Question

To test the performance of its tires, a car
travels along a perfectly flat (no banking) cir-
cular track of radius 564 m. The car increases
its speed at uniform rate of
a 101
al = 410 = 3.62 m/s2
dt
until the tires start to skid.
If the tires start to skid when the car reaches
a speed of 35 m/s, what is the coefficient of
static frictionsbetween the tires and the road?

Answer 1

The coefficient of static friction is 0.222

Explanation:

In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

$\mu mg=m\frac{v^2}{r}$

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where

$\mu$ is the coefficient of static friction

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the track

In this problem, we have:

r = 564 m

v = 35 m/s

$g=9.8 m/s^2$

And re-arranging the equation for $\mu$, we can find the coefficient of static friction:

$\mu = \frac{v^2}{gr}=\frac{35^2}{(9.8)(564)}=0.222$

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