The OJ Simpson Question: Magnitude and Direction of Accleration

Knowing what an acceleration vector means about motion is perhaps the biggest conceptual challenge in first semester physics.  No matter how many times I say "the direction of motion has nothing to do with the direction of acceleration," this misconception (among many others) remains.

I ask students to memorize:  

* Speeding up means acceleration is in the direction of motion.

* Slowing down means acceleration in the opposite direction of motion.

However, put these facts in the context of a velocity-time graph, or in the context of specific motion north and south, and heads explode.  And that's really all we can do -- ask about the meaning of acceleration in as many different contexts as possible until the class is sick of such questions.

In the first several kinematics assignments, I've displayed a position-time or velocity-time graph and asked for a description, in everyday language, of the represented motion.  In order to tease out the physical meaning of an acceleration vector, I switch up:  I present a description of motion, and ask students to make a velocity-time graph.

1.    In an alternate universe that still obeys our laws of physics, O.J. Simpson leaves a tollbooth in his white Bronco the morning after killing his wife.  Soon after, he sees a police officer flash his lights.  Hoping to get away, he slams the gas pedal to the floor, but then O.J. hits a concrete barrier and crashes.  (a) On the axes below, sketch a velocity-time graph of OJ’s motion.[1] 

---

[1] Sketch, according to the College Board’s course description, means to “draw a graph that illustrates key trends in a particular relationship, such as slope, curvature, intercept(s), or asymptote(s).  Numerical scaling or specific data points are not required in a sketch.”

Most of the class gets this essentially right on the first attempt; the rest get it after a quick conversation with a friend.  The real point of the problem comes next:

2. Describe in words the magnitude[1] and direction of O.J.’s acceleration as O.J. is leaving the tollbooth.

3. Describe in words the magnitude and direction of O.J.’s acceleration as O.J. is traveling along the road unmolested.

4. Describe in words the magnitude and direction of O.J.’s acceleration just after O.J. sees the officer.

5. Describe in words the magnitude and direction of O.J.’s acceleration while the Bronco slams into the wall.

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[1] “Magnitude” in this context means, how much acceleration does OJ have?  Answer relative to his acceleration at other parts of his motion.  No numbers are required, though you are welcome to make calculations if you so desire.

I can tell almost immediately upon reading these responses who understands acceleration, and who does not.  The ones who truly don't get it come in for a consultation, where we work on these concepts.  How can I tell?

Well, the response I'm expecting to parts (c) and (d) refers explicitly to the v-t graph and/or to definitions that we've learned:  "(c) When OJ sees the officer, the slope of the v-t graph is positive (a frontslash), so the acceleration is forward.  The slope of the v-t graph is steeper than when OJ calmly sped up from the tollbooth, so the acceleration has a larger magnitude here.  (d) When OJ crashes, the slope of the v-t graph is much steeper than anywhere else; so the magnitude of the acceleration is highest of all parts of the motion.  OJ's acceleration is backwards, because the car is still moving forward, but is slowing down."


The most common mistake is to state the direction of MOTION rather than of acceleration:  "(c) When OJ sees the officer, he speeds up rapidly.  So his acceleration is moving forward.  (d) When OJ crashes, he bounces back off the wall, so his acceleration is moving backward."  Anytime a student says that the acceleration is "moving," I know that he is conflating acceleration and velocity, so the answer is marked wrong -- yes, verbal skills are part of physics.


A less common mistake is to think that the acceleration must change if velocity changes.  "(c) After OJ sees the officer, his acceleration must change rapidly, because the accelerator pedal is on the floor.  (d) When he crashes, OJ's acceleration changes from a high value to zero."  No, constant acceleration means speeding up or slowing down; this student thinks acceleration must change in order for speed to change.


I do get about half the class writing clear, concise, and specific explanations that refer to the v-t graph or to the definition of acceleration.  I will show a fellow student's good explanation to someone who's struggling, to show the difference in the style of prose.  I'm teaching writing as much as I'm teaching physics, sure.  But the time I spend now demanding clear writing pays off tremendously later in the year, when an AP-style free response test requires one-minute justifications.


GCJ


P.S. Only about half of my class had ever heard of OJ Simpson.  That says something about pop culture in the post-internet era.  What it says, I have no idea.

Pole vaulting and the apparent weight at the equator

http://xkcd.com/852/

My first reaction when my colleague El Mole showed me the cartoon at the right was that 2 cm is way too significant a difference.

