Cheap-arse demonstrations: Goldfish bowl as refraction tank

Yeah, please feel free to pay $50-$200 for fancy refraction tanks from science supply companies.  I've gone that route myself -- twice -- but I discovered  that the flimsy pieces of junk lasted only a year or two performing as advertised.  You can get the same results with a goldfish bowl from the pet store.

I introduce the concept of refraction using the "4-wheel-drive vehicle* in the mud" analogy.  If one drives a pickup truck at an angle from a paved surface into mud, the truck's path bends simply because it slows down in the mud.  Similarly, light bends when traveling from one material to the next because its speed changes.  While light doesn't have tires, the muddin' analogy is sufficient to predict the direction of refraction, i.e. toward or away from normal.

* I teach at a boys' school in the South.  It's pronounced "vee-HICK-uhl."  And it has floodlights and a gun rack.

Next I introduce the index of refraction and Snell's law.  We imagine light incident from air to water at a 40 degree angle.  We calculate from Snell that the transmitted ray will be 30 degrees from the normal -- good, since the muddin' analogy suggests that since light slows down from air to water, the light should bend toward the normal, and thus the angle of transmission should be less than 40 degrees.

Thing is, all we've done so far is imagine and calculate.  I want to see this light bend, and measure 30 degrees from normal.  Here comes the fish bowl.  

The picture above shows a fish bowl half filled with water.  Key point:  A dry-erase marker writes and erases on glass just as well as on a dry-erase board.  

I have a student mark the water surface with the dry-erase marker.  Then he uses the protractor to mark the normal to the surface with a dotted line,* and a line at a 40 degree angle to the normal representing the incident ray.

*Watching otherwise brilliant AP physics students attempt to construct a perpendicular line segment with a protractor can be an amusing distraction.  You'd think I'd asked them to use nothing but a compass.  P.S. Do *you* remember how to use a compass to construct a perpendicular?  And how much is added to my nerdity score because I do remember?

The chalk erasers to the left of the fishbowl are clapped together so as to visualize the laser in air.  (A drop of coffee creamer allows the laser to be seen in the water.)  I hold the laser steadily so that in air the laser travels along the 40 degree line as drawn.  The student now traces the path of the ray in the water; the result can be seen in the picture.

Finally, a student measures the angle of the refracted ray with the protractor.  We always get within a few degrees of the predicted 30 degree angle.  Physics works.

The fish bowl works exactly like the expensive refraction tank.  And the fish bowl allows us to do the same experiment with the light originating in the water:  Since the tank is curved, I can just shine the laser in from the bottom, and I get light traveling from water to air.  

GCJ

Waves on a wave machine: quantitative iphone picture demo

Photo Credit Sam Russell, using my iphone in class.

I've got this ancient "wave machine," which consists of a set of iron* bars connected to one another atop a wooden base.  You can send a wave across the machine by pumping one of the bars up and down.  These waves are visually exciting, to the extent that students can never resist playing with the machine before and after class.  Good.

* Yes, iron, not some other metal:  the rust due to storage in my leaky classroom for decades is a bit of a clue.

I use the machine to perform a quantitative demonstration with v=lf.  

Step 1:  I ask the class how I might measure the speed of waves on the machine.  Someone will manage to suggest measuring the distance across the machine with a meterstick, and dividing that distance by the time measured on a stopwatch for a wave pulse to travel that distance.  We discuss how that method assumes constant  wave speed -- that's a good assumption, because wave speed depends on the material through which the wave moves, and that material is the same throughout the wave machine.*  

* I do not use the term "homogeneous," but y'all reading know that's what I mean.

Okay, so we do the experiment:  A student measures the wave machine to be 90 cm long.  We let a wave pulse go back and forth, so it moves 180 cm; they measure about 4.2 s for the wave to travel the 180 cm.  That gives a measured wave speed of 43 cm/s.  

Step 2:  I ask what I should do in order to put a wave with frequency 1 Hz onto the machine.  It takes a moment of thought, but generally the class figure out that this means I push the end up and down once every second.  Seems silly, but not everyone has internalized the meaning of a frequency.  Here's an opportunity to translate "1 Hz" into "one wave per second."

Step 3:  I ask the class to predict the wavelength of the 1 Hz wave on this machine.   The first answer that someone always spews:  "43."  That student looks sheepish for a moment as he realizes he didn't put units on.  He quickly computes the units:  in his mind, the frequency was "1", so 43 cm/s divided by 1 = 43 cm/s.  

Um, so a wavelength has units of cm/s now?  

[Don't think I'm being silly or hyperbolic here.  This happened in all three sections yesterday.]

So finally we all think carefully, and agree that 43 cm/s divided by 1 Hz gives 43 cm for a wavelength -- the units of seconds cancel in the improper fraction, and a wavelength has to be a length, anyway.

Step 4:  What good is a prediction without verification?  We must measure the wavelength.  

For years, and in my first section yesterday, we've had a student use his hands to estimate a peak-to-peak distance between the waves, and then use a meterstick to measure that distance between his hands.  

Yesterday, someone suggested taking a picture of the wave machine with a meterstick nearby.  "Just read the peak-to-peak distance right off of the meterstick in the picture," he said.  Certainly I've heard that suggestion before, and it's most definitely an appropriate method.  But until now, I've never had the technology to get this done in class.

I whipped out the iphone, and... handed it to a student.  What, you think that I can handle newfangled technology better than a whippersnapper?  Bah.  

I created the 1 Hz wave, and the student took the picture.  What he actually did was take a video, then play back the video to capture a still shot when the wave was in a good place.  He emailed the photo to me; I got the email, and put the photo up on the screen in my classroom.  No more than about 3-4 minutes elapsed.  That's still too long, but I'll bet with practice I can bring that time down.

The picture we took is at the top of the post.  Problem:  it's awfully blurry, so much so that we can't see the markers on the meter stick.  The iphone camera is convenient, but not so, um, good.  Nevertheless, we can make a quick estimate -- the peak to peak distance is clearly less than half a meter, but more than a quarter meter.  That's as good a measurement as I used to get with a student holding his hands above the wave.

