What Minimum Radius Prevents Cars from Leaving the Road on a Steep Hill?

[bfr]

[SOLVED] Centripetal Force at an Angle

Homework Statement

The design of a new road includes a straight stretch that is horizontal and flat but that suddenly dips down a steep hill at 22 degrees. The transition should be rounded with what minimum radius so that cars traveling 90 km/h will not leave the road?

Homework Equations

v=(2*pi*R)/T
F=ma
a=(v^2)/R

The Attempt at a Solution

90 km/hr=25 m/s. I tried setting up an equation where the centripetal force equaled the force of gravity:(25^2)/r=9.8 sin 22 , but that gives me a radius of approximately 170.25m, when the correct answer is 63.8m.

EDIT: Solved. (25^2)/r=9.8

 

[Orodruin]

To flesh out OP’s solution:

In the rounded segment the car will travel along a vertical circle. In order to do so, a centripetal accceleration of $mv^2/r$ is required. In the limiting case where the car just avoids lifting off, the only force supplying acceleration in the radial direction is the radial component of the gravitational force, which equals $mg\cos\theta$. It is therefore required that $v^2/r < g\cos\theta$ or, equivalently, $r > v^2/g\cos\theta$. This requirement is strictest for the largest value of $\theta$ in the problem, ie, $\theta = 22^\circ$. This gives $r > 68.8$ m.*

This assumes traveling at 25 m/s throughout the rounded segment. Assuming acceleration through the tangential component of gravity makes this somewhat worse, requiring larger radius of curvature. It is unclear why the problem author has only considered the requirement at the top of the segment.