You are correct that centripetal force does not affect speed. So let us consider tangential speed, which is the speed the turning car would have if centripetal force were removed. Linear speed = tangential speed = distance / time = (2 * pi * radius) / time. (To simplify, assume the car goes through one full circle at uniform speed.) The linear speed of the center of the car is proportional to the radius of the circle. Likewise the linear speed of each wheel is proportional to the radius of the respective circles they traverse.
When the front wheels turn, they place not only themselves, but also the center of the car, on the circumferences of circles of differing radii.
Each wheel and the center of the car will be moving at different linear speeds, and none of them will necessarily be at the same speed they had when they were on a straight line. Also, the rear wheels will be on different radii from the front wheels, from the center, and from each other.
If there were some way for you to measure the speed of the car's center independently of the speed of each wheel, you could intentionally keep it at 50 km/h by adjusting the rotation of the wheels. But if the wheels connected to the power train kept turning at the original number of revolutions, the center of the car would not continue at an instantaneous speed of 50 km/h, UNLESS you carefully chose to place it on the radius of a circle with speed component of the velocity vector equal to 50 km/h.
Alternatively, you could equate linear average speed with circumferential average speed:
v = r * w, where "v" is the linear speed, "r" is the radius of a circle, and "w" is angular velocity.
Incidentally, without friction there would be no centripetal force and neither the front wheels nor the center of the car could move in circles, as warhead pointed out.
They don't want to get in front of the rolling wheels -- at least, not intrinsically.
When you are turning, your front wheels stop going forwards so much, and start going sideways. As Newton taught us, a great way to think about such things is that an object in motion will tend to keep moving however it is moving. There are sideways forces on the front wheels created by the friction of the wheels with the ground, slowing down its forward speed, and transferring that into sideways speed. So that's how "objects keep moving" works out for the front side of your car: forces act on it to stop its forward motion and create sideways motion. Now let's examine how this principle works for the backside of your car.
Normally, there is an "easy direction" for the back of your car to move -- in the direction of the wheels -- and a "hard direction" to move -- perpendicular to that direction. This creates side-forces which keep your rear wheels "following" your front wheels. They have a less-steep turn than the front wheels but also they start acting a few meters before the front wheels do, so they tend to take turns tighter, which is why it's harder, e.g., to forwards-perpendicular-park on your right rather than your left (if you've just been succeeding unconsciously or always back-in: when you forwards-perpendicular-park to the right, you usually have to swerve left pre-emptively to give your rear wheels room to take the turn, because taking that turn tighter than the front wheels would otherwise mean hitting the car occupying the spot 1 before the one you want to park in).
When you pull the handbrake, or your rear wheels hit ice, hydroplane, or go airborne, you set these two directions on equal footing: your rear wheels then are not so well "guided" by these side forces, and mostly just want to keep going forward, with a slight pull induced by the front tires. Even worse, as @BillN notes, they now have a lower coefficient of friction than those stabilizing forces did, so slowing them down to rest becomes harder.
In the worst case, the front wheels stop their forward motion entirely while the rear wheels are still sliding. The ensuing driver panic usually means that they stop doing whatever they were doing, and brake: the rear wheels will then pull the car entirely around until the front wheels stop the slide. This is, you guessed it, why it turns 180 degrees and not some other number like 270 or such.[1]
In the case of a handbrake turn, the front wheels follow a curve that is so tight that would cause the back wheels, since they take the turn tighter, to hit the curb or worse. Allowing the rear wheels to slide in the forward direction of motion means that they instead "kick out" (keep going in a straight line rather than following the curve of the front wheels), effectively pivoting the car about its front wheels. So the mechanism is not special: it is just that things which have forward momentum need forces to guide them off of it, and the strong forces which keep the backside of the car guided towards the front have been exchanged for weaker forces which slow down the backside of the car without having a preferred direction.
- Yes, this happens on snow or when the rear wheels are underinflated and hydroplane. As an anecdote: When I was practicing for my driver's license we did some snow driving, because my Dad wanted me to be prepared for anything. I actually saw someone start fishtailing on slushy snow and slowed way the hell down myself, and the event indeed ended with traffic stopped and the poor dude was faced backwards, and we were looking face-to-face. He was ghost-white and I was pretty awestruck by the dangers of driving.
Best Answer
Let $v_c$ be the speed of the car (and thus the wheels) and $v_r$ the speed of the ramp.
Assuming the wheels roll without slipping, then (where $\omega$ denotes angular velocity):
$$v_c=R\omega_0$$
When the car starts moving up the ramp the relative speed of the car w.r.t. the ramp is:
$$v_{rel}=v_c-v_r$$
Still assuming the wheels aren't slipping as the car moves up the ramp, this causes a friction force $F_f$ to arise and the resulting torque will reduce the angular velocity $\omega_0$ of the wheel:
$$F_f\times R=I\frac{d\omega}{dt}$$ And: $$\omega(t)=\omega_0-\frac{d\omega}{dt}t$$
'CountTo10's' question "Is the car front or rear wheel drive?" is relevant here as with a rear wheel drive, the front wheels will only be free wheeling and their angular deceleration will not significantly affect the car's speed. But it will affect the car's speed once the rear wheels are in contact with the ramp.
The friction force $F_f$ is usually modelled simply as:
$$F_f=\mu F_N$$
Where $\mu$ is a friction coefficient (here the 'rolling resistance') and $F_N$ the Normal force.
As long as the car isn't fully on the ramp yet, $F_N$ is a bit hard to model but is approximately:
$$F_N\approx\frac14mg\sin\alpha$$
Once the car is fully on the ramp it will be subject to gravitational deceleration (regardless of any braking force applied, of course) of:
$$a=-g\sin\alpha$$ So that relative speed (again, without accounting for braking applied by the driver) is given by:
$$v_{rel}(t)=(v_c-v_r)-g(\sin\alpha) t$$
The reasoning remains much the same. Call the speed of the conveyor belt $v_{belt}$ and assume its vector points in the same direction as $v_c$ and $v_r$, then:
$$v_{rel}=v_c-v_r-v_{belt}$$
In the 'special case' where $v_c=v_r+v_{belt}\implies v_{belt}=v_c-v_r$, no friction force would arise. That would kind of be the 'ideal' way of 'parking' a car inside a truck, as it would be the same as driving uphill for a short distance.