Angular Momentum and the Moment of Inertia

Let me be clear—something isn’t working with my program.  However, I think the idea is solid. Also, by writing this I might be able to figure out my problem.

Two Forms of Angular Momentum

I’ll just just to the main point. I’m trying to make a connection between the two forms of angular momentum.  The first is the point-particle definition.  This says that the angular momentum of a point is defined as:

\vec{L} = \vec{r} \times \vec{p}

In this expression, L is the angular momentum, r is the position vector, and p is the linear momentum.

The other form of angular momentum is defined as:

\vec{L} = I\vec{\omega}

Here ω is the angular velocity vector for a rigid object and I is the moment of inertia tensor.

Ok, let’s get started.  I’ll begin with a simple case and move to more complicated stuff.

A free particle.

A particle has a mass m and moves with a constant velocity (no external forces).  What happens to the angular momentum?  Here is a python program for a free particle in which I also calculate the angular momentum.  This is just a picture – you need to go here to run it.

Trinket 2018-11-05 10-46-12.png

Here is a plot of the z-component of the angular momentum in this case.

Trinket 2018-11-05 10-48-10.png

Notice that it’s constant.  Oh, I assume you know about cross products—they are in vpython, so you don’t have to do it manually.

So, yes—angular momentum is conserved.  No big surprise there.  But what if you change the origin?  Is angular momentum still conserved?  Try it.

There is very little point looking at the moment of inertia.  I’ll do that next.

Single object constrained to circular motion.

I want to get an object moving in a circle—but I don’t want to make some crazy constraint.  This sounds like a job for THE SPRING.  Yes, I will model an object moving in circular motion by using a spring attached to the object and some stationary thingy.

Here is the code.  Here is what it looks like.

Nov-05-2018 11-19-46.gif

If you calculate the angular momentum (from the linear momentum), here is what you get.

Trinket 2018-11-05 11-21-20.png

It doesn’t look constant—but I think that’s just a rounding error.  Reminder: this is the angular momentum from \vec{L} = \vec{r} \times \vec{p}.  What if I use the moment of inertia and the angular velocity?  Do I get the same thing?

I can calculate the angular velocity as:

\vec{omega} = \frac{\vec{r} \times \vec{v}}{r^2}

For objects moving about a fixed axis, the moment of inertia is a scalar value that is calculated as:

I = \sum_i m_i r_i^2

Notice that if you put these two definitions together, you get (essentially):

I\vec{\omega} = \vec{r}\times \vec{p}

So it should work.

Two masses.

Using a stationary pivot point can cause some problems.  Since the pivot point is stationary, there must be some external forces on the system.  This means that calculating momentum and angular momentum can be difficult.

Here is the program.  Here is what it looks like.

Nov-05-2018 12-54-39.gif

Yes, these are two unequal masses but the center of mass is stationary.  Also, I will skip the graph, but angular momentum (the z-component) is constant.

Here is a different (but similar) version of the program in which I also calculate the moment of inertia version of the angular momentum.  Everything seems to work—until it doesn’t.

What happens if the center of mass of the system is not zero?  In this case, I need to redo the angular momentum calculation.  First, for the point model, it would be still be r cross p, but I can write it two ways:

\vec{L} = \vec{r}_1 \times \vec{p}_1 = \vec{r}_{com} \times \vec{p}_{com} + \vec{r}_{1r} \times \vec{p}_{1r} +\vec{r}_{2r} \times \vec{p}_{2r}

In this case, the 1r subscript means the position relative to the center of mass and the momentum relative to the center of mass.  The com subscript means center of mass.

For the moment of inertia method, I have:

\vec{L} = \vec{r}_{com} \times \vec{p}_{com} + I\vec{\omega}

But this is where I will stop. For some reason, I can’t get a constant angular momentum using the moment of inertia.  Here is the plot of the component of momentum for the case when the center of mass is moving.

GlowScript IDE 2018-11-05 15-08-36.png

I feel like I am making some silly mistake.  So, here are some notes and comments.

  • Maybe I am calculating the relative velocity incorrectly.
  • Maybe it has something to do with my definition of the angular velocity.
  • Note that the two masses can have slightly different angular velocities since this isn’t actually a rigid object—it’s just mostly rigid (stiff spring).
  • I feel like I have so many different programs, that I’m losing track of what works (that’s why I wrote this blog post).
  • What’s the next step?  Well, after getting this calculation to work—I have big plans.  The ultimate goal is to have a 4 mass rotator (4 masses connected by springs) and calculate the moment of inertia and the angular momentum.  I would be very happy if I could show that the angular velocity vector doesn’t have to be in the same direction as the angular momentum vector. That would be cool.

