Centripetal Acceleration - A Banked Turn - No Friction

In the previous blog I talked about the classic problem of a car going around a banked frictionless curve.  There were two web sites I used, that I will place in links below to make it easy to follow along with this blog post.

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In order to drive a car around a curve, there must be a frictional force between the tires and the road, or the road must be banked. Consider a 1250 kg car traveling at a speed of 25.0 m/s around a curve with a radius of 175 m.

b.) If the curve is banked and the road surface is frictionless, what must be the angle (with respect to the horizontal) of the road surface?

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At http://mtl.math.uiuc.edu/~t-anders/blog/road_slope_problem.htm is a dynamic free body diagram of this problem. (This diagram relies on your browser being able to handle Java, most newer browsers can.)

At Batesville.k12.in.us web site is a static sketch and the algebraic solution of this problem, resulting in the equations:

  

Equation 1

And therefore the needed angle::

Equation 2

The physics of the solution is that the gravitational attraction on the car shown by vector GF must be balanced by the upward component of the normal force shown by vector GC. We know that vector GF and GC must have the same size because the car is not accelerating in the vertical direction. Notice that both vectors GF and GC are the same length, and will remain the same length throughout the problem.  This argument corresponds to the Batesville equation N cos(theta) = m g, where m and g are constants.

Because the road slopes down as you move from point B to O, on the frictionless surface the car would want to slide towards point O. The component of the force pushing it to the right is the horizontal component of the normal force and is shown by vector GE. When done at the correct speed with the correct bank in the road, the car will make it around the bend despite it being a frictionless surface. Vector GE is the force that is supplying the centripetal force that when things are balanced, will keep the car coming around the curve in the road without sliding up towards B or down towards O, vector GE providing the force for the circular motion:

  

For additional understanding, and to help students check the reasonableness of answers they have arrived at, I like to have students look at the ends of the domain of the situation as well as in the middle of the domain to see if the solutions they got make sense. I will move along that line of questioning as we look at the correct solution given above.

Here would be some questions to ask your students:

1) If you drag point B down to point A so that the surface of the road is horizontal, what will happen to the car as it tries to round the curve? (Ans: The car will continue in a straight line, thus slipping off the road in the direction of point B.)

2) If you drag point B up to where angle AOB is a 45 degree angle, and then you drive very slowly around the curve, what will happen? (Ans: Due to the slope of the curve, gravity will drag the car downward towards point O, causing you to slip off the road towards the inside edge of the curve.)

3) If you drag point B up to where angle AOB is a 45 degree angle, what is it in the free body diagram that displays the centripetal force that is going to bring you around the curve? (Ans: As point B moves up to 45 degrees, the horizontal component vector GE becomes longer, indicating there is more centripetal force, which is the force needed to make the car follow a circular path.)

4) If you drag point B down to point A so that the surface of the road is horizontal, what is it in the vector diagram that supports the answer you got in question 1? (Ans: All of the centripetal force shown by vector GE disappears, so there is no force present to make the car follow a curved path.)

5) Does equation 1 also support the same result that you answered for questions 1? (Ans: Yes, when theta = 0, tan(theta) = 0, and v = 0 becomes the velocity needed to stay on the road. The car can only stay on the road if it is stopped. If it is moving, with no traction supplying a horizontal force, and being on a frictionless surface the car can not go around a bend. )

6) As you increase angle AOB by giving more slope to the road, does equation 1 reflect that the velocity would need to be higher to make it around the bend? (Ans: Yes, as theta increases, tan(theta) increases, r g tan(theta) increases, and that increases v.

7) If you increase the velocity in equation 2, how does that change the angle theta in equation 2? (Ans: As v increases the argument of the tan function, v^2/(g R), increases. As the argument increases the value of arctan increases, so theta increases.)

8) Sketch me a graph of arctan(x) to explain why your statement in 7) is true.

9) When you drag point B upward until you get an angle of 80 degrees, you see vector GE becomes extremely long, explain why this is the case in terms of what is happening with the car. (Ans: As the road becomes very steep, the car a large tendency to want to slip to the inside, moving down towards point O. The faster the car goes, the more of tendency it has to slip outwards towards point B. To overcome the extreme slope in the road created by an 80 degree angle, the car has to go extremely fast to get a large enough component of the tendency to want to slip out to B to overcome the slope of the road and gravity trying to push it toward O.)

10) How fast does the car have to go to stay on the road if the angle of the bank becomes 90 degrees? Support you answer with common sense, with the vector diagram, and with equation 1. (Ans: At 90 degrees there is no vertical component to the normal force being supplied by the road. In the vertical direction only gravity is acting, and the car will fall towards point O. The vector diagram is saying the horizontal component of the normal force would have to become infinite (notice how long vector GE becomes). In effect the vector diagram as it is drawn becomes inapplicable at 90 degrees because there should be no vertical component vector GC, and we can't have an infinite normal force. Equation 1 agrees with these results, because as theta becomes 90 degrees, the tan(theta) becomes infinite, and the equation says we would need an infinite velocity to be able to stay on the road.)