[Pause while you read the comic.  Good, ain't it?]

The fundamental principle is correct.  Because the linear speed of a point on the earth's surface is larger at the equator than at the poles, the "apparent weight" of a person, and thus the "apparent gravitational field g," will be smaller.  In advanced mechanics classes, Newton's Second Law is formulated in the rotating reference frame of the earth, and the effective g is reduced by a  centrifugal acceleration term equal to v²/r.

Why centrifugal and not centripetal?  In an INERTIAL reference frame, acceleration in circular motion is toward the center, i.e. centripetal.  "Inertial reference frame" means, in a sense, imagine that we observe the universe from a stationary camera placed above the rotating object.  Then the net force on the object is continually changing direction so as to push the object toward the circle's center.  However, if we instead observe the world from the eyes of the rotating object itself, then it seems like we are being pushed away from the center of the circle, i.e. in a centrifugal direction.  And if we consider a person rotating at the equator, it makes sense to consider the rotating reference frame; it's more interesting and useful to figure out what the rotating person feels than to figure out what would be observed by a stationary flying saucer over the north pole..

But to have the earth's rotation make a difference of nearly an inch?  An inch is significant in pole vaulting!  The last time the pole vaulting world record was broken, it was by Ukrainian Sergey Bubka over a ten year stretch from 1984-1994.  Each time he broke the previous record, he did so by just one or two centimeters.  The question that the comic begs is, should someone aspire to break records, should he compete exclusively in Ecuador rather than in, say, London?  Would the location make any difference at all?

I made my own order-of-magnitude estimate to check the comic.*  The gravitational field due to earth, without reference to rotation, is about 10 m/s². That term will be lessened by the "centrifugal" acceleration** v²/r. 

First, find v.  The radius of the earth is about 6000 km.  Multiply by 2π to get the circumference at the equator to be about 40,000 km, which is 40 million m.  We go around this circumference in 24 hrs = 80,000 s or so.  This gives a speed in the neighborhood of 500 m/s.

The "centrifugal" acceleration is then (500 m/s)²/(6,000,000 m) = 0.04 m/s². Compared to the gravitational g of 10 m/s², the centrifugal term is, say, four tenths of a percent.

Now, in the absolutlely simplest model, we might consider a pole vaulter as running at a fast horizontal speed, then launching himself as a projectile with that same speed.  The maximum vertical height the vaulter obtains is governed by vertical kinematics, with a known vertical launch velocity v_oy, final vertical velocity of zero, and acceleration of g downward.  This max height can be shown to be v_oy² / 2g.  Point is, the maximum height depends inversely on the first power of g.

So now we reach the end of the story:  what happens when we reduce g by a few tenths of a percent?  We increase the pole vaulter's maximum height by a few tenths of a percent as well. 

Bubka's record vault is 6.15 m.  Increasing that jump by four tenths of a percent would increase his vault height by... a couple of centimeters.  The comic is right. 

In practice, could Bubka have just gone to Indonesia to add two centimeters to his record?  Not exactly.  Four tenths of a percent is the difference in the apparent g between the pole and the equator.  If Bubka set his record at the 1994 Santa Claus's Merry Elves Invitational, then our analysis is sound.  But the farthest north city I can envision holding a major international track meet is, say, Oslo, Norway, at 60 degrees north latitude.  In Oslo, the effective g will be less than at the pole, but not an entire 0.4% less.  Since the linear speed of someone rotating on the globe drops off from equator to pole as the cosine of the latitude***, in Oslo the effective g is reduced by only 0.1%. 

The Oslo-Jakarta pole vault differential is more like 1.5 cm, not a full 2 cm.  Close enough.

Having read all this, my question for you is, who is the more complete nerd?  The xkcd author for carrying out this calculation and basing a comic strip on it, or me for checking the accuracy of the calculation?

GCJ
*This particular comic is generally quite good about its physics.  In fact, I'd be far more comfortable asking the xkcd writers to check me than vice versa.

** xkcd can explain this better than I ever could:  http://xkcd.com/123/

*** At the equator, cos (0) = 1, so his speed relative to Earth's center would be 500 m/s; at the pole, cos (90) = 0, so his speed is 0 m/s.  In Oslo, his speed is 500 m/s cos (60), or half his speed at the equator, and by the calculation above, the correction to g is one-fourth of the correction at the equator.