 

Video Analysis of Soyuz MS-10

There should be a grave yard for blog posts that start, but never get published.  Fortunately, I have this site.  Here I can share with you my failed posts.  Get ready.

It starts with this epic video from the Soyuz MS-10 failed launch.

That’s pretty awesome.  It’s doubly awesome that the astronauts survived.

Ok, so what is the blog post?  The idea is to use video analysis to track the angular size of stuff on the ground and from that get the vertical position of the rocket as a function of time.  It’s not completely trivial, but it’s fun.  Also, it’s a big news event, so I could get a little traffic boost from that.

How do you get the position data?  Here are the steps (along with some problems).

The key idea is the relationship between angular size, actual size and distance.  If the angular size is measured in radians (as it should be), the following is true L = r\theta where L is the length (actual length), theta is the angular size, and r is the distance.

Problem number 1 – find the actual distance of stuff on the ground.  This is sort of fun.  You can get snoop around with Google maps until you find stuff.  I started by googling the launch site.  The first place I found wasn’t it.  Then after some more searching, I found Gagarin’s Start.  That’s the place.  Oh, Google maps lets you measure the size of stuff.  Super useful.

Finding the angular size is a little bit more difficult.  I can use video analysis to mark the location of stuff (I use Tracker Video Analysis because it’s both free and awesome).  However, to get the angular distance between two points I need to know the angular field of view—the angular size of the whole camera view.  This usually depends on the camera, which  I don’t know.

How do you find the angular field of view for the camera?  One option is to start with a known distance and a known object. Suppose I start off with the base of the Soyuz rocket.  If I know the size of the bottom thruster and the distance to the thruster, I can calculate the correct angular size and use that value to scale the video.  But I don’t the exact location of the camera.  I could only guess.

As Yoda says, “there is another”.  OK, he was talking about another person that could become a Jedi (Leia)—but it’s the same idea here.  The other way to get position time data from some other source and then match that up to the position-time data from the angular size.  Oh, I’m in luck.  Here is another video.

This video shows the same launch from the side.  I can use normal video analysis in this case to get the position as a function of time.  I just need to scale the video in terms of size.  Assuming this site is legit, I have the dimensions of a Soyuz rocket.  Boom, that’s it (oh, I need to correct for the motion of the camera—but that’s not too difficult).  Here is the plot of vertical position as a function of time.sideviewsoyuzaccel.png

Yes, that does indeed look like a parabola—which indicates that it has a constant acceleration (at least for this first part of the flight).  The term in front of t2 is 1.73 m/s2 which is half of the acceleration.  This puts the launch acceleration at around 2.46 m/s2.  Oh, that’s not good.  Not nearly good enough.  I’m pretty sure a rocket has an acceleration of at least around 3 g’s—this isn’t even 1 g.  I’m not sure what went wrong.

OK, one problem won’t stop me.  Let’s just go to the other video and see what we can get.  Here is what the data looks like for a position of one object on the ground.

Data Tool 2018-11-04 13-31-41.png

You might not see the problem (but it sticks out when you are doing an analysis).  Notice the position stays at the same value for multiple time steps?  This is because the video was edited and exported to some non-native frame rate.  What happens is that you get repeating frames.  You can see this if you step through the video frame by frame.

It was at this point that I said “oh, forget it”.  Maybe it would turn out ok, but it was going to be a lot of work.  Not only would I still have to figure out the angular field of view for the camera, but I need to export the data for two points on the ground to a spreadsheet so that I can find the absolute distance between them (essentially using the magnitude of the vector from point A to point B). Oh, but that’s not all.  When the rocket gets high enough, the object I was using is too small to see.  I need to switch to a larger object.

Finally, as the rocket turns to enter low Earth orbit, it no longer points straight up.  The stuff in the camera is much farther away than the altitude of the rocket.

OK, that’s no excuse.  I should have kept calm and carried on.  But I bailed.  The Soyuz booster failure was quite some time ago and this video analysis wouldn’t really add much to the story.  It’s still a cool analysis—I’ve started it here so you can finish it for homework.

Also, you can see what happens when I kill a post (honestly, this doesn’t happen very often).

Actually, there is one other reason to not continue with this analysis.  I have another blog post that I’m working that deals with angular size (ok, I haven’t started it—but I promise I will).  That post will be much better and I didn’t want two angular size posts close together.

The end.

MacGyver Season 3 Episode 6 Hacks

Getting through a giant chunk of cement with acid.

Really, this isn’t a “Mac Hack” since MacGyver didn’t do it.  Instead it was someone else.  She used muriatic acid to help get through a tunnel that was plugged up with cement (or concrete—I always get those two confused).

But yes, muriatic acid will indeed “eat” through cement.  If the goal is to create a hole that will allow a human to get through, it’s not so bad.  You can use the acid to weaken enough of the structure that you can take it out in pieces.  You don’t have to dissolve all the cement.

