Question

A car goes on a horizontal circular road of radius
$\sqrt{4 + \div 3}$
m. m, the speed increasing at a constant rate of 2

m/s2. The friction coefficient between the road and the tyre is 0.4. Find the speed (in m/s) at which the car will skid. (g = 10 m/s2)​

Answer 1

As the car goes on a circular road with increasing speed, it has two accelerations - linear and centripetal - orthogonal to each other.

Hence the net acceleration of the car,

$\sf{\longrightarrow a=\sqrt{(a_l)^2+(a_c)^2}}$

where $\sf{a_l}$ is linear acceleration and $\sf{a_c}$ is centripetal acceleration.

The net horizontal force acting on the car is balanced by the frictional force when the car gets skidded.

$\longrightarrow\sf{f=ma}$

Since $\sf{f=\mu R}$ and $\sf{R=mg,}$

$\longrightarrow\sf{\mu mg=m\sqrt{(a_l)^2+(a_c)^2}}$

$\longrightarrow\sf{a_c=\sqrt{\mu^2g^2-(a_l)^2}}$

$\longrightarrow\sf{\dfrac{v^2}{r}=\sqrt{\mu^2g^2-(a_l)^2}}$

$\longrightarrow\sf{v=\left[\mu^2r^2g^2-r^2(a_l)^2\right]^{\frac{1}{4}}}$

In this question,

• $\sf{r=\sqrt{\dfrac{4}{3}}}$

• $\sf{a_l=2}$

• $\sf{\mu=0.4}$

• $\sf{g=10\ m\,s^{-2}}$

Then,

$\longrightarrow\sf{v=\left[\mu^2r^2g^2-r^2(a_l)^2\right]^{\frac{1}{4}}}$

$\longrightarrow\sf{v=\left[0.16\times\dfrac{4}{3}\times100-\dfrac{4}{3}\times4\right]^{\frac{1}{4}}}$

$\longrightarrow\underline{\underline{\sf{v=2\ m\,s^{-1}}}}$