Question

Obtain an expression for the maximum speed with which a vehicle can negotiate a curved smooth road banked at an angle theta with proper diagram...?

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Answer 1

Consider a car of mass "m" moving on a banked road of radius 'r'. The various forces acting on the car are:

(i) The weight of the car which acts vertically downwards

i.e., ω = mg ....(i)

(ii) The normal reaction R of the road acts perpendicular to the road.

Neglect the force of friction between the tyres of the car and the road. Now resolve the normal reaction R of the road in the two components:

(a) R cosθ which is equal opposite to mg

i.e., R cosθ = -mg ...(ii)

(b) R sinθ which acts towards the centre of the circular path and provides the necessary centripetal force ({mv2}/{r}) to the car.

i.e., R sinθ = {mv2}/{r} ....(iii)

Dividing (iii) by (ii) we get

tanθ = v2/rg ⇒ v = (rg tanθ)1/2

which is the safe speed of the car for given value of 'r' and 'θ' on a circular banked road.

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Answer 2

Consider a car of mass "m" moving on a banked road of radius 'r'. The various forces acting on the car are:

(i) The weight of the car which acts vertically downwards

i.e., ω = mg ....(i)

(ii) The normal reaction R of the road acts perpendicular to the road.

Neglect the force of friction between the tyres of the car and the road. Now resolve the normal reaction R of the road in the two components:

(a) R cosθ which is equal opposite to mg

i.e., R cosθ = -mg ...(ii)

(b) R sinθ which acts towards the centre of the circular path and provides the necessary centripetal force ({mv2}/{r}) to the car.

i.e., R sinθ = {mv2}/{r} ....(iii)

Dividing (iii) by (ii) we get

tanθ = v2/rg ⇒ v = (rg tanθ)1/2

which is the safe speed of the car for given value of 'r' and 'θ' on a circular banked road.

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