Derive the expression for the maximum velocity on a banked road whose angle of banking is theta . assume the radius of turn to be R and acceleration due to gravity to be represent by the symbol " g "

Asked by  | 9th Nov, 2012, 12:30: PM

Expert Answer:

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 angle?.

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

R cos? = Mg …(i)

R sin? = …(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 = ?R, acts in the direction AO.

The frictional force can be resolved into two components:

  • sin? in the downward direction

  • cos? in the inward direction

Since there is no motion along the vertical,

R cos ? = Mg + ? R sin? ……. (iii)

Let v max be the maximum permissible speed of the vehicle. The centripetal force is now provided by the components sin? and ? Mg cos?, i.e.,

sin ? + ? R cos? = …….. (iv)

From equation(iii),we have

Mg = R cos? (1?? tan?)…(v)

Again from equation (iv), we have

 = R cos? (? + tan?) …(vi)

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

Answered by  | 10th Nov, 2012, 02:02: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.