After class, I imported the picture into Microsoft Paint.  I measured several times, to get about 680 pixels from one end of the meterstick to the other.  I found the peak to peak wave distance (again, using several measurements) to average 270 pixels.  Dividing the scale out, I get a wavelength of... 40 cm.  Difference predicted to measured wavelength: 14%.  Not bad for a blurred meterstick.

Experimental evidence that brightness depends on power

A year or so ago, Michael Gray emailed me a wonderful quantitative demonstration idea to show that brightness of a bulb depends on power, not voltage.  Basically, he used a light probe to measure brightness directly.  When he doubled a bulb's voltage, the brightness didn't double -- the brightness reading quadrupled.  And that makes sense, since power is  V²/R .

I took the light probe approach a bit further the other day.  I asked the class to sketch a plot for the brightness reading in the probe as a function of the bulb's voltage.  After some discussion, the randomly chosen student sketched a parabola*on the board.  Yes -- since a bulb doesn't change its resistance, and since power is V²/R, a power vs. V graph should be quadratic.  And since brightness is correlated with power, the power graph should also be quadratic.

* Though he called it, of course, an exponential.  What is it with teenagers that any concave up, increasing function is labeled as "exponential?"  Have *you* ever seen an e^x in any physics B equation, at least since half-lives were taken off the exam a decade ago, and besides that was e^-x?  Should I stop ranting now?

And so I turned out all lights, held the probe about 10 cm above the bulb, and increased the bulb's voltage at a constant rate (by turning the dial approximately uniformly).  As you can see, I was a bit jerky in turning the voltage knob.  But the principle was well-verified -- the brightness vs. time graph was clearly curved.

A quantitative demonstration with springs... one that DOESN'T work!

All year long, the majority of my in-class time is spent doing quantitative demonstrations, in which the solution to an example problem is verified live in class via measurement.  Each demonstration seems to end the same way: I perform the measurement, we all see that the measurement matches the prediction within 10% or so... everyone exhales, and I say "Physics Works."

It's not a problem, per se, but by November or so the novelty of my approach has worn off.  Instead of bright shiny faces anticipating whether or not the experiment will match the prediction, I start to see resignation:  "Yeah, physics works, I know, we've done this a million times."  I've got to throw a changeup.

Once we've introduced the force and potential energy of a spring, I set up what looks like a routine quantitative demonstration.  I hang a mass from a spring, as shown in the diagram.  I determine the spring constant via the 5-second method that I discuss in this post.  This particular spring (the one on the left in the picture) has a spring constant of about 7 N/m.

Next, I hang 100 g of mass from the spring, and allow the spring to come to rest.  Using F=kx, I derive that the mass should stretch the spring by 14 cm; sure enough, a ruler placed alongside the spring shows 14 cm of stretch.  Physics works.

Now, I use the ruler to measure an additional 5 cm of stretch beyond the equilibrium position.  The question: After I release the mass from rest, what will its speed be when it passes through the equilibrium position again?

[Note to readers before you flame me about these next couple of paragraphs:  Keep reading.  Trust me.]

I use the force on a spring equation, F=kx, to calculate the force on the mass to be (7 N/m)(0.05 m) = 0.35 N.  Then, Fnet = ma, so the acceleration of the mass is (0.35 N)/(0.10 kg) = 3.5 m/s².

I take a brief interlude, if someone asks, to (correctly) explain that with a vertical spring, it's completely correct and most simple to treat it as if it were a horizontal spring, but with the x = 0 position at the place where the mass would hang at rest.  In other words, it is acceptable to consider that the spring force kx is the net force, as long as we define x = 0 at the resting position.


Now we have enough information to make a kinematics chart.  The mass starts from rest, its acceleration is 3.5 m/s² and it will move a distance of 5 cm.  Using the kinematics equation v² = v_o² + 2ax, the speed at the equilibrium position is 0.59 m/s, or 59 cm/s.

"How do we verify this prediction?" I ask the class.  They quickly and accurately tell me to place a motion detector below the mass, and to look at the velocity-time graph.  I do so, I read the v-t graph, and I see that it says about 40 cm/s.  Physics wor------  oops.  Physics DOESN'T work.


All year, I've conditioned the class to expect measurements accurate to 10 or 15%.  I'm not even getting within 30% this time... that's not a matter of "experimental error," especially when I try again and get the same result; or even more especially when I try it with the other spring, and I'm still off by more than 30%.  So, why didn't physics work?

Despite the desperate grasping at straws from some students, someone usually comes up with the right answer:  I used kinematics when acceleration was not constant.  The force of the spring was only 0.35 N immediately after release.  As the spring compressed, the spring force got smaller by F = kx.  So the acceleration got smaller.  So the mass didn't speed up as rapidly, and ended up going less than 59 cm/s.

How do we solve this problem correctly?  Using conservation of energy, of course.  Spring potential energy is converted to kinetic energy, 1/2 kx² = 1/2 mv².  Solving, I get v = 41 cm/s.  Now, physics works.

Side note:  Sure, occasionally a student will try to stop me from using kinematics with changing acceleration.  My response is a quick "shhh!!!," Dr. Evil Style, but with a smile.

GCJ

Static Fluids with Quantitative Demonstrations -- Detailed Class Notes

What's the buoyant force on a lionfish?  I ain't doing this
demo, but you can see the demonstrations I do do at  this link.

Regular readers are probably aware that my typical class period consists of one or more quantitative demonstrations -- I don't just solve an abstract example problem from a book, I physically set up the example problem as an in-class experiment.  Any calculation we make in class is verified by measurement. 

One of the questions I'm most asked is, "Do you have a list or writeup of all of your quantitative demonstrations?"  Unfortunately, I don't.  When a publisher (or the NSF) offers me a five figure advance, I'll consider writing a book.  

Until that glorious day, you ask, where do I look to find quantitative demonstration ideas?  Well, start by searching through this very blog, of course.  Sign up for one of my AP Summer Institutes (I'm doing four, I think, in 2012).  Every time you're choosing an example problem to use in your class lecture, think, "could I set this up in my classroom?"  Use a homework problem as the basis for a laboratory activity, and make your students create the setup.  Talk to other physics teachers, including those who are listed as followers of this very blog.  

