Question

A car goes on a horizontal circular road of radius R. The speed increases at a constant rate $a=\frac{dv}{dt}$. The friction coefficient between the tyre and road is $\mu$. What is the speed at which the car will skid?

• A. $(\mu g-a)R^2$
• B. $\left[\right({\mu }^2{g}^2-a^2)R^2{]}^{\frac{1}{4}}$
• C. $\left[\right({\mu }^2{g}^2-a^2)R{]}^{\frac{1}{4}}$
• D. $(\mu g-a)R^4$

Answer 1

Answer: (B)

$\left[\right({\mu }^2{g}^2-a^2)R^2{]}^{\frac{1}{4}}$

For uniform circular motion, the resultant acceleration is
$a_R=\sqrt{a_N^2+a_t^2}$
$a_N=\frac{v^2}{R},a_t=\frac{dv}{dt}=a$
From FBD of car on the rough road
$f=m{a}_R$
$\mu N=m\sqrt{{\left(\frac{v^2}{R}\right)}^2+{\left(\frac{dv}{dt}\right)}^2}$
$\Rightarrow {\mu }^2{N}^2=m^2\left(\frac{v^4}{R^2}+a^2\right)$
$\Rightarrow v^4=\frac{{\mu }^2{N}^2{R}^2}{m^2}-a^2{R}^2$
$=\frac{{\mu }^2{m}^2{g}^2{R}^2}{m^2}-a^2{R}^2$
$={\mu }^2{g}^2{R}^2-a^2{R}^2$
$v=\left[\right({\mu }^2{g}^2-a^2)R^2{]}^{\frac{1}{4}}$.