Banked Curve Formulas & Circular Motion

What is a Banked Curve?

Uniform circular motion driving problems can take place on unbanked (flat) or banked curves. So, what is a banked curve? A banked curve is a turn in which the driving surface is not horizontal. Banked curves in roads and racetracks are tilted inward (i.e. toward the center of the circle) in order to help vehicles get around the turn. Figure 1 shows an unbanked curve, while Figure 2 shows a banked track.

Figure 1: An unbanked curve is horizontal, so its normal force is vertical.

Figure 2: A banked curve is tilted, so its normal force is oblique.

In an unbanked curve, the centripetal force is only provided by the force of friction between the vehicle's tires and the road because the normal force (the reaction "push" force that is perpendicular to the surface) is vertical. In contrast, a banked curve provides a tilted normal force, so the normal force has a horizontal component that contributes to the centripetal force. Particularly when roads are icy (which decreases the force of friction between the vehicle's tires and the road), this horizontal component is crucial to having enough centripetal force to change the direction of the vehicle quickly enough for it to safely round the turn without slipping.

Formulas for Banking Circular Motion

The formulas (other than object height) for banking circular motion depend on whether friction is being included or ignored. In the formulas below, the Greek letter {eq}\mu {/eq} means the coefficient of friction, which is a measure of how strong the force of friction is between two particular surfaces. {eq}\mu {/eq} is unitless. When friction is being ignored, {eq}\mu = 0 {/eq} and the equations simplify accordingly.

With Friction

When the velocity of the object is more than the optimal velocity, the object will be prone to sliding up the incline; so the force of friction (which always opposes the direction of motion) will be down the incline. When the velocity of the object is less than the optimal velocity, the object will be prone to sliding down the incline; so the force of friction will be up the incline. Either way, friction will affect the net centripetal force and normal force experienced by the object. The force of gravity is always downward, whether the normal force is vertical or not.

In the equations below, the Greek letter {eq}\theta {/eq} means the angle between the banked surface and the horizontal g is the gravitational constant ({eq}9.8 \frac{m}{s^2} {/eq}), v is velocity in meters per second, h is the vertical height of the object in motion, s is the optimum speed in meters per second, d is how far diagonally up the track the object is, and r is the radius of the circle in meters.

H is vertical, r is horizontal, and d is oblique.

• banking angle formula: {eq}tan\theta = \frac{\frac{v^2}{rg} - \mu}{1 + \frac{\mu v^2}{rg}} {/eq} or {eq}\theta = tan^{-1}\left(\frac{\frac{v^2}{rg} - \mu}{1 + \frac{\mu v^2}{rg}}\right) {/eq}

• object height formula: {eq}h = dsin\theta {/eq}

• optimum speed formula: {eq}s = \sqrt{\frac{rg(tan\theta + \mu)}{1 - \mu tan\theta}} {/eq}

Without Friction

When friction is being ignored, treat {eq}\mu {/eq} as if it were 0. Doing so simplifies the equations significantly and makes the horizontal component of the normal force be the only source of the centripetal force. As before, {eq}\theta {/eq} is the banking angle, g is the gravitational constant ({eq}9.8 \frac{m}{s^2} {/eq}), v is velocity in meters per second, h is the vertical height of the object in motion, s is the optimum speed in meters per second, d is how far diagonally up the track the object is, and r is the radius of the circle in meters.

• banking angle formula: {eq}tan\theta = \frac{v^2}{rg} {/eq} or {eq}\theta = tan^{-1}\left(\frac{v^2}{rg}\right) {/eq}

• object height formula: {eq}h = dsin\theta {/eq}

• optimum speed formula: {eq}s = \sqrt{rgtan\theta} {/eq}

Banked Curves and Friction

Banked curve problems will either account for or ignore friction, so be sure to choose the correct version of the above formulas. For problems with friction, any necessary value of {eq}\mu {/eq} should be provided in the problem or in a reference table because that value changes based on the surfaces involved.

Remember that when the velocity of the object is more than the optimal velocity, the force of friction is down the incline. When the velocity of the object is less than the optimal velocity, the force of friction will be up the incline.

Banked Curve Formula Examples

Here are some examples of banked curve problems that use the equations described earlier in this article. These examples cover both with-friction and without-friction contexts.

Example 1 (banking angle without friction)

A highway engineer needs to figure out the banking angle for a curve with radius 210 m, at which cars of mass 2000 kg will be traveling at {eq}28 \frac{m}{s} {/eq} in icy conditions (i.e. friction will be negligible). What is the banking angle?

First, choose the relevant equation. The banking angle without friction is {eq}\theta = tan^{-1}\left(\frac{v^2}{rg}\right) {/eq}. Note that this equation does not contain m so, even though mass was given in the problem, it will not be used here.

Next, list the values needed for the equation: v = {eq}28 \frac{m}{s} {/eq}, r = 210 m, and g is the gravitational constant ({eq}9.8 \frac{m}{s^2} {/eq}).

Finally, insert those values into the equation: {eq}\theta = tan^{-1}\left(\frac{28^2}{210\cdot 9.8}\right) = 21^{\circ} {/eq}.

Example 2 (banking angle with friction)

What should the banking angle be for a racetrack curve with radius 0.20 km at which cars will be traveling at {eq}61 \frac{m}{s} {/eq} with a coefficient of friction of 0.70?

First, notice that r is not given in meters, so convert it: 0.20 km = 200 m.

Then, choose the relevant equation. The banking angle with friction is {eq}\theta = tan^{-1}\left(\frac{\frac{v^2}{rg} - \mu}{1 + \frac{\mu v^2}{rg}}\right) {/eq}.

Next, list the values needed for the equation: v = {eq}61 \frac{m}{s} {/eq}, r = 200 m, {eq}\mu {/eq} = 0.70, and g = {eq}9.8 \frac{m}{s^2} {/eq}.

Finally, insert those values into the equation: {eq}\theta = tan^{-1}\left(\frac{\frac{61^2}{200\cdot9.8} - 0.70}{1 + \frac{0.70\cdot61^2}{200\cdot9.8}}\right) = 39^{\circ} {/eq}.

Example 3 (object height without friction)

How high above the bottom of the track is a skateboarder who is going around a 45° banked track when the skateboarder is 12 m up the slope of the track? Ignore friction.

First, choose the relevant equation. The object height equation with or without friction is {eq}h = dsin\theta {/eq}.

Next, list the values needed for the equation: d = 12 m and {eq}\theta {/eq} = 45°.

Finally, insert those values into the equation: {eq}h = 12sin(45^{\circ}) {/eq} = 8.5 m.

Example 4 (object height with friction)

How high above the bottom edge of the road is a truck that is going around a 15° banked curve when the truck is 2 m up the slope of the road with a coefficient of friction of 0.75?

First, choose the relevant equation. The object height equation with or without friction is {eq}h= dsin\theta {/eq}. This equation does not contain {eq}\mu {/eq} so, even though coefficient of friction is given, it is not needed for this problem.

Next, list the values needed for the equation: d = 2 m and {eq}\theta {/eq} = 15°.

Finally, insert those values into the equation: {eq}h = 2sin(15^{\circ}) {/eq} = 0.5 m.

Example 5 (optimal speed without friction)