Q. 54.7( 3 Votes )

# If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking?

Answer :

Where,

R=Normal reaction

θ=Angle of banking

v=velocity of the car

m=mass of the car

g= acceleration due to gravity

r=radius

From the diagram, we can see that

R cosθ = mg

R sinθ = mv^{2}/r

On dividing the above equations, we get

Tanθ = v^{2}/rg

⇒ θ=tan^{-1}(v^{2}/rg)

We are given, v=10m/s

r=30m

⇒ θ=tan^{-1}(10×10/30)

⇒ θ=tan^{-1}(1/3)

∴ Angle of banking should be tan^{-1}(1/3)

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