Question 1099405
The centripetal acceleration {{{a[c]}}} needed to stay in circular motion is related to
circle radius, {{{R}}} , linear velocicty {{{v}}} and angular velocity {{{w}}} by
{{{a[c]}}}{{{"="}}}{{{v^2/R}}}{{{"="}}}{{{w^2R}}} .
If we cannot count on friction,
the centripetal acceleration needed will have to be supplied by the banking of the track.
 
The sketch below shows a vertical cross-section of the banked track,
an object upon that track, and a representation of some of the accelerations acting on that object.
The only force acting on the plane of the cross section is weight,
so the only acceleration is the acceleration of gravity.
We can consider it as made of components parallel and perpendicular to the track surface.
 
The acceleration of gravity and its component parallel to the track surface is shown in the sketch.
The acceleration of gravity has a (not shown) component perpendicular to the track surface,
which would make the car sink into the track,
but a reaction force from the track (called the normal force) keeps that from happening.
{{{drawing(600,350,-1,5,-2,1.5,
line(-1,0,5.5,0), triangle(0,0,0,1,4,0),
circle(0.524,0.972,0.1),
triangle(0.524,0.972,0.524,-1.914,1.203,0.802),
rectangle(0,0,0.07,0.07),
blue(line(0.524,0.972,0.524,-1.914)),
blue(line(0.524,0.972,1.203,0.802)),
blue(arrow(0.524,-1,0.524,-1.914)),
blue(arrow(0.524,0.972,1.203,0.802)),
red(arc(4,0,3,3,180,194)),
red(arc(0.524,-1.914,2,2,270,284)),
locate(0.4,-0.2,blue(g)),
locate(0.824,1.2,blue(a[c])),
locate(0.02,0.5,14.6m),locate(2,0.5,L),
locate(1.5,0.17,52m)
)}}}
{{{L=sqrt((14.6m)^2+(52m)^2)}}}{{{"="}}}{{{about}}}{{{54m}}}
For a car to stay on a circular path,
the acceleration component parallel to the track surface must be
{{{blue(a[c])}}}{{{"="}}}{{{v^2/R}}}
There are two similar right triangles in the sketch,
the wedge representing the cross-section of the track,
and the triangle formed by {{{blue(g)}}} and {{{blue(a[c])}}} ,
so, {{{blue(a[c])/blue(g)}}}{{{"="}}}{{{14.6m/"54 m"}}} .
Using {{{blue(g)="9.8 m / s^2"}}} ,
{{{blue(a[c])/blue(g)}}}{{{"="}}}{{{"9.8 m / s^2"(14.6m/"54 m")}}} ,
{{{v^2/R}}}{{{"="}}}{{{"9.8 m / s^2"(14.6/54)}}} ,
and {{{v^2}}}{{{"="}}}{{{(R)("9.8 m / s^2")(14.6/54)}}}
For {{{R=114m}}} ,
{{{v^2}}}{{{"="}}}{{{(114)(9.8)(14.6)/54)}}}{{{("m / s")^2=302}}}{{{("m / s")^2}}} ---> {{{v=highlight("17.4 m / s")}}}{{{"="}}}{{{(17.4m /"1s")(3600s/"1h")(1km/"1000m")}}}{{{"="}}}{{{"62.6 km / h"}}} .
For {{{R=166m}}} ,
{{{v^2}}}{{{"="}}}{{{(166)(9.8)(14.6)/54)}}}{{{("m / s")^2=440}}}{{{("m / s")^2}}} ---> {{{v=highlight("21.0 m / s")}}}{{{"="}}}{{{(21m /"1s")(3600s/"1h")(1km/"1000m")}}}{{{"="}}}{{{"75.5 km / h"}}} .
 
NOTES:
If I did not make a mistake, I believe those are the answers a teacher would want. (I am not a physics teacher. I am a chemist, but I did some substitute physics teaching 45 years ago, when I was in college).
 
If the track was a circle, and covered with ice, It would be a good idea, to keep your speed between 63 and 75 kph while entering the track and driving on it.
Lower speeds would cause the car to drift towards the inner lane, and eventually off the track.
Higher speeds would cause the car to drift towards the outside lane, and eventually fly off the track.
However, race tracks are not all curves, and drivers can go faster or slower than calculated by letting the car drift through the curve, and correcting that drift as they exit the curve. There is also friction (with track surface and hair) that can be harnessed to help.