Now, if you'd like an extraordinarily detailed description of what a class with quantitative demonstrations might look like, check out this piece I wrote for the College Board a few years ago.  They asked me to provide a "lesson plan" for AP-level static fluids.  I described each experiment, each check-your-neighbor question, each measurement that I make during class.  (If you've been to a summer institute, you'll recognize a few of these demonstrations.)

Do you have a quantitative demonstration that you use in class?  Tell me about it.  I'm open to guest posters on this blog...

AP-level Kinematics... In Just Two Weeks!

I got a nice note last night from New Yorker Scott Marzloff, who attended my AP Summer Institute at Manhattan College.  Scott noted that he started the year with equilibrium and torque using quantitative demonstrations, and that his approach was successful.  Awesome.  However, he asks:

I am getting ready to start kinematics and I am wondering how you get through all of 1-D kinimatics in what looks to be about 6 days?  I know I can get through 2-d projectiles in a week, but 1-D with graphs, equations, and freefall I have never come close to covering in less than three weeks.  Do you tie the equations in right away with motion diagrams and graphs?

Well, to be fair, it takes more like 8-9 class days to get through kinematics, including both one- and two- dimensions.  And, I'm not saying that every one of my guys is ready to take AP exam problems on day 9.  Nevertheless, I get through kinematics just that quickly, and we perform well above the national average on kinematics problems come May.

I begin with position-time graphs, demonstrating with a fan cart and Vernier motion detector.  We predict qualitatively what a couple of graphs should look like, show that the slope of an x-t graph is velocity, explain how to find displacement from the graph's axes.  On the second day I introduce velocity-time graphs.  Acceleration is defined as the slope of a v-t graph.  I take considerable time to get students arguing about how to use the fan cart to reproduce various straight v-t graphs.

Three hints about teaching motion graphs:

(1) Homework related to motion graphs is not purely quantitative.  With every graph, I require students to describe the motion represented in everyday language, without terms like "acceleration" or "negative."  For nearly a week, homework and quizzes hammer kinematics concepts rather than calculations.  One night's homework is a 21-question set of multiple choice questions about motion graphs, on which I require detailed explanations for any question they initially get wrong.  Make the class explain their thoughts in words.

(2) I don't (yet) stress over the difference between vector and scalar quantities.  Kinematics is confusing enough without the subtlety of distance vs. displacement, velocity vs. speed.  I use velocity and speed as synonyms on these first few days.  Heresy, you shout?  Perhaps, but it works.  Try it.

(3) In the same vein as (2), don't even discuss sidebars, no matter how interesting.  For example, curved v-t graphs simply don't exist during the two weeks of kinematics.  If a student asks about them, I explain that a huge set of moving objects produce straight v-t graphs, and that the things that don't have constant acceleration simply aren't relevant right now.  Other irrelevancies include a-t graphs, "jerk", anything involving calculus...

Graphs take at least half of my time teaching kinematics.  Why?  Because once we understand motion graphs conceptually, algebraic kinematics is straightforward.

I explain that all moving objects we consider will involve straight velocity-time graphs.  But we want to be able to make predictions without having an actual motion graph in front of us.  How can we do that?  By using a straight v-t graph to derive some algebraic formulas.  I show that taking the slope of a straight v-t graph produces the first of the constant-acceleration kinematics equations.  I show that taking the area under the v-t graph, along with some algebraic substitution, produces the second of the constant-acceleration kinematics equations.  (I simply state the third equation, without derivation -- they get the point.)

I do a set of quantitative demonstrations with the PASCO projectile launcher.  In each problem, we define a positive direction, and fill out a chart in which we identify the five basic kinematics variables (vo, vf, Δx, a, t). We learn that the problem is solvable as long as we know three of the five variables.  I don't spend any time on algebraic methods -- they figure out the most efficient algebraic techniques on their own by doing homework problems.  (Exception: I do one example in which I demonstrate that we do not ever need the quadratic formula.  There's always a simpler approach.)

At this point, one-dimensional algebraic kinematics problems become simple for the majority of students.  Free-fall doesn't need a separate unit, like most books give it; free fall is just accelerated motion where a = 10 m/s², down. We very, very quickly move on to projectiles.  Projectiles become simple, as well, once we learn to make two kinematics charts, one vertical, one horizontal.  Horizontal acceleration is always zero; time is always the link between the charts.  Since I taught equilibrium and free body diagrams first, no one is flummoxed when I show how to deal with an angled initial velocity using sines and cosines.  Projectile motion becomes more a reinforcement and solidification of kinematics concepts rather than a truly new unit.

That's it.

Most of the class is reasonably skilled, at this point, but they still need considerable practice.  But I don't give that practice in isolation.  Rather, the practice is integrated into our next few topics.  For example, next is Newton's second law.  Most problems require both a free body analysis AND a kinematics approach.  Well, when they have to use an acceleration they calculated from Fnet = ma to find how far an airplane goes on takeoff, they're practicing their kinematics skills.  Later, when two blocks collide at the edge of a cliff, conservation of momentum yields a projectile problem.  Practice in context is always more effective than rote practice.  (So, when did you really, truly learn to evaluate integrals quickly and accurately:  in your first three calculus courses where you were given explicit practice, or in your differential equations course, which took integration skills for granted in order to do more involved problem solving?)

Try moving along quickly in your simpler topics, so that either (a) you can cover more topics in the year, or (b) you can spend more time on the truly difficult topics later in the year.

GCJ

Equilibrium: quantitative demonstrations (or, how I teach vector math without teaching vector math)



That's a 2 N weight hanging by two strings.  The left-hand
rope passes over a pully to a 1 N weight; the diagonal rope
is attached to a digital scale at the top of the picture.


In Honors or AP Physics, I begin the year with equilibrium, not with motion.  In a few days, the class gets comfortable with free body diagrams, forces, and two-dimensional vector analysis. 

Next comes motion in one dimension (graphs first, then algebra), followed by projectile motion.  Finally, we cover Newton's second law.  Since we did equlilibrium already, students are comfortable with free body diagrams and writing a vector sum of forces; since we did motion already, they already have some idea of what acceleration is.  Rather than a tough *new* topic, the second law becomes a way to review and solidify the first two topics.

