speedway turn, with radius of curvature R, is banked at an angle θ above the horizontal.

Question

(a) What is the optimal speed at which to take the turn if the track's surface is iced over (that is, if there is very little friction between the tires and the track)? (Use any variable or symbol stated above along with the following as necessary: g.)
vzero friction =

(b) If the track surface is ice-free and there is a coefficient of friction μs between the tires and the track, what are the maximum and minimum speeds at which this turn can be taken? (Use any variable or symbol stated above along with the following as necessary: g.)
vmin =
vmax =

(c) Evaluate the results of parts (a) and (b) for R = 365 m, θ = 45°, and μs = 0.64.
vzero friction =
vmin =
vmax =

Answer 1

a) you want the resultant of the force vectors mg (vertical) and mv^2/r (horizontal) to be normal to Theta.

TanTheta=v^2/rg Or alternative, you want the component of force mg down the track equal to the sliding force centripetal force.

mv^2/r *cosTheta=mgsinTheta
and then tantheta=v^2/rg

b) Now one has an additional force, Parallel to the track,friction.

so, mgSinTheta-mu*mgCosTheta=mv^2/r cosTheta

or g Tantheta= mu*g+v^2/r
solve for v, in this case vmax
Now, vmin means friction is preventing it going down, or g TanTheta=-mu*g+v^2/r

Answer 2

is this how we solve for v for part a and b .
a ) v=sqrt ( g r / tan theta)

b) v min= sqrt ( r tan theta + r mu g )
v max= sqrt ( r tan theta - r mu g )

Answer 3

The book gives an answer for part a below.

v = sqrt (rg*cos theta*sin theta)

I assume that's incorrect?

Answer 4

Hey Bob, I think you need to check your algebra. As the student substituted 2 posts up vmin is bigger than vmax.