I like to solve physics problems.  Here is one for you (I just made it up).

A 4 meter ladder leans against a frictionless wall at a 30 degree angle.  The mass of the ladder is 10 kg.  A human stands 1 meter up the ladder and has a mass of 70 kg.  What is the minimum coefficient of static friction between the floor and the ladder so that the ladder doesn’t slip?

Here is the solution—in video form.

But wait! There’s more.  Let’s do another problem.

Suppose you have the same ladder and the same human.  However, in this case the coefficient of static friction between the ladder and the ground is 0.55.  How far up can the human move before the ladder slips?

I like this question because I don’t know how the answer will turn out.  That makes it fun.  So, let’s do it.

But what kind of problem is this?  I’ll make this a multiple choice question.  Here are your options.

1. A friction problem.
2. An equilibrium problem.
4. A work-energy problem.

Did you pick?  OK, I’ll tell you that there will be quite a few students that will say this is a friction problem because it has a coefficient of friction.  That’s not untrue—but it’s not a good way to classify the problem.  You could also say this is a “ladder problem”—again, not untrue but not helpful.

This is an equilibrium problem.  We are trying to find the point at which the ladder slips—the point it leaves equilibrium, but it’s still sort of in equilibrium.

For an object in equilibrium, there are two main ideas—especially for a rigid object.  First, it must have zero net force.  Second, it must have zero net torque about any point.  In two dimensions, I can write these two conditions as the following three equations.

$F_\text{net-x}=0$

$F_\text{net-y} = 0$

$\tau_\text{net-o} = 0$

Let’s talk about the torque stuff.  I don’t want to get into a whole thing about torque, so let me just say that torque is like a “rotational force” and it can be calculated as:

$\tau = Fr\sin\theta$

In this expression, F is the applied force, r is the distance from the point at which you are calculating torque, and $\theta$ is the angle between r and F.  This equation is equivalent to F-perpendicular times r or F times r-perpendicular.

Oh, torques that would make an object rotate clockwise are negative.

One last thing about torques—especially the sum of the torques.  If an object is in rotational equilibrium about some point (point “o”) then it is also in rotational equilibrium about any other point.  When you set up the torque equation, you can pick whatever point you like to sum the torques—it’s your choice (but choose wisely).

Now we can start setting up some equations for equilibrium.  I’ll start with a force diagram for the ladder.

I know that’s a little busy—but it will have to do.  Here are some comments.

• There are two normal forces.  One from the wall (labeled 1) and one from the floor.
• There are two gravitational forces—and this is a cheat.  There is a gravitational force on the ladder and it is as though it is one force acting at the center of mass for the ladder. If the ladder has a uniform density, the center of mass is the center of the ladder.
• The other gravitational force is fake.  This force $m_2\vec{g}$ is there to represent the weight of the human.  But that gravitational force acts on the human, not the ladder.  The ladder pushes UP on the human the same as the weight.  Since forces come in pairs, an upward pushing force from the ladder means there is a downward pushing force from the human on the ladder.  It’s equal to mg, but not mg.
• The friction force is parallel to the floor.  The maximum magnitude of this friction force would be: $F_f = \mu_s N_2$ where $\mu_s$ is the coefficient of static friction.

Now for some equations.  First, this is the sum of the forces in the y-direction.

$F_\text{net-y} = 0 = N_2 - m_1 g-m_2g$

Just a quick reminder.  These are not vectors.  These are components of vectors in the y-direction.  That’s why the two gravitational forces have a negative sign.  I guess I can simplify this a little bit.

$N_2 = (m_1 + m_2)g$

Next is the sum of forces in the x-direction.

$F_\text{net-x} =0=N_1 - F_f$

Since we are at the point of maximum friction, I can include the expression for the frictional force in terms of the coefficient.

$N_1 = \mu_s N_2$

Notice that I know the value of $N_2$ since it only depends on the mass of stuff—but I don’t know $N_1$.  I’m going to need another equation. That’s where the sum of the torques comes in.

In order to write down the sum of the torques, I need to pick a point about which I calculate the torque.  I’m going to go with the bottom of the ladder.  At this point, there are two forces applied.  Since they are applied at the point about which the torque is calculated, they contribute zero torque and they won’t appear in the equation.  Winning.

Here is the sum of the torques about point O (at the bottom of the ladder).

$\tau_\text{net-o} = 0 = m_2gs\cos\theta +m_1g\left(\frac{L}{2}\right)\cos\theta - N_1 L\sin\theta$

What do I want to solve for here?  I want the distance the human goes up the ladder.  This is the value “s” in the expression above.  Really, I have all the values I need in that expression to solve for s.  I just need to set $N_1$ to the maximum frictional force.  But that would be boring.