The picture shows my third or fourth demonstration of the school year.  We begin by equating horizontal tensions in ropes pulling on a stationary block.  Next, I hang a 200 g mass vertically to find the tension in the supporting rope.  Easy stuff, so far.

And then, I attach a horizontal rope over a pulley attached to a 1 N weight. (In the picture, the hanging weight on the left is below the table, and out of the frame.)  When we make the free body diagram, everyone's comfortable with the weight of the 2 N weight acting down, and the horizontal rope pulling leftwards.  We draw the arrow representing the diagornal rope's force at an angle, of course.  Then what?

The class has already been taught that in equilibrium, up forces = down forces and left forces = right forces.  I ask, which is this diagonal rope, an up force or a left force?  Someone always comes up with a reasonable answer:  "both."  I redraw the free body diagram, with the tension in the diagonal rope replaced by two arrows, one up, and one to the side.

The class is totally comfortable with the upward component being equal to the mass's weight of 2 N; and with the leftward component being equal to the weight hanging over the pully of 1 N.  (They're also totally comfortable with me using the term "component" without preamble.)  It only takes a suggestive diagram to get someone to suggest that the resultant tension in the rope itself will be not 2 N + 1 N, but the pythagorean sum of 2.2 N.

The clincher comes when  I call a student to the front to read the digital scale attached to the rope.  It reads... 2.2 N.  Physics works.  I can even predict (and then measure) the angle made by the diagonal rope. 

The key to this whole process is that I'm *not* just telling the class how to solve abstract vector addition problems.  I'm not telling them anything at all, really; I'm drawing diagrams and asking questions, getting someone in the class to suggest the next step wherever possible.  I model the correct problem solving method for equilibrium problems on the board, of course, but everything I do flows naturally from the fundamental principles of equilibrium: up = down, left = right.  I don't use any words like "vector" or "reference frame" or "coordinate system." The only technical term I'm introducing is "component," which was introduced organically.

The final demonstration with this setup involves adding a "mystery weight" to the stuff hanging over the pulley.  I measure the new angle that the diagonal rope makes, so I have a chance to suggest how to use sines and cosines to "break an angled force into components."  We predict the reading in the spring scale, and the amount of mystery weight that I added.  Once again, physics works. 

Electrostatics: Wayne Mullins' incredible demonstration

Photo credit to AP reader Teresa Walker -- Thanks!

I would contend that 3/4 of the ideas I use in my teaching have been inspired by, or sometimes directly copied from, the people I meet at the AP physics reading.  I arrived in Kansas City on Tuesday, ostensibly to grade 125,000 physics exams.  Within an hour of the opening of the physics teachers' lounge, I had heard Wayne Mullins' revolutionary thoughts about teaching electrostatics.

My own, obsolete, thoughts are detailed in this post.  Essentially, I treat the electric field as primary, teaching F = qE for days before even discussing the Coulomb's law force between charges.  This approach has worked quite well.  The enormous, gaping drawback is the lack of a quantitative demonstration.  I don't have an electric field measurer; the Vernier charge sensor doesn't really work so well, considering that static charge bleeds off an object (and the sensor itself) so easily; and I've never even gotten my Van de Graff generator to work correctly in my dungeon of a classroon.  I had given up on electrostatics demos.

Wayne brought his own electrostatics demonstration along -- see the picture above.  On the very first day of teaching electricity, Wayne sets up these two aluminum rods about a foot apart.  He connects the two rods to 25 V AC.*  In the tray he puts a thin layer of water, about a centimeter deep if that.

* Why AC?  It prevents corrosion.  And since the voltmeter set to AC reads an RMS value, the voltage readings still do what they're supposed to.

In Wayne's mind, voltage is the primary electrostatic quantity that students must understand.  He reasons that if students can get a literal feel for the measurement of voltage, then he can define electric field as related to changing voltage.  (Specifically, he defines electric field as the slope of a voltage vs. position graph.)  And THEN he can start talking about electric forces.

So Wayne takes a digital voltmeter set to ACV.  He clips the ground lead to the gounded metal plate.  He puts the tip of the other lead in the water, and moves the lead around.  The voltage reading is seen to change as the lead moves, but ONLY IF THE MOTION IS TOWARD A PLATE.  If he moves the lead parallel to the two plates, the voltage reading stays constant.  This is a beautiful, quantitative demonstration of E = V/d between two parallel plates -- and on the first day of the unit, even.  Wow.

But there's more!

Next Wayne places two finger tips (not more than two fingers!) right in the middle of the water between the plates.  (This process is shown in the picture above.)  He moves his fingers apart in a direction parallel to the plates -- nothing happens.  But when he slowly separates his fingers, one toward one plate and one toward another... he starts to feel a bit of a tingle that gets stronger with more finger separation.

What he's communicating is how electric field, and thus electric force on charges, depends not on the value of an electric potential but on the voltage DIFFERENCE between two positions.  Since the voltage is the same along a line parallel to the plates, two fingers separated on that line feel no electric field or force.  But when one finger is at a significantly higher voltage than the other, ooh, tingle!

I did ask the obvious question about safety concerns.  He's using 25 V, not a full 120 V outlet, but still, as his students say, we're told not to touch "electrified water" since we can understand English.  Wayne doesn't force any student to try the experiment, but he is sure that it is safe with the following three caveats:

(1) The fingers used should have unbroken skin.  I tried this yesterday, and all I got was a tingly sensation.  Wayne says that a papercut will hurt to high heaven.

(2) Only use TWO fingers.  Multiple fingers act as parallel paths for the flow of charge along the skin, meaning a much higher current experienced.

(3) Obviously, if a student has heart issues or a pacemaker or something like that, he shouldn't do this. 

The next follow-up to the parallel-plate setup is to place an aluminum-wrapped piece of PVC in the region between the plates.  What should you discover about the voltmeter reading inside the PVC?  

GCJ 

Brightness of a bulb -- quantitative demonstration

Pop quiz, hotshot:  The brightness of a light bulb depends on the bulb's

(A) voltage

(B) current

(C) resistance

(D) power

Answer:  POWER. 

