Physics, 30.07.2019 01:30 pinkdolphinc01

Highway curves are marked with a suggested speed. if this speed is based on what would be safe in wet weather, estimate the radius of curvaure for a curve marked 50km/h? the coefficient of static friction of rubber on wet concrete is .7, the coefficient of kinetic friction of rubber on wet concrete is .5 .

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Radial (aka centripetal) acceleration keeps cars on an unbanked road in a curve. This acceleration is given by

$a_{rad}= \frac{v^2}{r}$

The force required to keep in going in a circle is given by Newton's second law, F=ma. Here a is the radial acceleration from the above equation, so we get a radial force:

$F_{rad}=m \frac{v^2}{r}$

This force is mediated by the friction force between the tires of the car and the road. For friction force we use the coefficient of friction and the normal force of the road on the car (which will also be the force of the car on the road). On an unbanked road this is just the weight of the car, mg.

The question now is, which coefficient of friction to use? It might surprise you to find that although the car is moving relative to the road, we should use the static coefficient. To see why this is the case, think about what the wheel's surface is doing as it meets the road. As a wheel rotates around and comes into contact with the road surface, it's bottom section is moving in a direction opposite to the car's forward movement. The point that actually touches the road is moving backward at the same speed that the car moves forward and so it is at rest with the street. If this were not the case, your tires would always be skidding on the road. (If you're having trouble seeing this, there are plenty of animations of this phenomenon on the internet). Once a car is skidding it is no longer in control and cannot turn like one that is not skidding.

So now the friction force is just

$F_{fric}= \mu_s F_n=\mu_smg$

We now have to set these equal and solve for r, the radius of the circle we are traveling in. Note that mass will cancel and so we don't need to worry about the mass of the cars involved:

$m \frac{v^2}{r}=\mu_smg \\ \\ \frac{v^2}{r}=\mu_sg \\ \\ \frac{v^2}{\mu_sg}=r \\ \\ \frac{(13.89m)^2}{0.7*9.8 \frac{m}{s^2} } =28.12m$

Where I have converted the speed to m/s for unit consistency.

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none, its a machine

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