1.Derive a relation for the optimum velocity of negotiating a curve by a body in a banked curve.
            
2.Calculate the maximum speed with which a vehicle can travel on a banked road without skidding.
DO BOTH THESE QUESTIONS HAVE THE SAME ANSWER??
i.e., THE DERIVATION OF Vmax=root(mu.r.g)??
Experts plzz answer

Dear student,
Yes both have a same answer which is explained below,

Motion of a car on a banked road:

For the vehicle to go round the curved track at a reasonable speed without skidding, the greater centripetal force is managed for it by raising the outer edge of the track a little above the inner edge. It is called banking of circular tracks.

Consider a vehicle of weight Mg, moving round a curved path of radius r, with a speed v, on a road banked through angleθ.

The vehicle is under the action of the following forces:

• The weight Mg acting vertically downwards

• The reaction R of the ground to the vehicle, acting along the normal to the banked road OA in the upward direction

The vertical component R cos θ of the normal reaction R will balance the weight of the vehicle and the horizontal component R sin θ will provide the necessary centripetal force to the vehicle. Thus,

R cosθ = Mg …(i)

R sinθ = …(ii)

On dividing equation (ii) by equation (i), we get

As the vehicle moves along the circular banked road OA, the force of friction between the road and the tyres of the vehicle, F = μR, acts in the direction AO.

The frictional force can be resolved into two components:

• μ R sinθ in the downward direction

• μ R cosθ in the inward direction

Since there is no motion along the vertical,

R cos θ = Mg + μ R sinθ ……. (iii)

Let v_max be the maximum permissible speed of the vehicle. The centripetal force is now provided by the components R sinθ and μ Mg cosθ, i.e.,

R sin θ + μ R cosθ = …….. (iv)

From equation(iii),we have

Mg = R cosθ (1−μ tanθ)…(v)

Again from equation (iv), we have

 = R cosθ (μ + tanθ) …(vi)

On dividing equation (iv) by (v), we have

Regards