A car is moving along a banked road laid out as a circle of radius r . (a) What should be the banking angle `theta` so that the car travelling at speed v needs no frictional force from tyres to negotiate the turn ? (b) The coefficient of friction between tyres and road are `mu_(s) = 0.9` and `mu_(k) = 0.8` . At what maximum speed can a car enter the curve without sliding toward the top edge of the banked turn ?

Explanation

(a) x and y- components of reaction are
`N_(x) = N "sin" theta `
`N_(y) = N "cos" theta `
In our co-ordinate system , we apply Newton's law in component form .
`{:(sumF_(x)=ma_(x)),(Nsinthetahati=ma_(x)hati),(Nsintheta=ma_(x)):}:|{:(EF_(y)=ma_(y)),(Ncosthetahatj-Whatj=ma_(y)hatj),(Ncostheta-W=ma_(y)):}`
Since , the car travels along a circle `a_(y) = 0 , a_(x) = v^(2)//r`
`N "sin" theta = (mv^(2))/(r)`
`N "cos" theta = "mg" ` or tan `theta = (v^(2))/("rg")`

In formulating the above equations we have assumed friction to be absent (friction does not play any role is negotiating the turn) .
(b) If the driver moves faster than the speed mentioned above , a friction force must act parallel to the road , inward towards centre of the turn .

`(F_("fric"))_(x) = F_("fric") "cos" theta `
`(F_("fric"))_(y) = F_("fric") "sin" theta`
Now , `sumF_(x) = ma_(x)`
`N "sin" theta + F_("fric") "cos" theta = mv^(2)//r`
`sumF_(y) = ma_(y)`
`N "cos" theta - F_("fric") "sin" theta - W = 0`
Since , the tyre rolls forward without slipping , the tendency of slipping arises sideways . Hence , we get
`F_("fric") = (F_("fric"))_("max") = mu_(s)N`
`N "sin" theta + mu_(s)N "cos" theta = (mv^(2))/(r)`
`N "cos" theta - mu_(s) N "sin" theta = "mg" `
On eliminating N , we get `(("sin" theta + mu_(s) "cos" theta)/("cos" theta - mu_(s)"sin" theta )) = (v^(2))/("rg")`
or `v = sqrt(("sin"theta + mu_(s)"cos"theta)/("cos" theta - mu_(s) "sin" theta)"rg")`