Question

A cyclist riding at a speed of $14\sqrt{3}m{s}^{-1}$ takes a turn around a circular of radius $20\sqrt{3}$ m. what is his inclination to the vertical?

• A. ${30}^0$
• B. ${60}^0$
• C. ${45}^0$
• D. ${90}^0$

Answer 1

The correct option is B ${60}^0$

Cyclist rides the bicycle at speed, $v=14\surd 3m/s.$

radius of circular road , $r=20\surd 3m.$

Let inclination angle with vertical by cyclist is $A.$

a normal reaction acts between floor and bicycle.

then at equilibrium,

$NsinA=mv²/r...........\left(i\right)$

$NcosA=mg.........\left(ii\right)$

dividing equations (i) by (ii),

$tanA=v²/rg$

now put $v=14\surd 3,r=20\surd 3$ and $g=9.8m/s$

so , $tanA=\left(14\surd 3\right)²/(20\surd 3\times 9.8)$

$=196\times 3/196\surd 3$

$=196\surd 3/196$

$=\surd 3$

$hence,tanA=\surd 3=tan60°$

so, $A=60°$

Hence, the option $B$ is the correct answer.