Now for a bonus homework question—actually an estimation problem.  Suppose there is 10 feet of cement to get through.  If someone uses muriatic acid, how long would this hole take to make?  Go.

Thermite

Thermite is awesome—oh, and slightly dangerous.  Basically, it’s a chemical reaction between two metals in which one of the metals has the oxygen needed for the reaction (like iron oxide).  The key to getting this reaction to work is to have super tiny pieces of metal (like super super tiny).  Really, that’s the tough part.  But once you get that, the thermite gets really hot, really fast.  Hot enough to melt stuff.

Could it be used to close up gun ports in an armored vehicle?  Probably.

Pressure to open an armored vehicle

An armored car has armor.  That’s why they call it an armored car.  The primary role for the armor is to keep out things like bullets and people so that the stuff inside (probably money) is safe.

But what if you seal off the openings and then pump in some air?  The cool thing about air pressure is that even a small pressure increase can exert HUGE forces on a wall.  Let’s say that MacGyver doubles the atmospheric pressure inside the truck so that it goes from 10^5 \medspace \text{N/m}^2 to 2\times 10^5 \medspace \text{N/m}^2.  If you have a wall that has dimensions of 2 meters by 2 meters, there would be a net outward force of 4 \times 10^5 \medspace \text{N} pushing it outward.  For you imperials, that is like 90,000 pounds.

Oh, this is why submarines are cylindrical shaped.  Flat walls bend under great pressure.

DIY balance

I really don’t know how much detail to go over for this hack.  MacGyver builds a simple balance scale to find the weight of some money.  As long as the arms are equal, this should be easy.  Maybe in a future post (on Wired) I will show you a design for a symmetrical balance scale.

Pry Bar Lever

Again, there’s not too much to say here.  MacGyver uses a piece of metal to pry open a door.  This is why humans use crow bars.  In short, this is a torque problem.  The torque is the product of force and lever arm.  So a small force with a large lever arm (MacGyver pushing on the end of the bar) gives a small lever arm and huge force (from the tip in the door).

Ok, I lied.  Torque is way more complicated than that.  Really, it’s a vector product:

\vec{\tau} = \vec{F} \times \vec{r}

That’s a little better.

Oh, there were two good hacks that didn’t make it into the show.  Too bad.

 

 

Classical Mechanics: Newtonian, Lagrangian, and Hamiltonian

In classical mechanics, there are three common approaches to solving problems.  I’m going to solve the same situation three different ways.  It’s going to be fun.  Trust me.

Here is the problem.  A ball is at ground level and tossed straight up with an initial velocity.  The only force on the ball while it is in the air is the gravitational force.  Oh, the ball has a mass of “m”.

Newtonian Mechanics

First, let’s get this out of the way.  It wasn’t just Newton that did this stuff.  Also, please don’t lecture me on “Newton’s Three Laws of Eternal Motion and the Rules of the Universe”.  Yes, I think everyone spends too much time on “the three laws”.

In short, Newtonian mechanics works for cases in which we know the forces and we have a reasonable coordinate system.  Yes, it’s true that we can use unreasonable coordinate systems and still have this stuff work.  Also, it’s possible to deal with unknown forces (like the tension in a string with a swinging pendulum).  But Newtonian mechanics works best if we know the forces.

If you know the forces (or just one force), then the following is true (in one dimension):

F_\text{net} = m\ddot{x}

Honestly, I prefer to write the momentum principle—but let’s just go with this for now.  Oh, in case you didn’t notice, I am using the “dot-notation” for derivatives.  A single dot means:

\dot{y} = \frac{dy}{dt}

The double dot means a second derivative (with respect to time).

Back to the tossed ball.  The only force on the ball is the gravitational force.  Also, I am going to use “y” for the position.  This gives us:

-mg=m\ddot{y}

-g=\ddot{y}

That’s just a second order differential equation (and not a very difficult one).  If I integrate both sides with respect to time, I get:

\dot{y} = \dot{y_0} -gt

Integrating again, I get:

y=y_0 +\dot{y_0}t - \frac{1}{2}gt^2

Boom. That’s your equation of motion for a tossed ball.

Lagrangian Mechanics

I’m not going to go over the whole theory behind the Lagrangian.  Here is the short answer (super short).  If you define something called the Lagrangian as:

L = T-U

Where T is the “generalized” kinetic energy and U is the potential energy, then the path a particle will take will be such that the integral over time of L is stationary (which is like minimized).  When you need to find a function that minimizes an integral, you can use the Euler-Lagrange equation to get the following (in 1 dimension for simplicity):

\frac{\partial L}{\partial y} - \frac{d}{dt} \frac{\partial L}{\partial \dot{y}}=0

The best part of Lagrangian mechanics is that you don’t have to use normal coordinate systems.  If you have a bead moving along a wire, the coordinate system can be defined as the distance along the wire.  Also, since the Lagrangian depends on kinetic and potential energy it does a much better job with constraint forces.

