Question

A turn of radius 20 m is banked for the vehicle going at speed of 36 km/hr. if the coefficient of static friction between the road and the tyre is 0.4. what are the possible speeds of a vehicle so that it neither slips down nor skids up (g=9.8 m/s2)

Answer 1

Find tanθ=v^2/Rg and find cosθ and sinθ.

Then,

N=mv2/r sinθ + mgcosθ

mgsinθ - mv2/r cosθ = μN

(Difference symbol because if centrifugal greater then friction stops skid outward

where as if component of weight greater then friction stops slip inwards)

Solve above 2 equation 2 get:

V^2/gr=(tanθ-μ)/(1+μtanθ)         and            (tanθ+μ)/(1-μtanθ)

One for maximum speed and other for minimum speed .

Put value of g,r,μ,v,cosθ and sinθ

v=4.082m/s or 15m/s

Multiply by 18/5 to get kph

v=14.7km/h or 54 km/h

Answer 2

We know that,

Tana=v^2/rg

=100/20×10

=0.5

N1=mgcosa+mv^2/r sina

uN1=mv^2/r Cosa - mgsina

While solving both equations we get :::::

V1 = √rg(tana-u)/1+utana

= √200(0.1)/1.2

= 4.8m/s or 14.7 kmph

So, velocity is between 14.7 and 54 kmph.