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?

Respuesta :

We will have that the speed will b given by:

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

Now, we will have that:

[tex]\begin{gathered} \frac{sin(\theta)}{cos(\theta)}=\frac{v^2}{rg}\Rightarrow tan(\theta)=\frac{v^2}{rg} \\ \\ \Rightarrow v=\sqrt{tan(\theta)rg} \end{gathered}[/tex]

Then, the velocity will be:

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

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