy it. The problem is that the administration is not going to want to move from a 90 student course to 3 sections of 30 students.

I don’t know how to do the same stuff I am doing with the smaller class (meeting in a lab room) with a large lecture hall style course and setting. Guess that is why I am going to a workshop in San Diego.

Was the moonwalk fake? No, not the Apollo landings. I am talking about Michael Jackson’s moonwalk. You got to admit, he had a big impact on a lot of stuff and this is my way to give him respect – physics.

I am sure you know about the moonwalk. Maybe you can even do the dance move yourself, but how does it work? First, here is a clip of MJ doing his stuff.

As a side note, I can’t remember where I saw it but there was a great discussion of the history of the moonwalk. If I recall correctly, some were saying Michael didn’t create this move. One thing is for sure, he made it popular. Now for the physics.

The key concept here is friction. Friction is actually uber-complicated, but a simple model works for many cases. Static friction is a force exerted on an object when it is in contact with some surface but those two surfaces do not move relative to each other. Kinetic friction is a force exerted on an object when the two surfaces are moving. Suppose I have a block at rest on a table and I pull it with a slowly increasing force. This is what it would look like:

Two key things from this graph. As you pull on the stationary block, the block doesn’t move. If I pull with 1 Newton, and it doesn’t move then the frictional force is 1 Newton. If I then pull with 2 Newtons and it still doesn’t move, the frictional force is 2 Newtons. The static frictional force does what it can to make the thing not move – but not more than it can. This leads to the static friction model of:

In this model, the force is less than or equal to the product of some coefficient (that depends on the two types of surfaces) and the normal force (how hard the two surfaces are pushed together). The direction of this frictional force is parallel to the surface in the direction that prevents the object from sliding.

The other key feature in the graph is the small jump down when the thing starts to slide. This is because the coefficient of kinetic friction is typically smaller than that for static friction. Also, if the object is sliding, the frictional force is constant.

Back to Michael and the moonwalk. The key here is: how do you make one foot slide and the other not slide? If both feet are stationary, then this is dealing with static friction. I could make the frictional forces on these two feet different by changing my center of mass. Here is a free body diagram:

Since he is not accelerating up and down, the following must be true:

These are the forces in the y-direction. They must all add up to zero so that:

There is another condition that must be satisfied. Since he is not rotating, the total torque about any point must also add up to zero. If you want more info on torque, check out this post. But for this post I will just say that torque is like the ‘rotational force’. It depends on the point about which you want to rotate and is essentially the force applied times the perpendicular distance to the point of rotation. For the free body diagram of Michael, I have chosen one of his feet to be the point about which he is not rotating (I could chose any point). This makes 3 of the forces have zero torque (N₂, F₂ and F₁ have zero torque because the perpendicular distance to point O is zero). Here I labeled the other important distances:

The only two forces that exert torque about O are the weight and the N₁ force. They have opposite directions of torque because they would cause rotation in different directions. This along with the previous equation gives:

Eliminating mg, and solving for N₁, I get: (I know the indices for the forces and distances don’t match)

If his center of mass is in the middle, then r₂ – r₁ = r₁ and the two normal forces would be equal (as you would expect). If the center of mass is more towards the foot on the right, then r₂ – r₁ is less than r₁ and N₁ will be larger than N₂. This will make the frictional force on the foot to the right greater and the other foot slide.

Well, what if r₁ is greater than r₂? One of two things would happen. Either he would fall over, or there would have to be a force pulling the foot on the left down. This is similar to Michael Jackson’s trick in “Smooth Criminal”.

Here he used special shoes that connect to the floor so that he could do this. More details on this page.

Ok. So that is how Michael gets one foot moving. How does he keep one foot sliding and the other not sliding? It is really the same thing as above except that he can increase the force on the moving foot a little bit more since it is sliding. Sounds easy, but Michael could really make it look cool.

Finally, I just want to show another demo that is essentially the same idea.

Meterstick friction demo from Rhett Allain on Vimeo.

You can find more details on the meterstick demo in this blog post.

I made this screen cast for my algebra-based lab. Maybe you will find it useful also. This is a tutorial using Tracker Video Analysis (an awesome free program as I have said many times). In this tutorial I analyze a moving cart that shoots a ball up and lands back in the moving cart (called a Howitzer cart). The video is available at the LivePhoto Physics site.

The internet sure can bring some awesome stuff. Chad at Uncertain Principles has a great post about the vision of elves. He refers to a part of The Two Towers where Legolas can count the number of riders of Rohan from a great distance.

Not only is the post great, make sure you read the comments also. Very impressive discussion.

This was on my ‘to do’ list, but Tom at Swans on Tea beat me to it. Basically, this grocery store has these plates that when depressed produce electrical energy. Tom does a good job pointing out that this is not free energy (the original article says this also). Clearly, the energy comes from the cars. How much would this cost the cars?

As always, let me start with some assumptions.

