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Teaching Physics Blog

Centripetal Acceleration - A Banked Turn - No Friction

More Blog Posts

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:

image

Equation 1

And therefore the needed angle::

image

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:

image

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.

image

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.)

Content related to this blog posting can be found at HippoCampus under Uniform Circular Motion, Centripetal Acceleration, and Racetrack - Simulation.

Links
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