# velocity required when rounding a flat curve and banked track

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

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.