Question

A car drives straight down toward the bottom of a valley and up the other side on a road whose bottom has a radius of curvature of 115 m. At the very bottom, the normal force on the driver is twice his weight. At what speed was the car traveling?

Short answer

The speed of the car is $33.588\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Step 1. Understanding the equilibrium conditions of motion

In the equilibrium equations of motion, the net force acting on the car in the vertical and horizontal directions can be equated to zero. With the help of these equations, the required quantity can be evaluated.

Step 2. Identification of given data 

The given data can be listed below as,

• The radius of curvature is, $r=115\mathrm{m}$.
• The normal force on the driver becomes twice the weight at the bottom which is, $F_{\mathrm{N}}=2w=2mg$
• The acceleration due to gravity is, $g=9.81\mathrm{m}/{\mathrm{s}}^2$.

Step 3. Determination of the speed of the car

The diagram of the forces on the car can be represented as,

Here, $F_{\mathrm{N}}$is the normal force, mgis the weight of the car, m is the mass of the car.

From the above figure, apply the equilibrium equation in the vertical direction and use Newton’s second law. The equation can be expressed as,

$\begin{array}{c}\sum F_y=m{a}_{\mathrm{c}}\\ F_{\mathrm{N}}-mg=m{a}_{\mathrm{c}}\\ F_{\mathrm{N}}-mg=\frac{m{v}^2}{r}\end{array}$

Here, $a_{\mathrm{c}}$is the centripetal acceleration of the car, v is the car’s speed.

Then, the normal force on the driver becomes twice the weight of the car at the bottom.

So, the above equation becomes,

$\begin{array}{c}2mg-mg=\frac{m{v}^2}{r}\\ mg=\frac{m{v}^2}{r}\\ v^2=gr\\ v=\sqrt{rg}\end{array}$

Substitute the values in the above equation.

$\begin{array}{c}v=\sqrt{115\mathrm{m}\times 9.81\mathrm{m}/{\mathrm{s}}^2}\\ v=\sqrt{1128.15}\mathrm{m}/\mathrm{s}\\ v=33.588\mathrm{m}/\mathrm{s}\end{array}$

Thus, the speed of the car is $33.588\mathrm{m}/\mathrm{s}$.