Question

A cyclist riding the bicycle at a speed of $14\sqrt{3}m/s$ takes a turn around a circular road of radius $20\sqrt{3}m$ without skidding. What is his inclination to the vertical?

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${75}^o$

Answer 1

Answer: (C)

${60}^o$

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,

$NsinA=\frac{mv²}{r}$..........(i)

$NcosA=mg$.........(ii)

dividing equations (i) by (ii),

$tanA=\frac{v^2}{rg}$

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

so , $tanA=\frac{(14\surd 3{)}^2}{(20\surd 3\times 9.8)}$

$=\frac{196\times 3}{196\surd 3}$

$=\frac{196\surd 3}{196}$

$=\surd 3$

hence, $tanA=\surd 3=tan60°$

so, $A=60°$