Question

What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?

Final Answer

$4.14^\circ$

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Video Transcript

This is College Physics Answers with Shaun Dychko.
This car is accelerating around this curve. It's going at constant speed, but it is never the less accelerating since its velocity is changing direction. So we have centripetal acceleration in other words and that's caused by this component of the normal force which is directed towards the center of the curved path. So this angle here is

*theta*and we can use normal force multiplied by sine of*theta*to find this component in the x direction or radial component. We know this is*theta*because you can imagine this line here parallel to the ramp and then you can imagine another line here which is horizontal and this would be*theta*because these would be interior opposite angles. With that being*theta*and this being 90 because this is a normal force which is perpendicular to the surface of the ramp, this dot here plus*theta*has to add up to 90. But this dot here plus whatever this angle is here, let's assume we don't know it's*theta*yet, also has to make 90 since this*F n y*is pointing straight up and that makes a 90 degree angle there. So if this dot plus*theta*makes 90 and this dot plus the angle in here also makes 90, that means this must equal that and so it is*theta*in there. All right. So the y component of the normal force is the normal force multiplied by cosine*theta*because the y component is the adjacent leg of this triangle. We know that that has to balance gravity. So that's vertical, this*F n y*is vertically upwards and so it has to balance the*m g*force of gravity downwards. That means*F n*cos*theta*is*m g*. We can solve for*F n*by dividing both sides by cos*theta*. That gives*F n*is*m g*over cos*theta*and the reason that's useful is we can substitute that back into this formula replacing*F n*in the*F n x*formula in order to well, eventually solve for*theta*. But let's take it one step at a time. So re-writing*F n x*and noticing that it is the centripetal force, in which case it must be*m v*squared over*r*because that's the formula for centripetal force. Now we can say that*m v*squared over*r*is this expression we had before for*F n x*which is*F n*sine*theta*. But now we'll replace*F n*with what we figured out in this part here in green,*m g*over cos*theta*times sine*theta**. Now sine**theta*over cos*theta*can be written as tan*theta*and we'll divide both sides by*m*as well. We have*v*squared over*r*equals*g*tan*theta*. Then divide both sides by*g*and you get tan*theta*is*v*squared over*r g*. That means*theta*is the inverse tangent of*v*squared over*r g*. So that's the inverse tangent of 105 kilometers per hour converted into meters per second and then square that, and divide by 1.2 kilometers converted into meters, times 9.8 meters per second squared. This all gives 4.14 degrees is the ideal banking angle for this particular speed and this radius of curvature. This assumes there is no friction by the way.