Question

The centre of gravity of a car is $0.62m$ above the ground. It can turn along a track which is $1.24m$ wide and has radius $r$. If the greatest speed at which the car can take the turn is $22.02m/s$, What is the value of $r$? (in $m$)

• A. $49.48$
  
• B. $58.36$
  
• C. $21.36$
  
• D. $96.36$
  

Answer 1

Answer: (A)

$49.48$

assume that critical velocity is $v$, if speed of car is more than $v$ then car will topple due to torque of centrifugal force $F_c$ around point $P$ , so balancing the torque at point $P$

${\Rightarrow F}_c\times 0.62=mg\times 0.62\Rightarrow \frac{m{v}^2}{R}\times 0.62=mg\times 0.62\Rightarrow \frac{m{v}^2}{R}=mg\Rightarrow R=\frac{v^2}{g}=\frac{{22.02}^2}{10}=49.48m$