How Does Braking Compare to Turning in Avoiding Collisions?

[brotherbobby]

Homework Statement

A car driver going at some speed $v$ suddenly finds a wide wall at a distance $r$. Should he apply brakes or turn the car in a circle of radius $r$ to avoid hitting the wall?

Relevant Equations

Kinematic equations : (1) $v_x^2=v_{0x}^2+2a_{0x} (x-x_0)$ and (2) $x=x_0+v_{0x}t+\frac{1}{2}a_{0x}t^2$. Newton's 2nd Law : $\Sigma \vec F = m \vec a$. Kinetic frictional force supporting circular motion on a road of radius $r$: $f_k = \mu_k mg = \frac{mv^2}{r}$

1. For the car to apply brakes, we have $v^2=2ar⇒a=\frac{v^2}{2r}=μg\;\;[ma=μmg]⇒v=\sqrt{2μgr}$

2. For the car to go in a circle $\frac{mv^2}{r}=μmg\Rightarrow v=\sqrt{\mu gr}$.

We find from above that the maximum velocity $v$ possible to avoid a collision is $\sqrt{2}$ times as much if the car applied brakes than if it moved in a circle.

Which means that the car can afford to move with $\sqrt{2}$ times the speed if it applied brakes than if it moved in a circle in order to avoid collision.

Hence it would be better for it to apply brakes than take a circular turn. (My answer)

Am I correct?

 

[PeroK]

Looks good apart from braking and turning use static friction, not kinetic. Kinetic friction applies only when you skid.

 

[brotherbobby]

Looks good apart from braking and turning use static friction, not kinetic. Kinetic friction applies only when you skid.

Thank you, hadn't realized that. Always took friction to be kinetic when relative motion was there.

I can understand the use of static friction when you take a turn. The turn is what brings it on.

However, when you brake, don't your tyres go from rolling to skidding mode? Isn't it kinetic friction when you brake?

 

[PeroK]

However, when you brake, don't your tyres go from rolling to skidding mode? Isn't it kinetic friction when you brake?

No. Take a look at a car braking next time you get a chance. There's no skidding. Skidding makes a noise, burns and damages the rubber on your tyres and leaves rubber on the road. Braking reduces the speed at which the wheels turn and static friction causes the vehicle to slow without skidding.

That's the purpose of an anti-locking braking system. If you brake too hard, then the wheels can lock and skid. An ABS prevents that.

 

[brotherbobby]

No. Take a look at a car braking next time you get a chance. There's no skidding. Skidding makes a noise, burns and damages the rubber on your tyres and leaves rubber on the road. Braking reduces the speed at which the wheels turn and static friction causes the vehicle to slow without skidding.

That's the purpose of an anti-locking braking system. If you brake too hard, then the wheels can lock and skid. An ABS prevents that.

Fascinating point. Perhaps a bit too technical for physics, but thank you all the same.

How about a bicycle, just to simplify matters. On applying brakes on a bicycle, doesn't it stop rolling and go into skidding mode?

 

[PeroK]

Fascinating point. Perhaps a bit too technical for physics, but thank you all the same.

How about a bicycle, just to simplify matters. On applying brakes on a bicycle, doesn't it stop rolling and go into skidding mode?

Absolutely not! If you brake too hard on a bicycle, especially on a wet road, then you'll soon hear and feel if you start to skid. It's the same as a car, but without the ABS.

Why on Earth is this not physics? This is about as "physics" as it gets!

 

[brotherbobby]

Absolutely not! If you brake too hard on a bicycle, especially on a wet road, then you'll soon hear and feel if you start to skid. It's the same as a car, but without the ABS.

Why on Earth is this not physics? This is about as "physics" as it gets!

Aha, so let's clarify this point a bit further, if you don't mind.

Applying brakes, be it a cycle or a car, only reduces the speed at which it rolls without skidding (unless you brake too hard). In that case, however, the tyres are still purely rolling. We know that in the case of pure rolling, no static friction will act (point of contact at rest instantaneously). How then would static friction reduce the speed of the rolling vehicle?

 

[PeroK]

We know that in the case of pure rolling, no static friction will act (point of contact at rest instantaneously). How then would static friction reduce the speed of the rolling vehicle?

That only applies to constant velocity rolling. To accelerate or decelerate static friction provides the accelerating force.

You have the same thing when a car accelerates. Normally there is no skidding unless you try to go too fast, in which case the tyres may skid. This happens more easily on a wet, snowy or icy road, where the static friction is reduced.

Once you achieve constant velocity, there is no static friction involved.

 

[kuruman]

Here is some of the physics behind the motion of a car.

The accelerator increases the speed in the direction of motion defined by the selected gear. In physics terms it increases the magnitude of the velocity by providing a component of the acceleration in the same direction as the velocity.
The brake decreases the magnitude of the velocity by providing a component of the acceleration in a direction opposite to the velocity.
A turned steering wheel provides a centripetal acceleration, perpendicular to the velocity, that turns the car in a circular path the center of curvature being on the side towards which the wheel is turned.
A combination of the above affords the driver a continuous angle between acceleration and velocity that allows, for example, slowing down as you enter a turn, moving at constant speed at the halfway point and speeding up as you exit.

Specifically, the centripetal acceleration needed for turning is the result of static friction (no sliding) between the tires and the road surface. The magnitude of that force is inferred from the observed acceleration. For example, when a car of mass $m$ is moving with speed $v$ around a horizontal circle of radius $R$, the required force of static friction is $$f_s=\frac{mv^2}{R}.$$The car will skid if its speed is such that the static friction required for turning exceeds the maximum value $f_s^{max}=\mu_s mg$, where $\mu_s$ is the coefficient of static friction. Ice on the road reduces the coefficient of static friction which means reducing the speed when turning to avoid skidding.

 

[haruspex]

That only applies to constant velocity rolling.

... and then only if there is an external force acting on the vehicle to maintain that constant velocity. Even on a horizontal surface, a car needs static friction to maintain speed against wind resistance and rolling resistance.