Question

A circular curve of highway is designed for traffic moving at 92 km/h. When assumed the traffic consists of cars without negative lift ?

(a) If the radius of the curve is 150 m, what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at 92 km/h?

Answer 1

(a) `theta=23.9^@` rounded to first place of decimal
(b) `mu_(s(min))=0.44382`

Explanation:

(a) In the figure above
Let `mg` be weight of car,
`r` radius of curve it is traveling,
`theta` be the banking angle of the road,
`v` its maximum velocity with which it can travel safely while negotiating through the curve,
`N` be the normal reaction, and
`mu_s` be coefficient of friction between the road and the its tyres.

We know that while in circular motion the centripetal force acting on the car `=(mv^2)/r`
And force due to friction is given as`=mu_sN`

Assuming car to be situated at the origin, Force equations in the `x` and `y` direction at this maximum speed for the car are
`(mv^2)/r=Nsin theta+mu_sNcostheta`, and
`mg+mu_sNsin theta=Ncos theta`

For designing of banking angle of the curved roads, worst case is considered. `mu_s text{ is set } =0.`
Very low values of `mu_s` could be encountered in the event of rain, snow, mud, oil spillage on the road when road becomes slippery.

Above equations reduce to
`(mv^2)/r=Nsin theta`, and
`mg=Ncos theta`

Dividing first with second we obtain
`((mv^2)/r)/(mg)=(Nsin theta)/(Ncos theta)`
or `tan theta=(v^2)/(rg)`
Given `v=92km//hr`, converted in `m//s=(92xx1000)/3600=25.dot5`
Inserting all values in above expression
`tan theta=((25.dot5)^2)/(150xx9.81)=0.44382`
`theta=23.9^@` rounded to first place of decimal

(b) If curves are not banked, implies that `theta=0`. This makes `sin theta=0` and `cos theta =1.` Now the general expressions reduce to
`(mv^2)/r=mu_sN`, and
`mg=N`
Insert value of `N` in first and solve for `mu_(s(min)),` minimum coefficient of friction required between tyres and road that would keep traffic from skidding out of the curve:

`(v^2)/r=mu_(s(min))g`
`mu_(s(min))=(v^2)/(rg)`
We observe that this is equal to value of `tantheta` calculated in (a) above.
`mu_(s(min))=0.44382`