OK, let’s do this for the ball example.  I’m going to assume the ball has a position y as measured from the ground and a velocity \dot{y}.  That gives the following:

T = \frac{1}{2}m\dot{y}^2

U = mgy

L = T-U = \frac{1}{2}m\dot{y}^2 - mgy

To get the equation of motion, I need to first take the partial derivative of L with respect to y:

\frac{\partial L}{\partial y}=-mg

Also, I need that second term with the derivative of the partial with respect to \dot{y}.  It doesn’t matter in this problem—but here is where I caution students to be careful of the difference between a derivative and a partial derivative.

\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} = m\ddot{y}

Yes, I skipped a step in there because it was trivial.  Putting this together, I get:

-mg-m\ddot{y}=0

\ddot{y} = -g

Now we are at the same place as the Newtonian method.  I could integrate twice, but I would clearly get the same thing.

Hamiltonian Mechanics

I’ll be honest.  I sort of suck at Hamiltonian mechanics.  Oh sure—it’s super important.  However, for just about every problem in classical mechanics it’s going to be easier to use either Newtonian mechanics or Lagrangian.

Then why do the Hamiltonian?  The first reason is for quantum mechanics.  Yes, in quantum mechanics we use the Hamiltonian operator.  It’s probably a good idea to understand just what the heck that means.  The second reason is statistical mechanics.  The Hamiltonian turns up there too.  Oh, and other places.

Let’s get started though. I am again skipping the derivation of the Hamiltonian.  This is a blog post, not a textbook.

In one dimension (and for one particle) the Hamiltonian is defined as:

H = p\dot{y}-L

Yes, you have to find the Lagrangian first.  Oh, the p is momentum.  However, once you get the Hamiltonian you get the two following equations:

\dot{y} = \frac{\partial H}{\partial p}

\dot{p} = -\frac{\partial H}{\partial y}

OK, let’s do this.  I already have the Lagrangian.  I can write the Hamiltonian as:

H = m\dot{y}\dot{y} - \frac{1}{2}m\dot{y}^2+mgy =\frac{1}{2}m\dot{y}^2+mgy

Since the Hamiltonian really depends on position and momentum, I need to get this in terms of y and p.  Note: I am using p = m\dot{y} for the momentum.  This is not always the case—it depends on your choice of coordinate system.  But anyway, I will proceed.  This makes the Hamiltonian:

H=\frac{p^2}{2m}+mgy

Yes. In this case, this is the total energy.  It doesn’t have to be the total energy—it just works in this case.

Now I can use the two partials to get two equations:

\dot{y} = \frac{\partial H}{\partial p} = \frac{p}{m}

\dot{p}=-\frac{\partial H}{\partial y} = -mg

Here is a major point about Hamiltonian: using this method, I get two first order differential equations instead of one second order differential equation.  That might be important in some cases.

In order to get the equation of motion, I’m going to take the derivative of \dot{y}.

\ddot{y} = \frac{d}{dt} \left( \frac{p}{m} \right) = \frac{\dot{p}}{m}

Fortunately, I already have an expression for \dot{p}.  Putting this together, I get:

\ddot{y} = \frac{-mg}{m} = -g

Boom.  Back to that same equation as in the other two methods.

The end.

 

 

It’s Just One Semester at a Time

This is really for students—but maybe it applies to you also.  If so, I’m happy about that.

So there you are.  There’s still a month left in the semester (or quarter) and you are just plain burned out.  You have not motivation to study and your last test score wasn’t quite what you expected.  That thought creeps into your head—maybe you just don’t belong here.

NO. Don’t listen to that voice!

Yes, we all have that voice.  It’s in us all.  It’s the voice of doubt.  You can get through this—surely you can.

Let’s stop for a moment and consider something else.  Suppose you are in a race.  It’s a long race—maybe it’s a 10k.  You haven’t run this far before and you are worried about finishing last so you start off with a quick pace.

Oh, now it’s up to the 8 kilometer mark and you have lost it.  You have to stop.  You can’t keep up this pace anymore.

Has this happened to you in a race?  It has to me (and I hate races).  Of course the problem in this situation is the pace.  You can’t start off too fast or you will run out of energy.  You have to start off with a reasonable pace that you can keep up with the whole time.  It is indeed odd that starting off slower gives you a faster overall speed—but it’s true.

Back to studying.  You can see where this is going.  If you start off at a whirlwind pace at the beginning of the semester, you are going to run out of steam.

Here are some tips for taking care of business during the semester.