Sure, for *identical* bulbs, voltage and/or current will kind of relate to brightness.  For a fixed resistance, the equations I²R and V²/R show that with the same resistance, power, and therefore brightness, will increase as current or voltage increases.  However, you'll eventually get into trouble assuming that a bulb carrying more current is necessarily brighter than one carrying less current.

It's pretty straightforward to show experimentally that voltage does not necessarily correlate with a bulb's brightness.  Just get a bunch of miniature flashlight bulbs and holders from Radio Shack or Harbor Freight or somewhere.  Be sure to get bulbs with different voltage and current ratings, so that their resistances are different.  Connect them in parallel to a battery -- they will take the same voltage, but will NOT be just as bright as one another.

Michael Gray, a veteran of my 2010 AP Summer Institute and a frequent contributor to this column, came up with a much cleverer and more subtle demonstration.  He showed with a single light bulb that the bulb's brightness depends on the power, not the voltage or current.  How?  He measured the bulb's brightness directly with a Vernier light sensor.  Of course!  Brilliant.

He connected a bulb to a battery.  He showed that, by the equation V²/R, doubling the bulb's voltage should not just double the bulb's brightness, but quadruple the brightness.  He darkened the room, and placed a light sensor a fixed distance from the bulb.  He zeroed the sensor for the ambient light, and turned the bulb on.  When he doubled the bulb's voltage, the sensor reading quadrupled.  Physics works.

I will certainly use this demonstration next year.  Thanks, Michael!

Induced EMF caused by a falling magnet, and a qualitative demonstration

Helmholtz coils of the type useful for demonstrating
the answer to today's question.

I've seen the multiple-choice question from at least two different sources:  a bar magnet is dropped through a circular wire coil from above.  The north end of the magnet points down.  What will a graph of induced EMF (or current) in the coil vs. time look like?

Answer:

This question requires use of Lenz's Law.  I teach an approach to Lenz's law described here.  Let's consider first the time while the magnet approaches the wire loop, and then separately the time when the magnet recedes toward the ground.

As the magnet approaches, the magnetic field is down toward the ground because the B field due to a bar magnet points out of the north end.  The magnetic flux is increasing because the bar magnet approaches the coil, increasing the magnetic field at the location of the coil.  Becuase flux increases, I flip my right thumb opposite the magnetic field, and curl my fingers -- the current in the coil is counter-clockwise when viewed from above.

After the magnet passes through the coil and is receding, the magnetic field still points down toward the ground!  This is because the B field due to a bar magnet points into the south end.  The magnetic flux is now decreasing because the bar magnet recedes from the coil, producing an ever-smaller magnetic field at the location of the coil.  Becuase the flux decreases, I keep my right thumb pointing in the direction of the magnetic field, curl my fingers, and find the current in the coil is the other way -- clockwise as viewed from above.

I asked my AP classes this question as a "check your neighbor" exercise today in class.  At first, when working alone, most of the class (~2/3) didn't recognize that the current switches direction when the magnet passes through the coil.  Upon discussion, I found that the key misconception was the direction of the magnetic field when the magnet receded.  Either they thought that the magnetic field must have a different direction near the other pole, or they thought that since the field was decreasing, the field itself must switch directions.  Good arguments among the class changed some views, such that after discussion about 2/3 of the class had the correct answer.

Setting up the demonstration:

Long time blog readers know that I don't think it good enough to explain something like this conceptually and/or mathematically without doing the experiment.  Nature is the ultimate arbiter of arguments.  So I set up the problem.

I used a 10-cm diameter wire coil with a gazillion loops.  I've on other occasions used a bigger diameter set of coils scavenged from a 20 year old broken q/m device.  You'll get the best results if you use a coil with enormous numbers of loops, like one designed to produce strong, uniform magnetic fields -- a Helmholtz coil, like those shown at the top of the post.

I hooked a Vernier voltage probe to my labpro, and set the labpro to take 1000 data points per second for 2 seconds.  A student clicked "collect;"  When data collection began, I dropped a small (~5 cm length) bar magnet through the coil.

A bit of zooming produced the graph shown above.  As expected, the voltage switched polarity (i.e. the current switched directions) as the magnet passed through the coil. 

Though it's not brutally obvious, the right-hand hump is not as wide on the graph as the left-hand hump.  Why not?  Because the magnet speeds up in free fall, and so recedes from the coil faster than it approached the coil.  I dropped the magnet from about shoulder height.  The ~0.10 s duration of the voltage spikes are substantially less than the ~0.5 s time I calculate for the magnet to fall to the ground, as it should be. 

Any more follow up thoughts?  This demonstration can produce a wealth of interesting calculations and experiments suitable for an end-of-year independent laboratory investigation.

Mail Time: Bernoulli experiment troubles.

Michael Gray, of Georgia, writes: 

I have been preparing the Tennis Ball can experiment, but the water doesn't go quite as far as the math predicts. With a hole to top height of 18.5cm and a hole to floor height of 93.5cm I get a distance of 83cm. However, the stream is only reaching the high 70's of cm. Also, the water breaks up from a solid stream before it gets there making the exact landing point difficult to determine.

Any advice?

The stream breaking up is the problem, I think. In class, I've always hit the prediction dead on. But on Monday at a workshop in Alabama, I missed. See, I had quickly jury-rigged a gatorade bottle for this demo. I used my cheapo pocket knife to make a jagged hole. The stream was not particularly clean, but was breaking up. Sure enough, I missed -- I predicted 65 cm, but the stream only went 50 cm.

I suspect -- though I'll have to play around a bit to be sure -- that this demo depends on getting a "clean" stream out of the hole. After all, Bernoulli's equation explicitly is for "inviscid" flow, meaning no viscous drag.

Let me know if you try again, and I'll do the same.

 

GCJ

Mail Time: Vernier pressure sensor and pressure in a static fluid column

Jeff Bourne, from Northfield High School in John's Creek, GA, writes in:

I took your AP Physics workshop at Oglethorpe U in the Atlanta area several years ago. I was looking over your unit on fluids, and I wondered how do you use the vernier gas pressure sensor to measure the water pressure at a specific depth in a graduated cylinder? I can't find a lab manual that outlines how to do that. I would appreciate any help you can give me.

