Questions: What is the maximum speed (in m / s ) at which a car can round a corner with radius 25.0 m, if the coefficient of static friction between its tires and the road is 0.40?

Solution Steps

The maximum speed at which a car can round a corner is determined by the centripetal force, which is provided by the frictional force between the tires and the road. The frictional force can be calculated using the coefficient of static friction and the normal force.

The centripetal force $F_c$ required to keep the car moving in a circle of radius $r$ at speed $v$ is given by: $$F_c = \frac{mv^2}{r}$$ where $m$ is the mass of the car.

The maximum frictional force $F_f$ that can act on the car is given by: $$F_f = \mu_s F_n$$ where $\mu_s$ is the coefficient of static friction and $F_n$ is the normal force. For a car on a flat road, the normal force $F_n$ is equal to the weight of the car $mg$.

For the car to round the corner without slipping, the frictional force must be equal to or greater than the centripetal force: $$\mu_s mg = \frac{mv^2}{r}$$

Cancel the mass $m$ from both sides of the equation and solve for $v$: $$\mu_s g = \frac{v^2}{r}$$ $$v^2 = \mu_s g r$$ $$v = \sqrt{\mu_s g r}$$

Given:

• $\mu_s = 0.40$
• $g = 9.81 \, \mathrm{m/s^2}$
• $r = 25.0 \, \mathrm{m}$

Substitute these values into the equation: $$v = \sqrt{0.40 \times 9.81 \, \mathrm{m/s^2} \times 25.0 \, \mathrm{m}}$$ $$v = \sqrt{98.1}$$ $$v \approx 9.905 \, \mathrm{m/s}$$

Final Answer