Show the various forces acting on the vehicle negotiating a rough inclined circular turn. Hence find the maximum safe velocity of the vehicle with which it can negotiate the circular turn.
Answers
Explanation:
Dear Student,
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 =Mv2rMv2r …(ii)
On dividing equation (ii) by equation (i), we get
RsinθRsincosθ=MV2r/MgRsinθRsincosθ=MV2rMg
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 vmax 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 = Mv2maxrMv2maxr…….. (iv)
From equation(iii),we have
Mg = R cosθ (1−μ tanθ)…(v)
Again from equation (iv), we have
Mv2maxrMv2maxr = R cosθ (μ + tanθ) …(vi)
On dividing equation (iv) by (v), we have
v2maxgr=μ+tanθ1−tanθv2maxgr=μ+tanθ1-tanθ
⇒vmax=[gr(μ+tanθ)1−μtanθ]12/