Hey, Jeff... good to hear from you. I attach some hollow plastic tubing to the probe, and put a small rubber stopper on the other end of the tubing. The tubing is maybe 1.5-2 feet long -- the gas pressure sensor comes with some of this tubing and a stopper, as you can see in the picture.

First, I read the atmospheric pressure, which can range from 97 to 105 kPa depending on altitude, weather conditions, and the probe's internal calibration. Then I measure the depth of water in the graduated cylinder. I do the calculation with the class of what the pressure should be at the bottom using P = P_o + rgh. For a typical graduated cylender I use, I get that the pressure at the bottom will be 2-3 kPa above atmospheric pressure, which is easily measurable by this probe.

Finally, I start the data collection (taking about 20 points per second). I feed the tubing into the cylinder until the stopper hits the bottom.  Voila, the pressure jumps from, say, 99 kPa to 102 kPa.

GCJ

Two-cart elastic collision -- how to measure the speed of both carts

I've taught conservation of momentum with the same set of quantitative demonstrations for years. I get two carts and a motion detector, make the carts collide, and predict speeds of the carts before or after the collision. Pretty basic, without much room for creativity. Until today, that is.

The first collision problem I attempt is very straightforward.  Take a look at the setup in the picture to the right.  I keep the red cart at rest, and send the heavier blue cart toward the red cart.  A velcro strip on the carts causes them to stick together after the collision.  The motion detector measures the speed of the blue cart before and after the collision.  I tell the class what the motion detector said the blue cart's initial speed was; we predict the speed of the combined carts after the collision.  Generally we get this prediction correct within 5-10%.

 

In the next collision, I push the blue cart toward the stationary red cart without the velcro, so that the carts bounce off one another.  Again the motion detector records the speed of the blue cart before and after the collision; we predict the speed of the red cart after collision.  Problem is, how should I measure the speed of the red cart to verify my prediction?  I pose that question to the class.

 

Most quickly understand why the single motion detector can only read the blue cart's speed.  The detector can't "see through" the blue cart. 

 

The solution suggested by a majority of students is to place a second motion detector on the left side of the track to track the red cart.  And I've tried that before.  But I've never gotten two facing detectors to work properly -- I think the sound waves interfere, causing nonsense results.  (Anyone else have any thoughts on this issue?)

 

What I've always ended up doing -- until today, anyway -- is to have students measure the time between the collision and when the red cart hits the end of the track, 100 cm away.  Since the track is level, the speed of the red cart is 100 cm divided by the time the students measure.  This works, but stopwatch measurement uncertainty causes this to be accurate only to about 20-30%, if that.

 

In today's class, junior Will Choate made the suggestion that I've missed for the last 15 years:  "Can you pick up the blue cart after the collision?  Then the detector can read the red cart."  That's IT!

 

So, I collided the carts.  After the collision, I gave the detector a brief moment to read the blue cart's speed, then I picked up the blue cart.  Sure enough, the velocity time graph showed the initial speed of the blue cart (47cm/s); the final speed of the blue cart (31 cm/s); a bunch of nonsense where my hand briefly got in the way; and the final speed of the red cart (80 cm/s).  We used conservation of momentum to predict that the red cart should have been moving 84 cm/s, within 5% of the measured speed.

 

The class was pleased with Mr. Choate for his creativity, and because he earned a TWO Reese's Cup reward.  I think much of the applause and congratulations was kissing up to Mr. Choate in case he felt inclined to give away his candy, but nevertheless.

Quantitative demonstration -- Bernoulli's equation

Photo of my class by Frederic Lamontagne

A straightforward problem with Bernoulli's equation leads to a fun and dramatic demonstration, along with a fascinating discussion of the meaning of "negligible" and some use of the continuity equation.  Here we go...

I start with an empty tennis ball container.  With a pocket knife, I poke a small (<1 cm diameter) hole near the bottom of the container.  I show the class the container, and explain:  I'm going to cover the hole, and fill to the top with water.  The water clearly will shoot out of the hole.  What will the speed of this water jet be?

The class understands that a flowing fluid problem requires Bernoulli's equation.  So we write the equation:

P₁ + rgy₁ + ½rv₁² =  P₂ + rgy₂ + ½rv₂²

Consider point 1 to be the top of the water, and point 2 to be as the water exits the container.  We want to find v₂.  Let's see what terms can be cancelled or set to zero.

The water at the surface and at the exit is exposed to the atmosphere.  Any time a liquid, flowing or not, is exposed to the atmosphere, its pressure is atmospheric.  So the P terms cancel.

We can always take the lower y position to zero in Bernoulli or conservation of energy problems.  So get rid of the y₂ term. 

Now we have a class discussion / argument.  Is the speed of the water at the top zero?  Certainly not.  But the water at the top is moving very slowly relative to the jet at the bottom.  So let's try setting that v₂ to zero, and see what happens.

We measure the distance from the hole to the top of the container to be about 15 cm.  Solving Bernoulli's equation, we find the speed at the position of the hole to be 1.7 m/s.  Awesome!

But this is a quantitative demonstration.  We don't just solve the problem... we have to verify the answer.  How?  I can't just put a Vernier motion detector on the water jet.  The class brainstorms for a bit, and comes up with a great approach:  treat the water jet as a projectile.  Given the jet's initial horizontal speed of 1.7 m/s, we should be able to predict how far the jet will travel horizontally before hitting the ground.

What a great way to reinforce projectile motion problems, which we covered two months ago!  The horizontal acceleration is zero, the vertical acceleration is 10 m/s², the vertical initial velocity is zero (because the jet comes out of the tennis ball container horizontally).  I place a large beaker on the ground to catch the stream of water; we measure the vertical distance from the top of the beaker to the hole in the container for use in the projectile problem.

When all is said and done, we calculated a 65 cm horizontal distance for the jet to travel.  While I filled up the container (and kept my finger carefully over the hole!), the students measured 65 cm along the ground and placed the large beaker at the correct spot.  I released the water....

... and the jet landed right in the beaker.  Physics works.

