Consider a car taking a left turn along a road of radius r banked at an angle `theta` for a designed optimum speed v. Let m be the mass of the car. In general, the forces acting on the car are
(a) its weight `vec(mg)` , acting vertically down
(b) the normal reaction of the road `vec(N)`, perpendicular to the road surface.
(c ) the frictional force `vecf_(s)` along the inclined surface of the road.
If `mu_(s)` is the coefficient of static friction between the tyres and raod, `f_(s) = mu_(s)N`. If the car is driven fast enough, at a speed greater than the optimum speed v, it may skid up the incline so that `vec(f_(s))` is down the incline.
Resolve `vecN` and `vec(f_(s))` into (i) ` cos theta` vertically up and `f_(s) sin theta` vertically down (ii) `N sin theta` and `f_(s) cos theta` horizontally towards the centre of the circular path. So long as the car takes the turn without skidding off, the horizontal components `N sin theta` and `f_(s) cos theta` together provide the necessary centripetal force, and `N cos theta` balances the sum `mg + f_(s) sin theta`. If `v_(max)` is the maximum safe speed of the car (without skidding),
`(mv_(max)^(2))=N sin theta+f_(s)costheta+mu_(s)N cos theta`
` = N ( sin theta + mu_(s) cos theta) " "`...(1)
and `N cos theta = mg + f_(s) sin theta = mg + mu_(s) N sin theta`
`therefore mg = N(cos theta - mu_(s) sin theta) " "`...(2)
Dividing Eq. (1) by Eq. (2),
`(mv_(max)^(2)//r)/(mg)=(N(sin theta +cos theta))/(N(cos theta-mu_(s)sintheta))`
`therefore (v_(max)^(2))/(rg)=(sin theta+mu_(s) cos theta)/(cos theta-mu_(s) sin theta)=(tan theta +mu_(s))/(1-mu_(s)tan theta)`
`therefore v_(max) = sqrt((rg(tan theta +mu_(s)))/(1-mutantheta))" "...(3)`
At the optimum speed, the friction between the car tyres and the road surface is not called into play. Hence, by setting `mu_(s) 0` in Eq. (3), the optimum speed of a car on a banked circular road is
`v = sqrt(rg tan theta) " "...(4)`
