obtain an expression for maximum velocity with which a vehicle can take a turn of circular road?
Answers
Answer:
Banking of roads : To avoid risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inwards, i.e., the outer side of road is raised above its inner side. This is called 'banking of roads'.
Consider a car taking a left turn along a road of radius r banked at an angle θ for a designed optimum speed V. Let m be the mass of the car. In general, the forces acting on the car are:
(a) Its weight
mg
, acting vertically down
(b) The normal reaction of the road
N
, perpendicular to the road surface
(c) The frictional force
f
s
along the inclined surface of the road.
Resolve
N
and
f
s
into two perpendicular components Ncosθ vertically up and
f
s
sinθ vertically down, Nsinθ and
f
s
cosθ horizontally towards the centre of the circular path.
If v
max
is the maximum safe speed without skidding.
r
mv
max
2
=Nsinθ+f
s
cosθ
=Nsinθ+μ
s
Ncosθ
r
mv
max
2
=N(sinθ+μ
s
cosθ)....(1)
and
Ncosθ=mg+f
s
sinθ
=mg+μ
s
Nsinθ
∴mg=N(cosθ−μ
s
sinθ)...(2)
Dividing eq. (1) by eq. (2),
r.mg
mv
max
2
=
N(cosθ−μ
s
sinθ)
N(sinθ+μ
s
cosθ)
∴
rg
v
max
2
=
cosθ−μ
s
sinθ
sinθ+μ
s
cosθ
=
1−μ
s
tanθ
tanθ+μ
s
∴v
max
=
1−μ
s
tanθ
rg(tanθ+μ
s
)
...,.(3)
This is the expression for the maximum safe speed on a banked road.
At the optimum speed, the friction between the car tyres and the road surface is not called into play. Hence, by setting μ
s
=0 in eq. (3), the optimum speed on a banked circular road is
v=
rgtanθ
...(4)
∴tanθ=
rg
v
2
or θ=tan
−1
(
rg
v
2
)
From this eq. we see that θ depends upon v,r and g. The angle of banking is independent of the mass of a vehicle negotiating the curve.