Question

A car moves on a circular track of radius $r$ in a horizontal plane. The speed of the car is increasing constantly at a rate of $\frac{dv}{dt}=a$. The coefficient of static friction between the road and the tyre is $\mu$. Find the speed at which the car will skid.

• A. ${\left[({\mu }^2{g}^2+a^2)r^2\right]}^{1/4}$
• B. $\sqrt{\mu gr}$
• C. ${\left[({\mu }^2{g}^2-a^2)r^2\right]}^{1/4}$
• D. $\sqrt{ar}$

Answer 1

The correct option is C ${\left[({\mu }^2{g}^2-a^2)r^2\right]}^{1/4}$

Since the speed of car is increasing continously, it has tangential acceleration given by:
$a_t=a=\frac{dv}{dt}$
At any time, the centripetal acceleration of car will be directed towards the centre,
$a_c=\frac{v^2}{r}$
Net acceleration is given by:



$a_T=\sqrt{a^2+{\left(\frac{v^2}{r}\right)}^2}...\left(i\right)$
For car to just skid, considering the just slipping condition where friction will act at its maximum value to provide the net force.
$\Rightarrow f=m{a}_T...\left(ii\right)$
or $f=\mu N=\mu mg...\left(iii\right)$
(From the equilibrium of car in vertical direction, $N=mg$)

From Eq $\left(i\right)$,$\left(ii\right)$ and $\left(iii\right)$
$m\sqrt{\left[a^2+{\left(\frac{v^2}{r}\right)}^2\right]}=\mu mg$
Squaring both sides,
$m^2\left[a^2+{\left(\frac{v^2}{r}\right)}^2\right]={\mu }^2{m}^2{g}^2$
or ${\left(\frac{v^2}{r}\right)}^2={\mu }^2{g}^2-a^2$
$\therefore v={\left[({\mu }^2{g}^2-a^2)r^2\right]}^{1/4}$