Answer:[tex]\theta =57.37^{\circ}[/tex]
Explanation:
Given
Radius of curvature=20 m
Speed of cyclist=17.5 m/s
we know that velocity during a turn is given by
[tex]v=\sqrt{\frac{rg\left (tan\theta +\mu_s\right )}{1-\mu_stan\theta }}[/tex]
where v=speed
r=radius of curvature
[tex]\theta [/tex]=angle of banking
[tex]\mu_s [/tex]=coefficient of friction
g=acceleration due to gravity
here [tex]\mu_s=0[/tex]
thus
[tex]v=\sqrt{rgtan\theta }[/tex]
[tex]\tan\theta =\frac{v^2}{rg}[/tex]
[tex]tan\theta =\frac{17.5^2}{20\times 9.8}=1.562[/tex]
[tex]\theta =57.37^{\circ}[/tex]