  • You don’t have to be perfect in all (or any) of your classes.  That’s like assuming you are going to win in every race.  No one wins all the time—and this isn’t even a race.  It’s not a competition.
  • Take some breaks.  I’m not saying you should just sit around and chill, but if you work all the time your brain can’t process stuff.  Do something fun.  Go see something.  Hang out with friends.  These are the parts of college life that will have a huge impact.
  • Work with others in a study group.  This means you will help others and this means others will help you.  Both of these things are super useful.
  • Exercise.  Go for a walk or hit the gym.  Personally, I like to run—and I don’t use earphones.  Just use that exercise time to sort of meditate and let your brain unwind.
  • Need help?  Get help.  There are plenty of people to help you.  Go talk to your professor (they are most likely nice). Talk to your friends and family.  If you feel like things are getting out of hand, there are probably support services at your university.

Finally, maybe you like dogs.  Go find a dog and pet a dog.

Physics and Education Majors

There is this course.  It’s called Physics for Elementary Education Majors (PHYS 142) – maybe that’s not surprising.  Anyway, I really like this course – it’s awesome.  Let me tell you a little about the history and future of this course.

According my email archive, I think this course was created in 2003.  Ok, technically it was created before that but 2003 is when we started offering the course again.  Actually, the fact that the course already existed made it much easier to get it going.  If you have ever been part of a university curriculum committee, you know what I mean.

We created the course for the College of Education.  They needed a science course for their elementary education majors that satisfied some particular component of NCATE (the accrediting agency for Colleges of Education).  I honestly don’t know (or can’t remember) what specific thing the course was supposed to do – but there it was.  This course was perfect for them.

The first semester I taught this course, I used the Physics by Inquiry (McDermott) curriculum.  This curriculum was especially designed for education majors – and it’s quite awesome.  However, there was one problem – maths.  There isn’t a ton of math in PBI but there is enough to make students panic.  I think they should indeed work through their issues with math, but it was causing problems with the course. Note: I tell students that they shouldn’t say “I’m not a math person”.

After math troubles, I decided to switch to a new curriculum.  At the time it was called Physics for Elementary Teachers (PET) but was later changed to Physics and Everyday Thinking (also PET – by Goldberg, Otero, Robinson).  Here are some of the awesome features of PET.

  • Student learning based on evidence collected (not authority learning from the textbook or instructor).
  • Explicitly includes ideas about the nature of learning.
  • Emphasis on model building and the nature of science.
  • Includes children’s ideas about physics.
  • Math isn’t a barrier.
  • OH, the best part.  The new version of the curriculum is called Next Gen PET.  This version explicitly aligns with the Next Generation Science Standards.  This should be a huge win for the College of Education.

Honestly, it’s great stuff.  Oh, there are still problems.  Students get caught up in the whole “why don’t you just tell us the answer?” thing – but I can work around that.

But like I said – this is the course that we have been teaching for 15 years (wow – even writing that is incredible).  This course was designed for the College of Education.  We typically have been teaching three sections of the course each semester with an average of about 25 students per section.

PHYS 142 Today

I accidentally discovered something recently.  The education majors informed me that PHYS 142 is no longer required in the curriculum.  What? How can that be?  Yup, it’s true.  The new science requirements for elementary education majors have the following three courses:

  • Biology 1
  • Biology 2
  • Earth Science

That’s it.  I’m sure those are fine classes – but they miss a big thing.  They don’t emphasis the nature of science.  In fact, I suspect that these three classes might actually decrease the students’ understanding the nature of science.  Since these three courses have quite a bit of memorization elements in them, students might come away with the belief that science is about facts and not model building.

Yes, I’m not too happy about this.  Not only do I think this course is perfect for education majors (who will be the first to introduce science to children in many cases).  I also genuinely enjoy teaching this class. It’s great to interact with students and see them increase their understanding.  There’s nothing quite like being there when a student starts putting different ideas together.  It’s great.

On a logistical note, this course as some other huge impacts.  First – teaching load.  If we have 3 sections of this course, that would be 15 hours (it’s a 5 contact hour course).  Getting rid of the course will lose 15 contact hours for the department.  That’s one instructor position.  That sucks.

Oh, also I usually teach this course during the summer session. That’s going to suck to not have this.

 

 

MacGyver Season 3 Episode 5: Dia de Muertos + Sicarios + Family

Battery and Solar Powered Fridge

It’s not a big “hack” in the episode, but we see MacGyver walking through the build a refrigerator.  You could use solar power and a battery with a normal fridge, but there is also the peltier cooler option.  The peltier is a small solid state device – when you run current through it, one side gets hot and one side gets cold.  You can use this device to cool the inside of a fridge.  It’s not super efficient, but it’s very simple.

Actually, I started to build one (but I haven’t finished).  Here is my progress so far.

