Banked turn at constant speed and constant height with no slipping

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This article discusses a scenario/arrangement whose statics/dynamics/kinematics can be understood using the ideas of classical mechanics.
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This article discusses a scenario where a vehicle with wheels is making a banked turn with a banking angle of \theta without slipping and without any change in its altitude (i.e. it is not trying to go uphill or downhill).

Basic components of force diagram

The three candidate forces

Force (letter) Nature of force Condition for existence Magnitude Direction
mg gravitational force unconditional mg where m is the mass and g is the acceleration due to gravity vertically downward, hence an angle \theta with the normal to the incline and an angle (\pi/2) - \theta with the incline
N normal force unconditional adjusts so that there is no net acceleration perpendicular to the plane of the incline outward normal to the incline
f_s static friction no sliding adjusts so that there is no net acceleration along the incline it is along (up) the incline. Note that if the vehicle's speed were changing, then the friction force would have a component up the incline and a component opposite the direction of motion of the vehicle.


We choose to take components in the following three dimensions: the vertical direction, the horizontal direction that is perpendicular to the direction of motion of the vehicle, and the horizontal direction of motion of the vehicle.

It turns out that all the forces are in the plane spanned by the first two dimensions.

The vehicle has no acceleration in the vertical direction, so Newton's first law of motion gives us that:

\! mg = N \cos \theta \qquad (1.1)

The acceleration in the horizontal direction perpendicular to the motion of the vehicle is v^2/r where v is the speed and r is the radius of curvature at the point of the curve obtained by taking a cross section at that vertical height. If the cross section is circular, then r is the radius of the circle.

We thus get, by Newton's second law of motion, that:

\! N \sin \theta + f_s \cos \theta = mv^2/r \qquad (1.2)

We also have:

\! f_s \le \mu_sN \qquad (1.3)

Relation between limiting coefficient of static friction, radius of curvature, banking angle, and speed of turn

Plugging (1.1) into (1.2), we get:

\! f_s = m\frac{v^2/r - g \tan \theta}{\cos \theta} \qquad (2.1)

Plugging (1.1) into (1.3), we have:

\! f_s \le \mu_smg/\cos \theta \qquad (2.2)

Combining (2.1) and (2.2) we get:

\frac{v^2}{r} - g\tan \theta \le \mu_sg \qquad (2.3)

Rearranging, we get:

v \le \sqrt{rg(\tan \theta + \mu_s)}

In particular, in the frictionless case we would get v \le \sqrt{rg \tan \theta}.

Confusion with sliding motion along an inclined plane

The basic forces operating in the case of a banked turn are similar to the basic forces operating for sliding motion along an inclined plane, yet the analyses for these two situations are fairly different. We now examine how and why the analyses differ.

Direction of acceleration and no acceleration

In the case of sliding motion along an inclined plane, the block could potentially move up or down the incline. Thus, the direction of acceleration is potentially parallel to the incline in the up-down direction. There is no acceleration in the direction normal to the incline.

In the case of a banked turn, the body is moving along a curve at constant speed, hence its direction of acceleration is determined by the normal direction to the curve of motion. The curve of motion is (by assumption) at a constant height so the direction of acceleration is horizontal, perpendicular to the direction of motion (in our diagrammatic depiction, the direction of motion is perpendicular to the plane being depicted). There is no acceleration in the purely vertical direction.

Choice of axes to take components

Once we have the force equations as vector equations, it is possible to take coordinates (i.e. take components of forces) with any system of mutually perpendicular axes, and then set up and solve the equations. In most cases, though, there is a convenient choice of axes that makes the equation setup and solving as easy as possible. Usually, this is done as follows: one axis is the direction of possible acceleration, the second axis is an orthogonal direction where the forces have nonzero components but there is no net acceleration, and the third axis is the one orthogonal to both of these.

Based on our discussion above, we see that for the case of sliding motion along an inclined plane, the best choice of axes is along the plane (up/down) (the direction of possible acceleration), perpendicular to the plane (the direction of the normal force and a component of the gravitational force), and along the plane (horizontal) (no forces, no acceleration).

In the case of a banked turn at a constant speed and constant height, on the other hand, the best choice of axes is horizontal, perpendicular to direct of motion (direction of acceleration -- centripetal), vertical (direction of gravitational force, other forces), and horizontal, along motion (no force or acceleration).