A flat (unbanked) curve on a highway has a radius of 125 m. A car rounds the curve at a speed of 25.0 m/s. Consider an instant when the car is on the curve for which its center is... A flat (unbanked) curve on a highway has a radius of 125 m. A car rounds the curve at a speed of 25.0 m/s. Consider an instant when the car is on the curve for which its center is to the left of the car. (a) Draw a labeled free-body diagram of the car (based on the rear-view of the car). Indicate on your free-body diagram the force that allows the car to round the curve. (b) Calculate the centripetal acceleration of the car. (c) What is the minimum coefficient of static friction that will just prevent sideway skidding? (d) Suppose that the highway is icy and the coefficient of static friction between the tires and highway is one-third of what you found in part (c). What should be the maximum speed of the car so that it can round the curve safely? (e) As the car rounds the curve, does the centripetal force do work on the car? Explain your answer using the definition of work done. Read more

Steps to Solve

1. Draw the free-body diagram

From the rear view, the forces acting on the car are: - Weight ($F_g$) acting downwards - Normal force ($F_N$) acting upwards - Static friction ($F_f$) acting to the left (since the center of the curve is to the left). This is the force that allows the car to round the curve.

2. Calculate the centripetal acceleration

The formula for centripetal acceleration ($a_c$) is: $$a_c = \frac{v^2}{r}$$ where $v$ is the speed and $r$ is the radius of the curve. Given $v = 25.0 , \text{m/s}$ and $r = 125 , \text{m}$, we can plug in these values: $$a_c = \frac{(25.0 , \text{m/s})^2}{125 , \text{m}} = \frac{625 , \text{m}^2/\text{s}^2}{125 , \text{m}} = 5.00 , \text{m/s}^2$$

3. Determine the minimum coefficient of static friction

The centripetal force is provided by the force of static friction: $$F_c = F_f$$ $$ma_c = \mu_s F_N$$ Since the road is flat, $F_N = mg$. Therefore: $$ma_c = \mu_s mg$$ $$a_c = \mu_s g$$ $$\mu_s = \frac{a_c}{g}$$ Using $a_c = 5.00 , \text{m/s}^2$ and $g = 9.81 , \text{m/s}^2$: $$\mu_s = \frac{5.00 , \text{m/s}^2}{9.81 , \text{m/s}^2} = 0.510$$

4. Calculate the maximum safe speed under icy conditions

The new coefficient of static friction is $\mu_{s, \text{new}} = \frac{1}{3} \mu_s$. Therefore: $$\mu_{s, \text{new}} = \frac{0.510}{3} = 0.170$$ We know that $a_c = \mu_s g = \frac{v^2}{r}$, so $v = \sqrt{\mu_s g r}$. Therefore: $$v_{max} = \sqrt{\mu_{s, \text{new}} g r} = \sqrt{0.170 \times 9.81 , \text{m/s}^2 \times 125 , \text{m}} = \sqrt{208.6125 , \text{m}^2/\text{s}^2} = 14.4 , \text{m/s}$$

5. Determine if the centripetal force does work

Work is defined as $W = Fd\cos\theta$, where $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between the force and displacement vectors. In circular motion, the centripetal force is always directed towards the center of the circle, while the displacement is tangential to the circle. Therefore, the angle between the centripetal force and the displacement is always 90 degrees. Since $\cos(90^\circ) = 0$, the work done by the centripetal force is zero.