Modify boom box to pick up beacon signal

So, Oversight builds a homing beacon.  He sets the frequency to 457 kiolhertz.  MacGyver needs to modify the radio to pick up this frequency.  He needs to make the modification because a normal AM radio only goes down to about 540 kHz.

So, how do you change the tuning frequency of a radio?  Let’s look at a super basic radio (a crystal radio).

Photo Google Photos

There are two parts to tuning a radio – there is the capacitor (above that would be the tube with the aluminum foil) and the inductor (the tube with the wire wrapped around it).  The radio will amplify the signal with the frequency that matches that of the inductor plus capacitor combo.  So, just change on of those and you can change the minimum frequency of the radio.

Shock vest

I love the parts where MacGyver and Oversight argue about physics stuff.  Here are some notes about their discussion.

  • Does shortening the wires reduce the resistance?  Yes.  That’s true.
  • You can get a shock by storing electrical charge in a capacitor – that is true (but most taser type things don’t do it that way).  More capacitors means more charge and more shock.

50 foot drop calculation

How fast would you be moving after a drop of 50 feet?  Let’s go over this calculation really quickly.  I hate imperial units, so I am going to switch to metric.  Since it’s about 3 feet per meter, 50 feet would be about 15 meters (rough approximation).

When an object falls, it has a constant acceleration of – 9.8 m/s^2 (assuming no air drag).  That means that for each second that it falls, it will increase its speed by 9.8 m/s.  We can write this as:

-g = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{\Delta t}

Since the falling object starts from rest, the initial velocity is zero.  We can then solve for the final velocity.  Oh, this is all in one dimension.

v_2 = -gt

Oh, I am assuming the initial time is also zero. But we don’t know the time the object falls.  Let’s look at the two definitions of average velocity:

v_\text{avg} = \frac{v_1+v_2}{2} = \frac{\Delta y}{\Delta t}

I can use this with the previous equation to eliminate time.

v_2 = \frac{2\Delta y}{t}

t=\frac{-v_2}{g}

v_2 =\sqrt{-2g\Delta y}

So, the change in y is -15 meters and let’s just say g = 10 m/s^2.  That puts the final velocity at the square root of 300 or about 17-18 m/s.  That’s like 40 miles per hour.

Fan motor as a brake

MacGyver lets a rope wrap around a spinning fan motor.  He then uses that to go down a building (not slow – but slower than falling).  Yes, this is plausible.  It would be better if the motor is on since then there will be a resistance to spinning it.

I could probably say more about this – but it would get complicated quickly.  Oh, how about this – a motor is the same as an electric generator.  It just depends on how you use it.

 

 

Trip Report: Texas AAPT/APS Section Meeting

Since this is just a normal plain blog, I can do silly things like this report on my recent trip.  Why not?

Where and Why?

I was invited to give the keynote address as well as a workshop on python at the AAPT/APS section meeting at the University of Houston.  Since this isn’t too far away, I decided to just drive there – it’s about a 5 hour trip.  Not bad, plus I can bring as many pairs of shoes that can fit in my car.  I brought one pair of shoes.

I drove in on Friday and arrived Friday evening – I stayed at hotel on the outskirts of Houston.

A note regarding section meetings.

I really like section meetings.  They are smaller, cheaper, and it’s easier to get around and see everyone.  Oh, national meetings are cool too – but sometimes they are just too big.  Also, who likes paying 500 dollars just for registration?  Not me.

Python Workshop

For the workshop, I used my python material.  This is essentially the same stuff I used at the Chicago Section of AAPT.  Here are some notes.

  • The material basically this stuff on trinket.io.
  • I also have instructor materials and other files posted on the PICUP site.
  • It seems there were about 15 participants. The room had computers for people to use – that helps out a bunch.
  • There was an issue with the projector – it wasn’t quite working.  Someone brought in a backup, but it wasn’t bright enough.  It’s funny how small problems like this can make a big difference when people are learning.
  • Another issue for python workshops – variety of people.  Some people have never used python and some have experience. This makes it slightly difficult.
  • Other than that, I think the workshop went well.  I had one person ask me afterwards how to become an expert with python.  My response was to just keep practicing.  The best way to learn is to learn python to solve particular problems.  It’s pretty tough if you try to learn stuff without a purpose.  Oh, also – sloppy code is fine.

Keynote: Science Communication with MacGyver and MythBusters

Normally, I give a talk that focuses on physics of science fiction or video analysis or something like that.  I’ve talked about science communication before – but in this case I wanted to include a bunch of examples from MacGyver and MythBusters – so I had to make a new talk.

