Question

A roller coaster is designed such that riders experience "weightlessness" as they go round the top of a hill whose radius of curvature is 20 m. Find the speed of the car at the top of the hill.

Answer 1

Given data

1. The radius of curvature of the hill is $R=20m.$
2. The normal force on the rider is zero, at the top of the hill.

Centripetal force and normal force

1. The centripetal force on a body moving with a velocity $v$ in a circular path of the radius $R$ is defined by the form,$F_c=\frac{M{v}^2}{R}$.
2. Due to the weight a downward gravitational force is exerted on a body.
3. A normal upward force is acting on the roller coaster.

Finding the speed

In this case, three kinds of force are exerted on the rider. These are gravitational force (weight), centripetal force, and normal force.

At a particular point, the equation of force is,

$Mg-N=F_c$, where, $Mg$ is the weight, $F_c$ is the centripetal force, and $N$ is the normal force.

At the top of the hill, $N=0$.

So,

$$\begin{gathered} Mg=F_c=\frac{M{v}^2}{R} \\ or{v}^2=Rg \\ orv=\sqrt{20\times 9.8} \\ orv=14 \end{gathered}$$

Therefore, the speed of the car at the top of the hill $14m/s.$