accelerationcentripetal-force
When rounding a flat curve, the centripetal force is provided by the frictional force. I learned that when a car rounds a flat curve with a fixed radius $R$, it can be able to make a turn as long it is moving with a velocity equal to or less than $v_{max}$
$$v_{max}=\mu mg$$
However, in the case of a frictionless banked track with a fixed radius R, a car can only make the turn if it has a particular velocity… could someone please explain why that is the case
You are right in thinking that the car's acceleration is what keeps it in place, but it is important to remember that an object moving at a constant speed in a circle is accelerating (despite not speeding up). The reason for this is that acceleration is defined as a "change in velocity," and velocity is a vector quantity (i.e. it has magnitude and direction). Therefore, the car's centripetal acceleration must have a force corresponding to it through Newton's Second Law (that force is the radial component of the normal force). And by requiring the position of the car to behave in certain ways, we can calculate that force exactly.
So yes, the acceleration is "adding" to the normal force. The sloped track must apply a greater force in the radial direction than it would if the car were not moving because in that situation the car would begin moving along the radial direction (sliding down the ramp). Likewise, it is not moving in the vertical direction (sliding down the ramp) so the normal force in that direction must be greater as well (to exactly cancel its weight). Since the normal force is the vector sum of its radial and vertical components, you can determine the increased normal force.
The truth is that by stating the problem this way, we have required certain constraints on the motion of the car from which we can deduce the forces acting on it by using Newton's Second Law.
Frictional force opposes sliding motion, basically. Car tires produce centripetal force by changing their angle relative to the rest of the car's orientation. The tires do not slide in the direction of the tires' orientation: they roll. Friction in this direction rotates the tires, or if the engine is applying force to the wheels during the turn, friction prevents the tires from "burning rubber", and pushes the car in this direction.
Meanwhile, motion in the direction of the rest of the car's orientation is opposed by friction only to the extent that it is not motion in the direction of the tires' orientation. The velocity vector corresponding to the rest of the car's orientation can be understood in terms of these two orthogonal components. The component corresponding to the tires' orientation is basically not subject to friction for our purposes (ignoring whether one's foot is on the gas pedal). The component that does not correspond to that other component is orthogonal and opposed by centripetal friction.
Based on GIF by Droidmakr.
If tires' orientation $=$ rest of the car's orientation, basically no centripetal friction results.
Best Answer
The formula for a flat-track turn is incorrect; the required inward centripetal force must be supplied by the friction force: $$\frac{mv^2}{R}=\mu mg$$ which reduces to:$$v=\sqrt{\mu Rg}$$ Since the friction force used is the maximum friction force, it means that any slower velocity can be handled.
OTOH, with a banked curve, the centripetal force needed is the same; the only centripetal force available is the horizontal component of the normal force, leading to:$$\frac{mv^2}{R}=mg \tan \theta$$ which reduces to:$$v=\sqrt{ Rg \tan \theta}$$ But slower (and faster) speeds can be accommodated by getting help from friction.
Some NASCAR races are held at high speeds on highly banked tracks. But cars that hit a patch of oil still slide up into the outside wall, The drivers are driving so fast that they need both banking and friction to keep on the track. Lose friction, and they slide up the track... And when they slow down too much, through friction with the wall, they slide down again.