• The original article says that the bumps will generate 30 kW of energy every hour. That is an odd thing to say. I am going to interpret that as 30 kW of power for all hours (every hour). They couldn’t have meant 30kW/hour. That wouldn’t be power OR energy. 30 kW of power for 1 hour would be 30 kW*hr of energy. Ok, so power is 30 kW.
• I am going to assume that 1 car passes over this every 5 minutes. That is a complete guess. I would imagine the average rate that a car goes over it be anywhere from 1 a minute to 10 a minute.
• The efficiency of the device. This is a wild guess, but I am going to say 20% to 50%.
• The energy density of gasoline is 1.21 x 10⁸ Joules/gallon.
• The price of gasoline is $2.50 per gallon.
• A car engine is 15%-25% efficient at converting energy into mechanical energy

Let me start off with the assumption that it is just one car doing all this driving over the device. If the car gets 25 mpg, how much extra gas would be expended from this device being in the way? First, I can calculate the estimated amount of energy per use it would get from the car.

If that is the energy loss by the car, how much gasoline would that take?

That is the extra gas needed for one car going over once. Let me put this all together in my normal spreadsheet reader changeable form.

Something doesn’t look quite right. This says it would take about 9 gallons of gas an hour. Ok, that may be ok. If I wanted to run a 30 kWatt generator, it would take some stuff. But what if people came in their cars and just hit the device while coasting? Then the electrical energy would come from the decrease in kinetic energy of the car. How fast would a car be going so that it is stopped after hitting that device? I added this in my calculation sheet. From that, I get 120 m/s? No way. That is with 30 cars hitting it an hour. Let me look at each piece and see if I can find a mistake.

In 1 hour, the energy from the device would be: (assuming a power of 30 kW):

Of course, the device is not 100% efficient. If it is 50% efficient, it would need twice that amount of energy coming in to get that much out. That means that during that hour, there must be a loss of 2.16 x 10⁸ Joules of KE from the cars. If there were 30 cars during that hour, that would be 7.2 x 10⁶ Joules per car. If this is all KE then:

I don’t see where I made my mistake. I estimated 1000 kg for the mass of the car, that seems ok.

Maybe this thing doesn’t really generate 30 kWatts of power?

Note: This is an old post from the time before my blog was in wordpress. I noticed there was some incoming link for this, and I never moved it over. Here it is in it’s unaltered (except for this part) format.

I don’t know why I even suggest a new energy source. Fusion power is only a few years away in the future (just like it as always been). This will replace any other sources of energy that we could come up with. But, I can’t help myself, I need to share my idea and save the world. It’s what I do. (call me a superhero is you want).

We can get all of our energy from the rotation of the Earth.

The basic idea is to use the rotation energy of the Earth to power our Nintendos and computers and stuff. How much energy could we get out of this and what would we lose? First – what we lose. If we use the rotational energy of the Earth, it would spin at a slower rate. I will start off assuming we make the day 1 second longer.

Currently, the Earth takes 23.9345 hours to rotate. It take the 24 hours for the Sun to be back in the same location. This is the difference between sidereal day and synodic day. I am concerned with the rotational rate of the Earth on its axis, so I need to use the sidereal day. Check the wikipedia link for a great diagram that shows the difference between the two days.

This leads to an angular speed of:

Now, suppose I want to increase the length of the sidereal day by 1 second, this would give a new angular speed of:

How much energy would this produce? (assuming it could all be turned into useful energy). The energy of motion for an object rotating is:

Where (1/2) is (1/2), I is the moment of inertia about the axis you are rotating and ? is the angular speed. The moment of inertia is sort of like the “rotational mass”. An object with a higher moment of inertia is more difficult to change its rotational motion. In this case, we are dealing with a spherically-shaped object (the Earth is mostly spherical). The moment of inertia for this can be approximated by assuming it is a uniform sphere, but it isn’t. The density in the center of the Earth is much greater than on the surface. So, for this calculation, I will use someone else’s determination of the moment of inertia of the Earth – here is Wolfram’s Research value –

From this, I can calculate the change in energy by slowing the Earth down.

and putting in the values from above:

This is how much energy the Earth loses, so we could use this for other stuff. Is this enough energy?

Energy Usage:

Let me just look at the energy usage by the U.S.A. because they would be the ones to harness the rotational energy of the Earth (but really, because I found the data for US energy usage first). http://tonto.eia.doe.gov/ask/electricity_faqs.asp#home_consumption – this site has the data that I started with. It has a spreadsheet with average monthly usage for residential, commercial and industrial:

• Residential: There were 122,471,071 consumers that used an average of 920 kilowatt hours per month.
• Commercial: There were 17,172,499 users that used an average of 6,307 kilowatt hours per month.
• Industrial: There were 759,604 users that used an average of 110,946 kilowatt hours per month.
• NOTE: by “users” I mean companies or places or whatever the spreadsheet meant.

A kilowatt is a unit of power – or the rate at which energy is used. A kilowatt-hour is a unit of energy. Since 1 watt is a Joule per second, 1 kilowatt-hour would be:

and the monthly usage in the US would be:

that is per month. So how many months (and years) would this energy last for the US assuming a steady energy usage? Actually, I will also assume that only 50% of the rotational energy of the Earth goes to useful stuff (like nintendos) and the rest is wasted. This would mean the amount of useful energy would be 2.5484 x 10²⁴ Joules. So

This is a long time. So the length of the sidereal day would only increase by 1 second over this time period (that way wouldn’t have to store all this energy, but rather generate it as we use it).