Instead, let me make a plot of the frictional force as a function of human distance up the ladder.  That will be more fun, right?

Here’s what I get.

From this plot, the human can go up 1.16 meters before the required frictional force exceeds the maximum.  Oh, here is the code for that plot.

Now that you have the code, you can change stuff—like the angle of the ladder or the mass of the human or whatever.

The end.

# Thanksgiving Physics

I am honestly not quite sure how many blog posts I have about Thanksgiving.  It’s probably about 1 per year for 8 years.  I’m going to guess it’s 8.  Here goes my internet search.

This is what I found.

# Intro to Chaos in Mechanics

This is really just for me so that I won’t forget.  I mean, I will forget—but then I can look back at this post and remember stuff.  Here’s to you Future Rhett.

What is a chaotic system?  Really, that’s the question—isn’t it?  There is the classic example of the double pendulumHere is some code for a double pendulum. And this is what it looks like.

But this isn’t the best system.  The problem is that there are two coordinates—the angle for the top bar and the angle for the bottom bar.  Sure, it’s cool—but what if you want to plot angle vs. time or something.  You have to plot both angles vs. time and that’s a bummer.

OK, how about a model of bounded population growth?  That’s just one dimensional, right?  Actually, it doesn’t even have to be population, it’s just an equation—something like this.

$x_{n+1} = 4rx_n(1-x_n)$

In this expression, r is some parameter—it really doesn’t matter what.  Let’s just model this expression for different values of r.  I’ll use a starting x value of 0.1 and r values of 0.7 and 0.9.   Here is the code.

Notice that when r = 0.7, the population reaches some stable value—but this is not true for r = 0.9.

Bifurcation Diagram

Now for another way to look at a chaotic systems—the bifurcation diagram.  Honestly, I didn’t really understand these things until I made one.  Here’s what we are going to do.

• Start with some initial value of x (just pick something—I’m going to use 0.5).  Pick a value for r also.  Let’s just start at 0.1.
• Run the model for 200 iterations and throw out that data.  This should allow us to look at the long term behavior for that particular value of r (throws out the transient behavior).
• Now run the model for 100 additional iterations and save these.
• Create a plot of these final x values vs. r.
• Next increase the r value a little bit (I will increase it by 0.001)
• Repeat until you get bored.

So if the model is stable after the initial stuff, then it will just keep plotting the same value of x after the first 200 iterations and you will just get a dot.  If it’s not stable after the first stuff, then you will get a bunch of dots with different x values.

OK, let’s do it.  Here is the code.  Oh, I made a function to iterate the model.  I probably should put more comments in there.

This is what it looks like.

Up to an r value of about 0.75, you only get one final x value.  After that, you get two different values . With r over 0.9, it gets crazy.

OK, that’s enough for now.  I just want to make sure future Rhett knows how to make a bifurcation diagram.

# 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 &omega; 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.

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

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.

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

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.

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.

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.

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.

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.

# 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.

# Update on Python Physics Curriculum

So here is the deal.  I had this idea.  The plan was to include numerical calculations into the intro physics curriculum by writing a sort of online textbook.  Or maybe just redo my Just Enough Physics ebook to include more numerical calculations.  Anyway, this is what I came up with. It’s written with trinket.io – an online implementation of python that pretty much rocks.

Here is my curriculum (it’s incomplete – but totally free).

Introductory Physics with Python

Here are some of my own thoughts on this curriculum (including using trinket.io):

• It’s free and online.  That’s mostly good – but I don’t know if online is the best format for physics.
• There is one thing about trinket.io that makes this rock.  There is python RIGHT IN THE PAGE.  Readers can view and run code – no logging in, no saving, nothing.  Just edit and run.  No barriers.
• It has the same idea as Just Enough Physics in that it goes over the basic stuff – but doesn’t overload the student with tons of different ideas (no fluid dynamics, waves, buoyancy, sound…).  It’s not that those are bad topics, it’s just too much.  Too much.
• Homework.  Students want homework questions.  I sort of added those in – but students seem to want traditional homework questions.

Now for the part that needs work.  Well, all of it needs work – it’s not complete.  But I made an error – I figured I would finish this curriculum as I was using it to teach the summer session of physics, but the pressure was too much.  In the end, I think I made it too much like the traditional format of a textbook (with the traditional order of topics).  Really, I started along the best path – but went off the rails when I wanted to do a problem that involved new physics.  So, I just added that new stuff in there.