Check out the venue (maybe it’s difficult to see from this pic though):

This is the “club level” of the University of Houston football stadium.  No, there wasn’t a game going on at the time (but that would have been funny).  It was a nice place – the screens were in a weird position, but still it was nice.  Oh, I did make one fairly big mistake.  I was having trouble with the projectors and I ended up with “mirroring” on my computer.  This means that I didn’t see the next slide and and I didn’t have a clock. I really like seeing a clock.

For the talk, I focused on 4 “rules” of science communication:

  • You can’t be 100% correct, but you can be 100% wrong
  • Build a bridge from the science to the audience (complicated, conceptual, or shiny physics).
  • Science fiction is still fiction.
  • Use mistakes as a foot in the door to talk about what you want.

Overall, I think it went well.  Oh, there was one great question at the end.  “How do we use science communication to help people understand climate change?”  My response: we need to focus on the nature of science and understanding of what exactly science is all about.

Finally, here is another picture. This is me on the football field (which was kind of cool).

MacGyver Season 3 Episode 4

Pressure Lift Bag

This one is pretty awesome.  MacGyver needs to lift up a truck to get it un-stuck.  So, he takes a rubber bladder (not sure where he got it – it could be part of a shock) and connects it to the exhaust (or maybe he connects it to the liquid oxygen).  Anyway, he fills the bladder with an expanding gas.  The bladder fills up and lifts the truck.  This would totally work.

Check out this version you can do yourself.

How does this even work?  Ok, so you have a trash bag.  When you blow air into it, you can approximately get a pressure of 2 atmospheres (just a guess).  The force from this pressure depends on the surface area using this formula:

F = PA

If you want to lift a human (mass of 75 kg) with a gauge pressure of 10^5 \text{ N/m}^2, how big of an area would you need?  Solving the above equation for A:

A =\frac{mg}{P} = \frac{(75\text{ kg})(9.8 \text{ N/kg})}{10^5 \text{ N/m}^2} = 0.00735 \text{ m}^2

That might seem like a tiny area – but that would be a square about 9 cm on a side.  So, this is clearly possible (as you can see in the video that I actually did it).

Liquid Oxygen

We normally think of oxygen as a gas – and at room temperature it is indeed a gas.  Actually, it’s a molecular gas of O2 – two oxygens bound together.  I guess we should first talk about air and oxygen.  Yes, we need air to breath – but air is more than just oxygen.  It’s approximately 21 percent O2 and 79 percent nitrogen gas.

If you decrease the temperature of oxygen gas – it will turn into a liquid.  Yes, it has to be super cold at negative 183 C.  How cold is this?  Here is a video that shows how cold this is (and liquid nitrogen) along with some of the cool things you can do with super cold stuff.

High Pressure Air

Humans can survive under very high pressures.  However, there is a problem with breathing high pressure air.

The nitrogen in high pressure air can be absorbed into your tissues and stuff.  When the human then goes back to a lower pressure, this nitrogen comes out of the tissues.  If the change in pressure happens too fast, this nitrogen can bubble and cause all sorts of problems.  This is basically what we call decompression sickness.

The other problem is oxygen.  At 21 percent oxygen at normal atmospheric pressure, everything is fine for humans (since we live in this stuff).  However, as the pressure increases, the partial pressure of oxygen also increases.  At normal cases, the partial pressure of oxygen is 0.21 atm (atmospheres).  If you have 50 percent O2 at atmospheric pressure, this would be 0.5 atm.  The partial pressure is the current pressure multiplied by the fraction of gas.

Here’s the deal.  If the partial pressure of oxygen gets over 1.6 atm, bad stuff happens.  Stuff like convulsions.  Oxygen is bad stuff.  How do you get a partial pressure of 1.6 atm?  If you increase the pressure, the partial pressure of 21 percent O2 is 1.6 atm.

OK, now back to the show.  MacGyver can survive in high pressure one of two ways.  Method number 1: don’t breath air.  If he breathes a gas mixture that has a lower concentration of oxygen, This is what deep divers do when they breath mixed gasses like trimix.  Method number 2: use a constant atmosphere suit so that he stays at 1 atm pressure.  That’s what he does in this case.

What happens if MacGyver pulls out his air hose? Yup.  That would work.  Even at super high pressures.  Oh sure, his lungs would get super small because of the external pressure – but that’s just fine.  This is exactly what happens when a free diver goes deep (breath holding).

Oh, he would have to equalize his ears just like a free diver.

MacGyver Season 3 Episode 3

Transparent Explosives

Yes, this is probably real – http://www.guns.com/2016/10/21/army-working-on-high-tech-see-though-explosives/

Liquefaction of Sand

This is real.  You can make a simple version of this yourself.

 

 

 

Or you could make a crazy huge version like this.