FOLLOW UP:  We all saw obviously that the speed of the water at the top was NOT zero.  But our prediction of a 1.7 m/s speed was accurate!  What's up?

We go back through the problem, this time not assuming that the speed at the top v₁ is zero.  Instead, we use the continuity principle ( A₁v₁ = A₂v₂ ) to get another equation relating the speeds at top and bottom.  We can solve the two-variable system for the speed at the bottom.  We need to measure the cross-sectional area of the hole, and the area of the tennis ball can.  We find that the can has a diameter of about 7 cm, while the hole only is 1 cm wide.  Therefore the area of the can is 49 times that of the hole.  We resolve for the speed... and find that we still get 1.7 m/s.  Any difference would be in a later, very insignificant digit.  So, the speed of the water at the top is negligible compared to the speed of the water jet.

Buoyant Force problem and demonstration with a crazy answer

A fluid mechanics problem I often assign is based on a problem from, I think, one of the older Serway editions.  It shows a beaker full of oil sitting on a platform scale.  A bar of iron is suspended in the oil from a rope which is attached to a spring scale.  The problem asks:

A 1.0 kg beaker containing 2.0 kg of oil (density = 916 kg/m3) rests on a platform scale. A 2.0 kg block of iron is suspended from a spring scale and is completely submerged in the oil.

(a) Which scale reading should be larger (or should they be the same)? Explain conceptually.

(b) When the iron is in equilibrium, what is the reading in the spring scale?

(c) When the iron is in equilibrium, what is the reading on the platform scale?

Most everyone gets the idea that the platform scale reads a bigger force -- after all, even without considering anything tricky (like fluid mechanics), the spring scale seems to read just the 20 N weight of the iron, while the platform scale seems to read the 30 N weight of the beaker/oil.  A bit more logic with buoyant forces convinces the students that the spring scale must read LESS THAN 20 N, because of the upward buoyant force on the iron.  No problem.

 

Part (b) is similar to a demonstration from class, and numerous example and practice problems in texts.  They know to draw a free body, calculate the buoyant force using Archimides' principle, and use the free body to calculate the tension in the string connected to the scale.  The only halfway tricky part is finding the volume of the iron, which is easily done once the density of iron is looked up.  The buoyant force is about 2 N in this case.

 

Part (c) is the part that causes trouble.  Most of the class, at least initially, says that the reading on the platform scale is just 30 N -- the weight of the oil plus the weight of the beaker.  Others get the right answer of 32 N, but for crazy reasons.  Some come to the conclusion that since the oil "lost" the 2 N buoyant force, that we must return these 2 N to the oil through the reading on the platform scale by conservation of force.  Others simply draw the buoyant force acting down directly on the beaker.  Many make no argument whatsoever, but just add in 2 N, presumably because their friends told them to and they couldn't quite explain it. 

 

I'm glad that so many students have the physics instincts to recognize that 30 N can't be right.  A few will say they made a lucky guess, but I consider such a guess good physics intuition.  However, only a very few students get the justification for why the platform scale reads 32 N.  Do you know?

 

It's Newton's Third Law.

 

The buoyant force is the upward force of the oil on the iron.  Therefore, there must be a downward force of the iron on the oil.  When we consider the oil-beaker system, the downward forces sum to 32 N, including the weights of the oil and beaker, and the third law companion force to the buoyant force.

 

Do you believe me?

 

My students don't, at least not if they didn't get the answer right in the first place. So I set up a similar situation.  The picture at the top (credit to Frederic Lamontagne, WFS class of '11, for the photography) shows a beaker containing a submerged aluminum weight, just like in the problem.  When I remove the weight from the water, the reading in the spring scale increases, but the balance scale goes out of balance!  I have to rebalance the scale to make up for the removal of that downward force of the aluminum on the water.  Since the spring scale reading increased by 0.2 N, I had to add about 20 g to the balance scale reading.  Physics works.

 

Circular motion demonstration: turntable

When I arrived at Woodberry Forest eleven years ago, I discovered several 1970s-vintage turntables in the storeroom.  I turned one into a nice, quantitative, circular motion demonstration.

In the picture you see the turntable, with five identical brass masses placed on top.  Before I start the demonstration, I show the class that a brass mass placed right near the center stays put when the table rotates; but the same mass placed near the outer edge flies off.  The goal of the demonstration is to predict the maximum radius for which the mass will not fly off.

The friction force on the brass mass acts toward the center of the rotation, and is a centripetal force equal to mv²/r.  The variable r represents the radius of the circle -- that's what we're looking for here. 

Problem is, the speed v itself depends on r, because a mass near the edge covers a larger distance in the same time as a mass near the center.  So we write v in terms of the period of revolution, T:  v = 2πr/T.  The period of revolution is measured with a stopwatch.

The friction force can be set equal to μFn.  The normal force on the mass is simply its weight.  The coefficient of (static) friction must be measured... but I can do that with a spring scale or a force probe.  I pull a stack of brass masses with the scale.  I divide the reading in the scale just as the masses start to move by the weight of the stack of masses. 

(Here the class can be asked whether the coefficient of friction for a single mass will be greater, less than, or the same as that for the stack of masses.  Answer:  the same, because we still have brass in contact with the same surface.  I only use a stack of masses to make the spring scale reading easier to obtain.)

So now we know everthing in the relevant equations necessary to solve for the radius r.  We find that radius to be in the neighborhood of 4-5 cm.  That's in the middle of the second mass, as measured from the center.

What does this mean experimentally?  When I turn the turntable on, the outer masses should fly off, but the first one or two should stay put.  And sure enough, that's what happens -- physics works.

Follow-up:  Next I replace the 20 g brass masses with 10 g brass masses from the same set.  Which ones should fly off now?  Answer: still, the masses more than 4-5 cm from the center will fly off, the others will stay put.  The coefficient of friction still hasn't changed, because we're using the same surfaces.  Mass cancels out when solving in variables, so the mass doesn't matter.  Sure enough, the 10 g masses fly off outside of 4-5 cm, as predicted.

