Physics, asked by tejasurya3229, 4 months ago

Derive an expression for safety speed of vehicle along bank curved road

Answers

Answered by Anonymous
1

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.

Answered by Anonymous
2

Explanation:

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.

solution

Answered By

toppr

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