11) When the carnival used to come to town they had a silo with a sloping base and vertical walls. As the viewer, you would go up a platform to the top of the silo, about 15 feet up, and look down into the silo. A motorcycle rider would start riding on the sloping base, increase his speed, and finally circle inside the silo on the vertical walls. How could he do that based on the previous ten answers? (Ans: His surface wasn't frictionless! I doubt that he would try this on a rainy day.)

Using Multiple Representations in Physics

The National Council of Teachers of Mathematics has placed a lot of emphasis on using multiple representations when teaching mathematics.  By that they mean students should be able to deal with mathematics by writing about it, talking about it, viewing it visually such as when graphing or as in geometric drawings, and by viewing it algebraically as in equations and functions. 

One of the classic problems in first year physics is to calculate the amount of bank needed to prevent cars from sliding off the highway when they round a curve at a given speed.  This problem can benefit from multiple representations.

This problem comes up in my AP Physics C course from Monterey Institute, it is problem 2 part b) on the end of Chapter 5 FRQ.  Here is how it is stated there:

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In order to drive a car around a curve, there must be a frictional force between the tires and the road, or the road must be banked. Consider a 1250 kg car traveling at a speed of 25.0 m/s around a curve with a radius of 175 m.

b.)  If the curve is banked and the road surface is frictionless, what must be the angle (with respect to the horizontal) of the road surface?

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If you search the Internet, you will also find a good drawing and a good explanation of the solution to this problem at the Batesville.k12.in.us web site , explaining why the relationship between the velocity, the radius of the curve, the acceleration of gravity, and the need angle will be given by:

        Equation 1

And therefore the needed angle will be given by:

       Equation 2

In addition to looking at the problem in terms of equation, I also want to look at this problem graphically in a free body diagram, and I want to suggest how we might relate these either by asking the students to write about their solutions, or to discuss their solutions in class.    When looking at the free body diagram, there is now software out there that will allow you make a dynamic free body diagram that can be manipulated to see how things change with the amount of bank in the road.

The equations themselves, and where they come from, are nicely covered at the above Batesville.k12 site.  By looking at this graphically we can go farther, getting an even deeper understanding of the physics that is happening in the problem, getting a feel for the underlying mechanics.

I have added a "dynamic sketch" of the situation at http://mtl.math.uiuc.edu/~t-anders/blog/road_slope_problem.htm .  (This diagram relies on your browser being able to handle Java, most newer browsers can.) 

Line segment BO represents the bank on the road.  If you drag point B you can change the angle of the bank on the road (angle AOB).  The car is located at point G, and vector GD represents the normal force exerted on the car by the road.  As you move point B, try and relate what you are looking at in the free body diagram, to what you are seeing in the Batesville equations, and to the description of the original problem.  See if what you are looking at visually makes sense.  Think about questions you might ask your students as they look at the multiple representations. 

In the next blog, I will talk about the underlying physics, and suggest some questions you might ask your students, either in class, or in a written exercise, to cause them to think about whether a particular solution really makes sense.

Finding Your Way Around the Winter Sky

There is nothing that is more awe inspiring to me than to go out on a very clear night and look at the stars, realizing that each of them is the equivalent of our sun. It puts things in perspective for me. I believe an appreciation for the magnificence and immensity of the universe is something every parent should share with their child and every physics teacher with their students.

In June of 2009 I did a blog posting regarding finding your way around the summer sky. As we move into December 2009, this posting is going to be about finding your way around the winter sky (if you live north of the equator).

In The Stargazer's Bible by W.S. Kals that I referred to in the June posting (I believe it is out of print), the author presents a mnemonic for finding your way around the winter sky, it reads:

"CAPtain, ALL DE RIGging SEEms PRoperly POLished."

This stands for the names of the six bright stars that make up the winter hexagon, Capella, Aldebaran, Rigel, Sirius, Procyon, and Pollux. In Figure 1 below, I've placed a wide view of the night sky to place the winter hexagon in context, I followed that by Figure 2, where I have enlarged just the winter hexagon.

Figure 1

Figure 2

The chart above is what the sky will look like at midnight near the middle of December, facing south. How high the pattern is in the sky will depend on your latitude, the curved arc you see in Figure 1 is the path the sun would have taken through the sky that day, known as the ecliptic. Knowing how high the sun was during the day will give you an idea how high up to look in the southern sky at night for these stars. This is a very large pattern in the sky spanning about 60 degrees of arc from one side of the hexagon to the other.

Each of the major six stars in it is associated with some of the best known constellations, the names of which are highlighted in yellow above. The three stars that form the belt in Orion, are one of the easiest objects to spot in the winter sky, and you can use those as a jumping off point to find the six very bright stars named above. From those six bright stars, you can more easily locate each of the constellations.

As the map shows, the constellations are near the ecliptic. I've highlighted the names of the constellations and the stars in the above map. Capella is the bright start in the constellation Auriga the Charioteer, Aldebaran is the bright star in the constellation Taurus the Bull, Rigel is the bright star in the constellation Orion the Hunter, Sirius is the bright star in the constellation Canis Major the Big Dog, Procyon is the bright star in the constellation Canis Minor the Small Dog, and Pollux is the bright star in Gemini the Twins.

