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29 September 2010

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.


17 September 2010

Most of you missed the point of #1 last night...

I assigned the question above for homework one night last week.  Students were asked to justify their answer.  I'd say that 2/3 of the class got this wrong.  So, here's what I posted that night to the class folder.  Use the problem and/or the discussion, if you wish.

The graph shows something going 20 m in 4 s at constant speed, stopping for a couple of seconds, then continuing at the same speed for another 3 s.

That means an average speed of 5 m/s in the first 4 s. That's like 12 mph (to convert from m/s to mph, just mutiply by 2 and a bit.)

So, youall are telling me that a baby can crawl at 12 mph? Maybe young Jor-El... or, are you telling me that a car on the freeway is going 12 mph? He'll be rear-ended, or, hopefully, pulled over for being either drunk or stupid.

Point is, you do need to be aware of the meaning of speeds. 10 m/s isn't just some old number. 10 m/s is like 24 mph, like a car at Woodberry or on a small street. But only the fastest sprinters like Usain Bolt can run this fast.

13 September 2010

Email discussion group and an excellent conversation

I set up an internal email conference accessible only to my classes.  Each class has a separate folder, to which I post assignments and commentary.  Students are also encouraged to ask specific questions of each other and of me on the folder.

Getting the students to use this folder properly and usefully is difficult.  Most are too timid to post anything.  Then when someone does post, they say "I can't do #4.  Help."  I have to tell such a student to try again with a SPECIFIC question:  say, "On #4, I drew the free body diagram, but I can't figure out how to get the components written."  Specific questions can provoke excellent e-discussions.

I also use this conference myself to hammer home points that I don't want to make in class; or, points that I have made in class, but need reinforcement.  For example, yesterday when I graded homework I saw some ridiculous free body diagrams; posted the picture above with the subject line "What are the two things wrong with this free body for the boat?"  The body of the message included the statement, "First to post the answer gets candy!"

A few minutes later, I found a student's homework contradicting Newton's second law.  Instead of screaming, I merely posted to the folder, offering another piece of candy:  "What's wrong with the statement 'Since the resistive force exerted by the water is less than the forces exerted by the ropes, the boat can move in the direction of the ropes.' "

I got excellent answers.  To the latter question, student Peter Chen replied: 

"You don't necessarily need to exert greater forces than the resistive force to make it move. What you really need is the same quantity of forces on the opposite direction to the resistive force. If you don't fulfill this requirement, no matter how great the forces are, you cannot move the boat.  eg: Even if you exert 1 billion Newton of forces to both sides of the boat, which are perpendicular to the direction of the resistive force, you will never move the boat. Because the quantity of forces on the opposite direction to the resistive force is 0.

To the question about the free body diagram, I reposted James Moyler's answer with commentary:

1.You shouldn't write 600N on the free body diagram. It is a rope so it should say T1 and T2. You don't put quantitative values on a FBD.

2. F isn't enough. F of what? I believe it should be Fr for Force of Resistance.
In this particular problem, the resistive force was DEFINED as "F." So the plain old label "F" is okay here. But generally, Mr. Moyler is correct -- put a subscript or words on the label on a free body.

Another student, Frederic Lamontagne, followed up:

Firstly, there is no normal force when the object is on water. The force on water is Buoyant Force.  Secondly, [this force] would not be acting to the left (as it is indicated in the free body diagram) would act perpendicular to the water.
Absolutely correct.

Thing is, everyone in the class read that exchange.  Everyone in the class had done those problems just the day before, so the information was current and relevant.  The five minutes of my time it took to post and reply were worth 20 minutes of lecture time.

11 September 2010

Mail Time: Weird Fundamentals Quiz Questions and Answers

Georgian Michael Gray asks about a couple of questions I ask on a kinematics fundamentals quiz:

I am looking at your fundamentals quiz 1 and I don't quite understand what answer you are looking for in these two Questions.

1.       An object has a negative acceleration.  Explain briefly how to determine whether the object is speeding up, slowing down, or moving with constant speed.

If an object has a negative acceleration... how could it move with constant speed?

Exactly, Michael.  A correct student response says that a positive velocity means slowing down, a negative velocity means speeding up, and the object can NOT move at constant speed.*

(* Just so everyone understands, in this portion of the course we are explicitly discussing straight-line motion only.)

2.       Given the velocity of an object, how do you tell which direction that object is moving?

I mean, velocity has magnitude and direction... so if v is positive then you see which way you defined to be positive... and that way.

Yup... you're two for two.  I'd make it an even simpler response:  "The object moves in the direction of the velocity vector."
These seem like silly questions. I'm trying to get my students to articulate their understanding in a variety of situations. They think they know what acceleration means... then they'll say that something moving at a constant speed of 10 m/s must be falling near earth's surface. Or, they'll get so caught up in the mathematicalness of the words "positive" and "negative" that they think too hard... when the direction of motion IS the direction of velocity. The more different ways they're asked the same kind of question, the more they'll understand.

08 September 2010

Why I grade homework thoroughly

Background:  I work at a boys boarding school.  I'm on duty in one of the upperclass dorms every week or so.  This was the conversation I overheard last night, the night after the first day of classes.  The entire conversation took at least five minutes.  (Names have been changed to protect the innocent, and the silly...)

Plaxico:  So, Mr. Greer doesn't grade our homework, right?

Jim:  Yes he does, he collects it every three days.

Plaxico:  Every three days regularly, or approximately every three days?

Jim: I don't know.  He just said "Every third day."

James:  I think he said "Every third day or so."

Plaxico: *I* think he said "Every third day."  I'll go with that.

[Pause for thought.  much walking around by all.]

Plaxico:  Does he just collect the homeworkevery three days, or does he grade it?

Joe:  He looks at it every day.  He just collects it every three days.

Plaxico:  But does he grade it every day?

Joe:  I don't know... he doesn't pick it up every day.

Plaxico:  On the days when he doesn't pick it up, does he give us a grade?

John:  I'm not sure.

Plaxico:  Okay, when he picks it up, does he grade it for completion or correctness?

John: Why are you asking ME this?

Plaxico (increasingly frustrated and panicked):  Becuase I NEED TO KNOW.  Does he grade for completion or correctness?

John: I am not sure.

Plaxico (soliloquy, to audience, with hands in the air): Already!  It's just the first day, and already I'm confused.  I don't know what to do!

George:  So, Plaxico, why don't you just do it right?  Then you won't have to worry.

Plaxico: [Glares at George.]

Just letting you know, folks, setting up rules about homework IS but a game, and we teachers are merely players.  Set up the rules of the game so the work gets done.  Don't underestimate a student's willingness to play games.

04 September 2010

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.)