The coefficient of friction must be 0.196
Explanation:
For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:
[tex]\mu mg = m\frac{v^2}{r}[/tex]
where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:
[tex]\mu[/tex] is the coefficient of friction between the tires and the road
m is the mass of the car
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
v is the speed of the car
r is the radius of the curve
In this problem,
r = 750 m is the radius
[tex]v=85 mph \cdot \frac{1609}{3600}=38.0 m/s[/tex] is the speed
And solving for [tex]\mu[/tex], we find the coefficient of friction required to keep the car in circular motion:
[tex]\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196[/tex]
Learn more about circular motion:
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