Answer to Question #141551 in Physical Chemistry for lauryann

Question

A car travelling around a banked (tilted) highway turn goes out of control, leaves the road, travels through the air, and lands on the ground a distance (d) from the launch-point, after dropping a certain distance (h) below that point. (A sports equivalent would be a ski jumper taking off at an angle θ above the horizontal.)
An investigator needs three pieces of information in order to make a model of the trajectory, mostly for the purpose of finding the initial speed of the car. Why are three pieces of data needed and not two, four, or some other number?

Answer 1

Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.

Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net F = ma. For uniform circular motion, the acceleration is the centripetal acceleration—a = a_c. Thus, the magnitude of centripetal force F_c is F_c = ma_c.

By using the expressions for centripetal acceleration a_c from \(a_c=\frac{v^2}{r};a_c=r\omega^2\\\), we get two expressions for the centripetal force F_c in terms of mass, velocity, angular velocity, and radius of curvature: \(\text{F}_c=m\frac{v^2}{r};\text{F}_c=mr\omega^2\\\).

You may use whichever expression for centripetal force is more convenient. Centripetal force F_c is always perpendicular to the path and pointing to the center of curvature, because a_c is perpendicular to the velocity and pointing to the center of curvature.

Note that if you solve the first expression for r, you get \(\displaystyle{r}=\frac{mv^2}{\text{F}_c}\\\).

This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.