Question

A curve in a road has a radius of curvature of 112 m and is banked at a 14.35° angle.What is the speed, in m/s, at which a car can drive through the curve if the road is frictionless?

Answer 1

We will have that the speed will b given by:

`\begin{gathered} \sum F_x=Nsin(\theta)=(mv^2)/(r) \\ \\ and \\ \\ \sum F_y=Ncos(\theta)=mg \end{gathered}`

Now, we will have that:

`\begin{gathered} (sin(\theta))/(cos(\theta))=(v^2)/(rg)\Rightarrow tan(\theta)=(v^2)/(rg) \\ \\ \Rightarrow v=√(tan(\theta)rg) \end{gathered}`

Then, the velocity will be:

`\begin{gathered} v=√(tan(14.35)(112m)(9.8m/s^2))\Rightarrow v=16.75693972...m/s \\ \\ \Rightarrow v\approx16.76m/s \end{gathered}`

So, the safe velocity will be approximately 16.76 m/s.