I wish to acknowledge that the above star charts came from the "Free Star Charts" link of the web page for The Department of Physics and Astronomy at Stephen F. Austin State University .

Making use of Mathematica in Physics

It's never easy to know how much technology to use when teaching science and math. I'm coming over to the philosophy that an appropriate aspect of the process of thinking about that is to factor in what I do myself, when I'm working the problems sets for myself. I don't balance my check book without reaching in the drawer and getting out a calculator. If I have "modeling" to do for personal finances, I'll use a spreadsheet. If I'm working a physics problem set, I'll often turn the mathematics over to a piece of software called Mathematica.

When I'm trying to explain the physics to a student, I want to focus on the physics concepts, and not spend a lot of time going though algebra steps or graphing steps the student already knows, but we all know can take quite a bit of time. Mathematica allows me to use my time to focus on the physics.

Some Basics of Mathematica

To solve an equation, we use the command Solve, the syntax of which is

Solve[the equation written with a double equal sign, the variable to solve for]

For example the equation for the time it takes an object dropped from rest to fall 100 meters, under the influence of gravity would be:

The work above is the physics, solving for t is the mathematics, and that we can turn over to Mathematica by using the Solve command.

Here is this equation solved in Mathematica, what you input is shown in the line called In[1], the solution you get back is shown in the line called Out[1].

And here is a quick and easy plot of the distance an object falls under the influence of gravity in time t.

The command is Plot. One way to handle the syntax is to define the function using an underscore after the variable name on one line, and then use the Plot command in the next line. The syntax of the Plot command is:

Plot[ name of the function to plot, the domain you want it plotted over]

I did leave off the labels for the axis to make the syntax easier and to save time.

With a graph quickly in place, now you can talk about the physics that is going on in the graph. Does the object fall the same distance during each one second interval? And then on to another question about the physics, does the velocity change by the same amount during each one second interval?

Recall that velocity is the derivative of the position, so let's make a graph.

And what is happening to the acceleration, which is the second derivative of the position?

Or maybe you have a more complicated problem where you want to focus on conservation of momentum and not have the focus be on the mathematics. Two objects of mass m1 and m2, each 10 kg collide. The initial velocity of m1 was 20 m/s, and m2 was sitting at rest. After the collision m2 is observed to leave at an angle of 30 degrees above the horizontal with a velocity of 6 m/s. What will the velocity and angle of departure for m1 be after the collision.

From the conservation of momentum, the physics is:

Turning the algebra over to Mathematica looks like this:

I'd be interested in hearing if other physics teachers agree or disagree with these thoughts.

Linking in the MIT Physics Video Presentations

As you might have guessed, my "errant" student, mentioned in the previous blog, didn't email me back.  Given the amount of arm twisting I had to do last year to get assignments turned in, my prediction would have been "no reply".

In this blog, I'll finish off describing how I have linked the MIT content into my course and also explain why I did it the way I did.  If you do a Google search on (MIT open courseware physics) you will get in excess of 200 hits. 

My initial thoughts were that I would link to the "YouTube" version, certainly the students would relate to that.  However, what I found is that while the video was there, it still needed to be selected out of a group along the right side of the screen.  Of course I could drill down further, and get the URL to the specific lecture I wanted, but I didn't find the material at this site to be organized in the topic order that reflects the order concepts are typically taught in most physics courses.  I also found I  had to watch much of the video to identify the topics that were covered in the video and I needed to know those topics to link it correctly into my course as an enrichment activity.

I found it much more convenient to go directly to the MIT web site for the course I wanted.  At that site, the content was laid out in a way that made it easy to identify what the lecture was about, and then to slot it in as an appropriate link in my course (graphic of the MIT site below).

Doing it this way, when I click on one of the above links, I am able to give my students several choices in how they might view the lecture (shown below).

My course is based on the National Repository of Online Courses (NROC) and the free online HippoCampus content.  Because this MIT web site had the traditional naming of the topics, as well as the traditional ordering, it made it easy to link specific videos into my course at appropriate locations. 

My course as originally supplied, has its own set of videos, as well as examples, and practice problems.  After the material has been presented, students take a Self Check Quiz over what they learned, I use the Self Check Quiz solely as a practice area.  But following the Self Check Quiz is the real Quiz that becomes part of their grade.  I wanted my students to do the Self Check right after they had gone through the original material.  But then I felt that after the Self Check would be a good time to give them the opportunity to also see a second, but somewhat different presentation of the same material.  If they choose to watch the MIT lecture, my expectation is that this would help them on the Quiz. Below is a graphic of the slotting into my course:

Before leaving this topic, I also want to add a more general link to the MIT content.  Here they give more details of what all is on their site that pertains to physics.  I've circled in green the course title of the material I'm linking in as enhancement/supplemental materials to my AP Physics C - Mechanics course.

 

The MIT materials, as you should expect, is written, selected, and presented at a level that targets the average MIT student.  That makes it a notch above what an average AP level course would be taught at, and that's OK.  We still have students in our classes that can benefit from this as an enrichment/supplemental activity.