A cyclist riding the bicycle at a speed of $14\sqrt 3 ms^{-1}$takes a turn around a circular road of radius $20\sqrt 3 $ m without skidding. Given $g = 9.8 ms^{-2},$ what is his inclination to the vertical ....... $^o$
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(d)$\theta = {\tan ^{ - 1}}\left( {\frac{{{v^2}}}{{rg}}} \right) = {\tan ^{ - 1}}\left[ {\frac{{{{(14\sqrt 3 )}^2}}}{{20\sqrt 3 \times 9.8}}} \right] = {\tan ^{ - 1}}[\sqrt 3 ]$$ = 60^\circ $
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