There are 3 possible scenarios depending on the speed of the car, $v$.

I) For certain speed $v_o$, there is no friction force between the car and the ground

II) For $v>v_o$, frictional force would act down the incline as shown

III) For $v<v_o$, frictional force would act up the incline as shown

So, let’s begin with establishing $v_o$ for which frictional force, $f_s = 0$

In this scenario, $N \cos \theta = mg$ and $N \sin \theta = m \cfrac{v_o^2}{r}$. Dividing the two equations we get $v_o = \sqrt{gr \tan \theta}$.

For $v> v_o$, $N \sin \theta$ will not be able to provide the needed centripetal force and the car would have tendency to slip outwards (up the incline) and as such a frictional force would kick in. Now, the two equations in the vertical and horizontal direction would become

$N \cos \theta = mg + f_s \sin \theta$ and

$N \sin \theta + f_s \cos \theta = m \cfrac{v^2}{r}$

Solving the two we can find the desired unknown. Note, to find $v_max$ at which car is about to slip, use $f_s = \mu N$

We trust that you will be able to write equations for the 3rd scenario on your own.

With that, now let’s explore the motion of a daredevil’s bike tracing a vertical circular path at a constant speed