# Car travelling around arc of circle

1. Jul 22, 2007

### boatman

This is not a homework problem, but a friend and I were discussing this and have not come to an agreement as yet.

Supposing that we are in a car which is traveling a left curve in the road which is of constant radius, and we still have some way to go before the road becomes straight again. While traversing this curved section of road the driver senses that the car is moving so fast that it is in danger of sliding off the road.

Here is the question; If the driver wishes to reduce the chance of the car sliding sideways the best thing to do is which of the following?

A. Immediately apply the brakes to reduce speed of the car whilst still in the curved part of the road.

B. Put the car's transmission in neutral and allow the car to slowly reduce speed due to air resistance and rolling resistance, but do not apply brakes.

The two ideas being discussed here are:

1) Braking would be good because a slower car is less likely to slide when traversing a turn in the road.

2) Braking would be bad because it would increase the force vector to a point which may exceed the static friction of the tire to the road, the tire would then slide on the road and when sliding would have less friction on the road than if not sliding.

2. Jul 22, 2007

### marcusl

Answer b is correct. People who drive in icy and snowy climates know that applying brakes in a turn when the tires are near their limit of adhesion can start a skid.

3. Jul 22, 2007

### gabee

Hmm, this is an interesting problem...

If you have two forces acting on the car--a static frictional force toward the center of the circular path of the car (the centripetal force) and a braking force applied backwards on the car (tangential to its direction), then the magnitude of the forces combined cannot exceed the maximum force of static friction.

$$\sqrt{F_{braking} + F_{centripetal}} = F_{s\,max}$$

$$\sqrt{m a_{braking} + \frac{mv^2}{r}} = \mu_{s\,max}mg$$

$$m a_{braking} + \frac{mv^2}{r} = \mu_{s\,max}^2m^2g^2$$

$$a_{braking} = \mu_{s\,max}^2mg^2 - \frac{v^2}{r}$$

This tells us at what deceleration the forces become too large and overcome the static frictional force (i.e. the car will slide).

But is this right? Plugging in values like $$\mu_{s\,max}$$ = 0.15, m = 1300 kg, v = 30 m/s, r = 10 m gives enormously large answers...with these numbers, a_braking = 2070.9 m/s^2. Of course this is only a very very rough interpretation so you can't really apply this to the real world, but it's strange that you should get such a large maximum deceleration for such a small, fast turn on ice....

Last edited: Jul 23, 2007
4. Jul 25, 2007

### gabee

EDIT: this is the correct solution, in the above post I forgot the Pythagorean theorem *doh*

Given a certain coefficient of friction $\mu_{s\,max}$, velocity v and circular path radius r such that

$$\mu_{s\,max} g > \frac{v^2}{r}$$ (if it isn't, you'd be sliding already!),

your maximum deceleration while braking is given by:

$$a_{max} = \sqrt{\left(\mu_{s\,max}g\right)^2 - (\frac{v^2}{r})^2}$$

And this solution gives reasonable answers...for $\mu_{s\,max}$ = 0.15 (rubber on ice), v = 5 m/s (about 11 mph) and r = 30 m, your maximum deceleration during braking is about 1.211 m/s^2. My car decelerates due to friction (on its own, no braking) at these low speeds at about 0.45 m/s^2.

I think I would definitely choose B.

Last edited: Jul 25, 2007