Question

A bike successfully made a turn around a circular path whose radius is 13.0 m. At what angle is the road banked if its velocity was 12.0 m/s/ 

Answer 1

$r = 13m$

$v = 12\frac{m}{s}$

$m -\text{bike weight}$

$\alpha - \text{ banked angle}$

$\text{Forces acting on a bike:}$

$F_c-\text{the centripetal force}$

$F_c = ma$

$a =\frac{v^2}{r}$

$F_c= m\frac{v^2}{r}$

$N - \text{road reaction force}$

$F_g -\text{gravity}$

$F_g = mg$

$\text{Friction force is not specified in the problem statement}$

$\text{Vertical force projection}$

$N\cos\alpha-F_g=0$

$N\cos\alpha= mg(1)$

$\text{Horisontal force projection}$

$F_c-N\sin\alpha=0$

$N\sin\alpha=m\frac{v^2}{r}(2)$

$\text{From }(1) \text{ and }(2)$

$\tan \alpha= \frac{v^2}{rg}= \frac{12^2}{13*9.8}\approx1.1303$

$\arctan(1.1303)=48.5\degree$

$\text{Answer: } 48.5\degree$