Question

A circular road of radius r in which maximum velocity is v, has angle of banking
(a) $\tan^{-1}\bigg \lgroup \frac{v^{2}}{rg}\bigg \rgroup$
(b) $\tan^{-1}\bigg \lgroup \frac{rg}{v^{2}}\bigg \rgroup$
(c) $\tan^{-1}\bigg \lgroup \frac{v}{rg}\bigg \rgroup$
(d) $\tan^{-1}\bigg \lgroup \frac{rg}{v} \bigg \rgroup$

Answer 1

To make the vehicle turn safer on a curved road, the outer edge of the road is  often raised above the inner edge, allowing some inclination with the horizontal. This is known banking of road. Thus, when the road is banked, then the inclination of the surface of the road with horizontal is known as the angle of banking.

As per the question -

Radius - r

Maximum velocity - v

Considering the force and angles the equations derived will be

N cos θ = mg   (1eq)                

N sin θ = mv²/r  (2eq)                    

Dividing equation (2) by (1), we get

Tan θ = v²/rg

Answer 2

Answer:

The answer is (a).

Explanation:

Consider a vehicle of mass 'm' moving with velocity 'v' on a banked road of radius 'r'. Let '∅' be the angle of banking.

Now, the velocity 'v' on a curved bank road derived by the expression:

                 v = √[rg (tan∅+u)/ (1-utan∅)]        where

 u is the coefficient of friction between the road and the wheels.

and            u = tanΔ                                         where Δ is angle of friction.

 therefore v = √(rg tan (∅+Δ)

when road is horizontal ∅ = 0°.

hence        v = √(urg).

When the frictional force between the road and wheels of the vehicle is negligible u = 0,

                    v = √rg tan∅

              tan∅ = v²/rg

                    ∅ = tan⁻¹[v²/rg].

therefore Angle of Banking is  ∅ = tan⁻¹{v²/rg}.