Question

To test the performance of its tires, a car
travels along a perfectly flat (no banking) circular track of radius 368 m. The car increases
its speed at uniform rate of
at ≡d |v|dt = 3.3 m/s2
until the tires start to skid.
If the tires start to skid when the car reaches
a speed of 31.5 m/s, what is the coefficient of
static friction between the tires and the road?
The acceleration of gravity is 9.8 m/s2

Answer 1

The Newton's second law states

$$ma=F_{\rm net}$$

In this case

$$ma=F_{\rm fric}=\mu N=\mu mg$$

The total acceleration

$$a=\sqrt{a_n^2+a_{\tau}^2}=\sqrt{\left(\frac{v^2}{R}\right)^2+a_{\tau}^2}$$

Hence

$$a=\mu g$$

Finally, the coefficient of static friction

$$\mu=\frac{\sqrt{\left(\frac{v^2}{R}\right)^2+a_{\tau}^2}}{g}$$

$$=\frac{\sqrt{\left(\frac{31.5^2}{368}\right)^2+3.3^2}}{9.8}=0.435$$