Answer to Question #271431 in Mechanics | Relativity for أبرار

Question

A bike successfully made a turn around a circular path whose radius is 13 m. At what angle is the road banked if it’s velocity was 12.0 m/s?

Answer 1

Let's apply the Newton's Second Law of Motion in projections on $x$- and $y$-axis:

$$Nsin\theta=\dfrac{mv^2}{r},$$ $$Ncos\theta=mg.$$

Dividing the first equation by the second one, we get:

$$tan\theta=\dfrac{v^2}{gr},$$ $$\theta=tan^{-1}(\dfrac{v^2}{gr}),$$ $$\theta=tan^{-1}(\dfrac{(12.0\ \dfrac{m}{s})^2}{9.8\ \dfrac{m}{s^2}\times13\ m})=48.5^{\circ}.$$