Question

The curved section of a horizontal highway is a circular unbanked arc of radius 740m. If the coefficient of static friction between this roadway and typical tires is .40, what would be the maximum safe driving speed for this horizontal curved section of highway

Answer 1

Coefficient of static friction = tan(a) = 0.4
r = 740 m
g = 9.8 m/s²
$v \: = \: \sqrt{gr \tan( \alpha ) }$
v = √(9.8 × 740 × 0.4) m/s
v ≈ 53.85908 m/s

Answer 2

The maximum speed at which the car makes the turn without slipping is $\boxed{54\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$.

Further Explanation:

Given:

The coefficient of static fiction between the road and tyres is $0.40$.

The radius of curvature of the road is $740\,{\text{m}}$.

Concept:

As the car turn around the curved road, the centripetal force acting on the car should be balanced by the friction force acting on the car to avoid the slipping of the car on the road.

Write the force balancing equation for the forces acting on the car.

${F_{centripetal}} = {F_{frcition}}$      …… (I)

                                                                                           

The centripetal force acting on the car is given by.

${F_{centripetal}} = \dfrac{{m{v^2}}}{r}$  

Here, $m$ is the mass of the car, $v$ is the velocity of the car and $r$ is the radius of the curvature of path.

The friction force acting on the car is given by.

${F_{fricton}} = \mu mg$  

Here, $\mu$ is the static friction coefficient and $g$ is the acceleration due to gravity.

Substitute $\dfrac{{m{v^2}}}{r}$ for ${F_{centripetal}}$ and $\mu mg$ for ${F_{friction}}$ in equation (I).

$\begin{aligned}\frac{{m{v^2}}}{r} &= \mu mg\\ v&= \sqrt {\mu rg}\\\end{aligned}$  

Substitute the values in the above expression.

$\begin{aligned}v&= \sqrt {0.40 \times 740 \times 9.8}\\ &= \sqrt {2900.8}\\ &= 53.85\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&\approx 54\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$  

Thus, the maximum speed at which the car makes the turn without slipping is $\boxed{54\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$.

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Answer Details:

Grade: High School

Chapter: Circular motion

Subject: Physics

Keywords:  Curved section, circular, unbanked road, static friction, maximum safe driving, radius 740m, without slipping, centripetal force, friction force.