Question

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding?

Answer 1

Answer:

Coefficient of static friction between the road and the car's tire is 0.81

Explanation:

It is given that,

Velocity of the car, v = 40 m/s

Radius of the curve, r = 200 m

We need to find the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding. Let it is equal to μ.

The centripetal force of the car is balanced by the force of friction as :

$\dfrac{mv^2}{r}=\mu mg$

$\mu=\dfrac{v^2}{rg}$

$\mu=\dfrac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}$

$\mu=0.81$

So, the coefficient of static friction between the road and the car's tire is 0.81 Hence, this is the required solution.

Answer 2

The minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding is 0.816.

Given the data in the question;

• Velocity of car; $v = 40.0m/s$

• Radius of curve; $r = 200 m$

For a not car not to slide of the road, the frictional force and centripetal force should balance each other.

That is; Frictional force = Centripetal force

$Frictional\ Force = uF = umg \\\\Centripetal\ Force = \frac{mv^2}{r}$

So,

$u_{min}mg = \frac{mv^2}{r} \\\\u_{min}g =\frac{v^2}{r}\\\\u_{min} =\frac{v^2}{gr}$

Where $u_{min}$ is the minimum coefficient friction, v is the velocity, r is the radius and g is acceleration due to gravity( $g = 9.8m/s^2$)

We substitute our values into the equation

$u_{min} = \frac{(40.0m/s)^2}{9.8m/s^2\ *\ 200m}\\\\ u_{min} = \frac{1600m^2/s^2}{1960m^2/s^2} \\\\u_{min} = 0.81$

Therefore, the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding is 0.816

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