GCJ

Two-body problem quantitative demonstration

Consider the classic two-body problem, in which a cart on a horizontal low-friction track is attached via rope-and-pulley to a hanging mass.  The cart has mass 340 g, and the hanginging mass is 100 g.  I release the cart, which speeds up.

Is the tension in the rope...
(A) greater than 1 N
(B) less than 1 N
(C) equal to 1 N?

"Equal," say a majority of the students -- because the rope is attached to the 100 g hanging mass, which has a weight of 1 N.

"Less than 1 N," the cleverer ones respond.  The hanging mass has a downward acceleration, because it is moving down and speeding up.  So the down forces must be GREATER THAN the up forces, meaning the tension is less than the 1 N weight.

"Let's see," I say.  The picture shows my cart with a Vernier force probe taped inelegantly on top.  The string is attached to the force probe, is run over a low-friction Pasco pulley, and connected to a 100 g hanging mass.  I tell Vernier's Logger Pro software to collect force probe data.  Before I let go of the cart, the probe reads 1.0 N.  As soon as I let go, the force probe's reading very obviously dips, to something like 0.7 - 0.8 N.  Looks like the cleverer ones were right.

Next, I use free body diagrams and Newton's second law to predict the acceleration of the cart and the tension in the rope -- I get 2.3 m/s² and 0.77 N.  Sure enough, that's just about what the force probe and a motion detector read.  (I get the acceration from the slope of a velocity-time graph, which I make with the vernier motion detector.)

Next question:  Instead of letting the cart go from rest, I give the cart a shove to the right, away from the rope. 

After I let go but as the cart is still moving to the right, is the tension...
(A) greater than 1 N
(B) less than 1 N
(C) equal to 1 N

"Greater than 1 N," say the majority.  "The hanging mass is now moving upward, so the up forces must be greater than the down forces."

"Nonsense," I say.  The mass is slowing down while it moves upward.  Slowing down means that acceleration and velocity are in opposite directions.  Thus, the acceleration must still be in the downward direction, and the tension must still be less than the weight.

In fact, the entire set of free bodies is unchanged from the previous problem -- the force that I pushed with doesn't act once I let go!  So, all calculations are the same, and the acceleration and tension should be unchanged.

Of course, I finish the class by doing the experiment -- sure enough, the tension and acceleration readings are the same as before.

GCJ

Pushing a truck and Newton's Second Law

At my Manhattan College Summer Institute, participants can earn graduate credit by submitting a "final project."  As one focus of the course is the development and use of quantitative demonstrations, an option for the final project is to create a couple quantitative demonstrations for use in participants' classes.

Alex Tisch, a colleague of mine starting this year, took this institute and submitted a clever but involved demonstration he had brainstormed.  Before I give the details, I ask you to BE CAREFUL with this one -- it involves a live, full-sized, moving pickup truck.  I generally scoff at the dire warnings that textbooks present with their lab ideas ("Drop a tennis ball and time its descent.  But WEAR GOGGLES!!!  And a cup!")... but don't do this particular experiment unless you're sure you can accomplish it safely.  If you're worried, have the class watch while a few faculty members carry it out.

Alex envisions his pickup truck in neutral on a level surface.  One person is in the driver's seat, just to keep the truck going straight.  One person sits in the back with a turkey baster full of paint; maybe another person is back there with a stopwatch.

A big strong person pushes the resting truck, trying to apply a relatively constant force.  Once he starts pushing, the painter drops one dollop of paint off of truck every second.

The class can now measure the distance from the starting point to each dollop of paint on the road.  From this data, a position-time graph can be made; analysis of this graph can lead to a calculation of the acceleration of the truck.  Knowing the mass of the truck + occupants, the average force applied by the pusher can be calculated using Newton's second law.

Neat, but not awesome yet.  Alex's big brainstorm was to MEASURE the force of the pusher while he's pushing.  He envisioned a bathroom scale with a nearby observer to record the force every second.  I suggested a force plate.  (A force plate is essentially a bathroom scale that connects to Vernier or Pasco data collection softward, so can make a live graph of force vs. time.)  Vernier's force plate even comes with handled attachments that make pushing pretty easy.

The force calculated from Fnet = ma can be compared to the force measured directly from the force plate.  Awesome.

(Picture from Leatherneck magazine.)

Using Red AND Green Lasers

As soon as we get back from spring break, I will be discussing "physical optics" -- double slits, diffraction gratings, thin films, and so on.  I always start with the double slit -- why light through a double slit produces a diffraction pattern, and the equation for the location of the minima and maxima. 

Here the best demonstration, I think, is to shine a red laser through a diffraction grating.  (No, I don't start with a real double slit... a diffraction grating produces sharp maxima which don't invite arguments about whether we're really seeing bright and dark spots; and, the mathematics for diffraction gratings are identical to those of double slits.  Later, after the class is comfortable predicting the location of the bright spots, I explain the qualitative differences between diffraction gratings, double slits, and single slits.)

This demonstration can start off quantitatively.  Using the known slit spacing of the diffraction grating, the wavelength of the laser (printed on there somewhere), and the distance to the screen, predict the distance between the first two bright spots using dsinθ = mλ. 

Next, ask the qualitative question: if the red laser is replaced by a green laser, what happens to the distance between bright spots?  Let the class argue it out amongst themselves:  the wavelength of green light is smaller than the wavelength of red light.  (Yes, you have to know that fact for the AP exam.)  By the above equation, reducing the wavelength while keeping d and m constant reduces sin θ as well.  As the sine of an angle decreases, the angle itself decreases.  So, the angle to the bright spot decreases, and the bright spots will be closer together. 

Note that it is legitimate to argue using the small angle approximation equation x = mλL/d.  You'd find that reducing the wavelength reduces x, the distance from the central maximum to the mth spot.

Then, if you have different diffraction gratings, you can ask how the distance between spots will change using these other gratings.

"But I don't have a green laser," you say.  I bought one a few years ago for $200.  But now both green and red lasers have come way down in price.  Search around.  I just bought a pack of twelve keychain red lasers for about $2 each; I found green lasers for about $10 each.  Considering its other uses in geometric optics (for ray tracing in an aquarium, say) or in an astronomy unit (to point out stars in a nighttime observing session), the $10 is worth it.