I need to rethink just what I want to cover – and here is my new plan.

• Kinematics in 1-D and 2-D. I like starting with kinematics because students can model motion and this works great with numerical calculations.  The one problem is that you have to use acceleration instead of change in momentum – and this messes up with my momentum principle.  Actually, maybe I will just do 1-D motion so that I don’t need vectors.
• Forces. I don’t really want to focus on forces and equilibrium, but the students need this to do more stuff.  In this, I need to do the following.
• Vectors.  Boom – need vectors.
• Special forces: gravity, real gravity, maybe Coulomb force.
• What about friction, and forces of constraint (like the normal force)?  Here you can see how it gets out of hand.  Friction is super crazy if you think about it – so are normal forces.
• What if I just did simple forces – like pushing with your hand or rockets?
• Momentum Principle.  Here I need to make a connection between forces and motion.  Since I used acceleration before, I need to make a connection between the momentum principle and $\vec{F}_\text{net} = m\vec{a}$.  Honestly, I hate calling this Newton’s Second Law – it seems wrong.
• But what about circular acceleration?  How do you deal with that?  I don’t know.  Maybe just avoid it for now.
• Work Energy Principle. I think this is mostly ok – except I need to introduce the spring force and spring potential energy.
• Angular Momentum Principle.  My initial idea was to cover “Three Big Ideas” – momentum principle, work-energy, angular momentum principle.  However, there is SO MUCH baggage associated with angular momentum principle.  Much of this stuff is just beyond intro-level students.

I think I have a new plan.

• Forces – but simple stuff.  No friction.  No normal forces.  All the examples will be in space or something.
• Momentum Principle and acceleration. Again, normal stuff.  No forces of constraint.  Mostly space stuff because that will be fun.  Projectile motion stuff too.
• Work-Energy Principle.  Springs, gravity, dropping objects.  Orbits.
• Special cases.  Instead of Angular Momentum, I’m going to go over forces of constraint, friction, normal forces, circular acceleration.

The end.  Oh, I need to make sure there are plenty of exercises for students.  Rewrites coming.

I need to redo all my physics labs.  They are terrible.  I want to make them even MORE about model building.

With that in mind, I saw this:

One sentence labs.  Leave the procedure up to the students.  I think I will need some type of turn in sheet for these labs though.  What about informal lab reports?

# The worst high school physics question EVER

Here is a multiple choice question from an online high school physics question.  It’s bad, but it’s probably not actually the worst ever.

It goes something like this:

You have three objects that start at the same temperature.  Which one cools off the fastest?

1. A dry bean
2. Toast
3. Water

I’ll be honest, I answered this question incorrectly – well, I should say that my answer didn’t agree with the key.  Let’s go over the options.

Water

I’m starting with water because this is the answer I chose.  Why would water cool off the fastest?  My assumption was that the water would evaporate and cool off the liquid more than the other two objects.

Of course the evaporative cool depends on several things:

• The water temperature
• The air temperature and humdity
• The volume of water
• The surface area of water.

If I take some water and pour it into a very shallow pan with a large surface area, this stuff is going to cool off quick.  Note: here is an older post about evaporative cooling.

Toast

This was my second answer.  What is special about toast and why would they choose it?  In my mind, toast is special because it has lots of holes.  Lots of holes means that it has a large surface area to volume ratio.

Since things radiate thermal energy through the surface area, things with high surface area to volume ratios cool off faster.  This is why small objects cool off faster than large objects.  This is also why the moon’s core is cooler than the Earth’s core (the moon is smaller).

Oh, this is also how a heat sink works.  Large surface area to volume ratio.

Dry Bean

A dry bean could cool off the fastest because it is small (high surface area to volume ratio) and it is low density.  I assume if it has a low density it has a low specific heat capacity.  This means that with a low specific heat capacity, the dry bean has a small amount of thermal energy even though it has the same temperature as the water and the toast.

This is essentially the same reason that you can put pizza on aluminum foil in the oven.  Once it is hot, you can touch the aluminum foil, but not the pizza.  Although they are at the same temperature, the aluminum foil has less thermal energy to burn you (because of the low mass).

This was the correct answer (according to the people that wrote this dumb question).

Writing questions isn’t so simple

I think what the author really wanted to ask was “which has the lowest thermal energy?”  But even then, you have to take mass and specific heat capacity into consideration.

It’s really just a super bad question.  Super bad.  Oh, but it’s probably not the worst one.  I saw some others that were just as bad if not worse, but I have blocked them from my memory.