 

Weather balloon pop

MacGyver needs to get a thermal camera down from a balloon.  The balloon (it’s not actually a weather balloon) is tethered down by multiple lines.  So Mac uses the jumper cables from the car and connects them to the car battery.  Then he connects ONE cable to the wire and the current causes the balloon to burst.

OK, let’s step back for a moment.  Remember that this is a show – this is not real life.  I just want to make sure we are all on the same page there.  So, there’s a small mistake here (you can blame me if you like).  In order to get an electric current from the car battery to go through the balloon, you would need to make a complete circuit.  One jumper cable connected to the line is a start, but there needs to be a path for the current to get back to the battery to make a complete circuit.

One way you could get this to work is to take another line going to the balloon and connect the other jumper cable to that one.  If you look close, it seems like the other cable isn’t connected to anything (in the show).  Of course, that mistake is better than connecting both wires to the same line.

This is sort of the same problem as this double spark in Iron Man 3.

Thermal camera

Yes.  Thermal cameras are indeed real.  Yes, the heat signature of an electric car would be different than an internal combustion engine car.  Actually, I need to see how hot they get in real life (electric cars).  I’m going to test this the next time I see a Tesla.

Oh, and here is an overview of seeing stuff in infrared (also called “thermal image”).

Just for fun, here is a visible and infrared image of me with a bag over my head.

X-Rays from a Vacuum Tube

MacGyver needs to find the transparent explosive.  One of the tools he needs for this is a source of x-rays.  This seems to be real – but it appears you can make x-rays from a vacuum tube, a lighter (the long kind) and a diode.

Here are the instructions from hackster.io (I need to build one of these).

There are so many cool parts of this hack, I could probably write a book on just this one thing – maybe I will write a separate post.  This x-ray device does the following:

  • Uses a vacuum tube from an old radio.  Historically, the vacuum tube was used where transistors are used now.  These things are awesome.
  • The lighter has a piezoelectric in it.  When you apply a pressure to these devices, it produces a voltage – the voltage can get high enough to make a spark in air which lights the gas from the lighter.
  • When you connect the piezoelectric to the vacuum tube, you can make a super high voltage inside the tube.  This can accelerate electrons such that they crash into the other electrode.  This crashing electrons is exactly how you create x-rays.
  • X-rays are just like normal visible light except that they have super small wavelength.  This can make them interact differently with matter.  For instance, they can pass through some materials (like human skin).
  • What is the x-ray used for in this hack?  X-ray fluorescence.  This is essentially the same as glow in the dark (kind of) material except get’s “activated” with x-rays instead of other visible light.

Oh wait! I already have a video on x-ray fluorescence.

 

 

One final note.  In the show, MacGyver says something about shooting ions.  That’s not really what happens here.  X-rays are not ions.

Hydrogen balloon from a trash bag.

Can you fill a trash bag with hydrogen?  Yes.  Will it lift stuff?  Yes.  Could it lift a trash can?  Maybe…just maybe.

Here is my super short introduction to buoyancy.

Suppose you take a box of air – the box is 1 meter on a side such that the volume of this air is 1 x 1 x 1 = 1 m3.  Assuming there is no wind or breeze, this “box of air” will stay in the same location.  Since the box is at rest, the total force acting on the air must be zero.

OK, there is obviously a downward gravitational force on the air puling it down.  Yes, air has weight.  If something has mass, it has a gravitational interaction with the Earth.  Everyone likes to think of air as being weightless – but that’s probably because it has a low density and it’s normally “floating”.  But if there is a downward gravitational force on the air, there must be an upward force pushing to counteract the weight.  This upward force is the buoyancy force.

Since the box of air floats, we know the buoyancy force has to have the same magnitude of force as the weight of the air.

Now let’s suppose I take away that “box of air” and replace it with a sealed cardboard box (it could be a box made out of anything, but in my mind it’s a cardboard box).  The air around this box is going to interact with it in the same way as it did with the box of air (because air is dumb and doesn’t know any better). This means the cardboard box has the same buoyancy force as the box of air – it is equal to the gravitational weight of the air the box displaces – this is essentially Archimedes’ principle for floating stuff.

Oh, this buoyancy force is still the same no matter if the object is floating or not – it just has to displace air.  You can also do this with water or really any substance –  like pudding.  Not sure why you would float something in pudding.

But what if you want to calculate this buoyancy force?  In that case, you need to know the density of the air (which is around 1.2 kg/m3) and the local gravitational field (9.8 N/kg).  With that, the buoyancy force would be:

F_\text{buoyancy} = V_\text{object} \rho_\text{air}g

Finally, we are getting somewhere.  Now you can calculate the size (solve for V) of a balloon needed to lift a trash can.  If you want a simple estimate – you can ignore the mass of the hydrogen in the balloon (but it does indeed have both mass and weight just like the air).  I’m leaving the rest of this as a homework assignment for you.