Question

21. For traffic moving at 60 km/h, if the radius of
the circular curve of a highway is 0.1 km, what
is the correct angle of banking of the road? Take
8= 10 m s. Given tan- (0.278) = 15.5​

Answer 1

see free body diagram as shown in figure,

at equilibrium,

centripetal force = component of normal reaction along centre of circular road ,

or, mv²/r = Nsinθ .......(1)

weight of body = component of normal reaction along vertical upward direction

or, mg = Ncosθ .....(2)

from equations (1) and (2),

(Nsinθ/Ncosθ) = mv²/r/mg = v²/rg

⇒tanθ = v²/rg

putting, v = 60km/h = 60 × 5/18 = 50/3 m/s

r = 0.1km = 100m and g = 10m/s²

so, tanθ = (50/3)²/(100 × 10)

= 2500/(9 × 1000)

= (5/18)

= 0.278

⇒ θ = tan^-1(0.278) = 15.5°

hence, angle of banking is 15.5°