Now for the details:

How exactly do I propose that this rotational energy be harnessed? I will leave that as an exercise for the reader – but I will give a hint: Magnets and wire (I have already said too much).

Also, this method produces no greenhouse gases.

Maybe this isn’t the best video to analyze, but it sure is funny. I am not sure why it is so funny – maybe you should just watch it first.

I don’t really know why these guys are putting explosive on sledge hammers. Sure it looks fun, but I think I would pass. Anyway, here is the part I was curious about.

Did the exploding hammer lift him off the ground, or did he jump as a reaction to the explosion. It kind of looks like he was lifted, but I am not sure that is possible. Video analysis to the rescue. (using Tracker Video) So, I assumed the sledge hammer was 1 meter long (if not true, it is close enough to being true). I used that to scale the video. I tried to look at the velocity of the heavy part of the hammer after the explosion, but I could only get two frames of data. That is enough for a speed. Here is the y-position of the hammer with a linear fit.

This gives the hammer a velocity (in the y-direction) of 10 m/s. What about the handle end of the hammer? That is in the video a little longer. Here is a plot of it’s y-motion.

This gives a velocity of 7 m/s. So, 7-10 m/s seems to be the speed of the hammer while the guy is holding it. How much energy did it have after the explosion? It would probably going even faster than that, but not too much. I have no idea how fast the hammer was going have he let go, but I can still do some simple calculations.

Did the hammer have enough energy to lift the guy up? If I break this into a before and after case, before the hammer is moving with some speed. After the interaction, the hammer is still moving up, and the guy has increase in gravitational potential energy (and so has the hammer). This would give a conservation of energy equation as:

So, here are my starting values (estimates)

• Mass of sledge hammer = 5 kg
• Mass of guy = 65 kg
• Initial velocity of hammer = 8 m/s
• Final velocity of hammer = 4 m/s
• Final height of hammer = 1 m
• Final height of guy = 0.15 m

Well, that last one was a measurement and I am going to calculate what the estimates give me to compare. Solving the above energy equation for height of the guy:

Well, with my estimations, I get 11 cm. So, I guess it could lift him up. I am glad because it didn’t really look like he jumped. Plus, it makes the whole video funnier (also that it does not seem like anyone got hurt – that would make it un-funny).

I have been wanting to look at this whole curved bullet thing, but I wasn’t sure how to approach it. In case you are familiar with the myth, this is from the movie WANTED (which I did not see). Apparently, some people learn how to make bullets curve by moving their gun. Here is a shot of a bullet curving in front of someone.

Maybe the picture doesn’t do the clip justice, but it is enough for you to get an idea. Before I do an analysis, this reminds me of a great educational activity. In the activity, you give groups of students a full sheet of paper with lines that look something like this:

You also give the students a ball, like a racquet ball or something. The goal for the students is to make the ball roll through the curve without hitting the lines. This CAN BE DONE. Students then proceed to try a whole bunch of different things, none of them work. It shows the common idea students have that an object can somehow “remember” motion. Give it some curving motion and it will kind of continue with that.

But I said it could happen! That you can get the ball through the curve without hitting the lines. Yes, there is a trick. Maybe you know the answer, but I am not going to give it away. I don’t know where this activity originally came from, but I first saw it done by Bob Beichner as part of the SCALE-UP project.

Ok, back to the bullet. So, why would a bullet curve? For something to change direction of motion, there must be a force acting on it. Instead of calculating the motion of a spinning bullet or something, I am just going to determine the force needed to make a bullet curve. That force can then be compared to other forces like gravity and air resistance.

How does a force make an object curve? If you want to think about it in terms of force and acceleration, the acceleration of an object moving in a circle is:

The negative and the r-hat indicate that the acceleration is towards the center of the circle. So, what do I need to calculate this force to move in a circle?

• The radius of the circle. I will talk about this in a moment.
• The speed of the bullet
• The mass of the bullet
• Maybe the cross sectional area of the bullet and the coefficient of drag if I want to compare this force to air resistance.

For the circle, first I am using a circular path just because it is a little easier. I know the bullet might be able to do something else weird, but I am ok with that. How do I estimate the radius for a bullet? Well, in the WANTED scene, the bullet actually does two things. First it curves away from the person in the middle and then curves back to the target behind her. If I just look at the first curve, I can estimate the distance from the person and the deflection. From this, I can calculate the radius of the circle. Here is a picture.

So, if I estimate d and s I can get the radius of the circle. Here is the location of the two points on a circle that is centered at the origin.

So, the coordinates of the two points would be:

And these would have to satisfy the equation for a circle:

The first point clearly works. Plugging in the second point for x and y:

Now, what about the speed and mass of the bullet? There are so many different types of guns and bullets, I will just pick something. I will just use the info from the bullet in a barrel