Question

A vehicle takes a turn on an unbanked road,the radius of the turn being $42m$. If the coefficient of friction between the tyres and the road surface is $0.6$, then the car travelling at which of the following velocities will slip while taking the turn? (in m/s)

• A. $15.7$
  
• B. $17$
  
• C. $14$
  
• D. $15$
  

Answer 1

The correct option is B $17$

Given that $\mu =0.6$ and $R=42m$

assume that mass of vehicle is $m$

as the diagram(FBD) shows, if centrifugal acceleration $a_c$ applying on cycle is more than acceleration $a_f$ due to static friction then the cyclist will slip down while taking turn

$N=mg...\left(1\right)f=\mu N...\left(2\right)$

from above two equations

$\Rightarrow f=\mu mg=0.6\times m\times 10=6m$

$\Rightarrow a_f=\frac{f}{m}=\frac{6m}{m}=6m/s^2$

and $\Rightarrow F=m\frac{v^2}{R}$

$\Rightarrow a_c=\frac{F}{m}=\frac{\left(\frac{m{v}^2}{R}\right)}{m}=\frac{v^2}{R}=\frac{v^2}{42}m/s^2$

if$a_c$$\ge$$a_f$ so the cyclist will slip down

$\Rightarrow \frac{v^2}{42}\ge 6\Rightarrow v^2\ge 252\Rightarrow v\ge 15.87m/s$

so the answer should be $\left(B\right)$.