A Banked Turn - No Friction

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Conceptual:

A Car on a Level Surface All forces on the car are vertical, so no horizontal force can be generated.	A Car on a Banked Turn The normal force on the car due to the road is no longer vertical, so a component of the normal force acts in the horizontal direction.	The Centripetal Force The horizontal component of the normal force is shown in blue in the diagram above. This force can supply a centripetal force to turn the car.

If a car is on a level (unbanked) surface, the forces acting on the car are its weight, mg, pulling the car downward, and the normal force, N, due to the road, which pushes the car upward. Both of these forces act in the vertical direction and have no horizontal component. If there is no friction, there is no force that can supply the centripetal force required to make the car move in a circular path - there is no way that the car can turn.

On the other hand, if the car is on a banked turn, the normal force (which is always perpendicular to the road's surface) is no longer vertical. The normal force now has a horizontal component, and this component can act as the centripetal force on the car! The car will have to move with just the right speed so that it needs a centripetal force equal to this available force, but it could be done. Given just the right speed, a car could safely negotiate a banked curve even if the road is covered with perfectly smooth ice!

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Mathematical:

A free-body diagram for the car on the banked turn is shown at left. The banking angle between the road and the horizontal is (theta). The normal force, N, has been resolved into horizontal and vertical components (the blue vectors).

In the vertical direction there is no acceleration, and:

so:

In the horizontal direction:

Since F_net = F_centripetal:

Solving for v gives:

A car moving at velocity v will successfully round the curve!

Note: Your initial thought might have been to resolve the weight vector parallel and perpendicular to the road - after all, that is what we did for all of those lovely inclined plane problems, remember? The difference is that we expected the object to accelerate parallel to the incline, so it made sense to have the vectors pointing parallel and perpendicular to the incline. Here, though, the acceleration is horizontal - toward the center of the car's circular path - so it makes sense to resolve the vectors horizontally and vertically.

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Example 1:

A curve has a radius of 50 meters and a banking angle of 15^o. What is the ideal, or critical, speed (the speed for which no friction is required between the car's tires and the surface) for a car on this curve?

Solution:

• radius of curve, r = 50 m
• banking angle, = 15^o
• free-fall acceleration, g = 9.8 m/s²
• no friction speed, v = ?

From the free-body diagram for the car:

If the car has a speed of about 11 m/s, it can negotiate the curve without any friction.

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Example 2:

 A turn of radius 100 m is being designed for a speed of 25 m/s. At what angle should the turn be banked?

Solution:

• radius of turn, r = 100 m
• speed of car, v = 25 m/s
• free-fall acceleration, g = 9.8 m/s²
• bank angle, = ?

From the free-body diagram for the car:

So, the banking angle should be about 33^o. Whoops! That's a pretty extreme angle, even for a race track (see example 3).

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Example 3:

Talladega Motor Speedway in Alabama has turns with radius 1,100 ft. that are banked at 33^o (source). What is the "no friction" speed for a car on these turns?

Solution:

We can use the free-body diagram and derivation from example 1, and get:

So, a car going about 100 mph could negotiate the turns at Talladega without any friction between its tires and the pavement. During a NASCAR race, however, the cars go through the turns at about twice that speed...

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last update April 21, 2008 by JL Stanbrough