A curve in a highway has a radius of curvature of 110m and it banked at 3.40 degrees.?

Question

on an icy day, what is the safest speed to go around the curve?

Answer 1

Consider the forces and directions of the forces on the car:

The force of gravity pulling the car directly downward
The centrifugal force pulling the car directly horizontal
The force of the car on the road, which is banked
The friction of the road on the tires.

Since it is icy, let's set the friction to zero

Then the forces parallel to the road are the parallel component of the centrifugal force:

m*cos(3.4)*r*w^2

and the force of gravity towards the center
m*g*sin(3.4)

When these force balance, there is no force of friction parallel to the road surface. So

m*cos(3.4)*r*w^2=m*g*sin(3.4)
w=sqrt(g*sin(3.4)/(cos(3.4)*r))
The linear velocity, or what the driver will read on the speedometer
is w*r

8 m/s

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