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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

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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.

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That's because the radius of its path is tied to its velocity in the absence of any other force.

Compare it with for example a satellite. In order to maintain orbit they have to have a certain speed that needs to be higher when the radius of their orbit is smaller.

If they travel at a slower velocity they will fall back; if their velocity is greater their orbit will expand (at least their avarage orbit will - they'd probably have an elliptical orbit).

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