Question

find the expression for the safest speed of a car while negotiating a banked road?​

Answer 1

Negotiations:

(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.

Explanation:

I hope it helps you

Answer 2

To Find,

The expression for the safest speed of a car while negotiating a banked road.

Solution,

Consider a vehicle of mass m traveling at v down a curved banked road at an angle. Let f denote the frictional force between the vehicle's tires and the road surface.

The forces operating on the vehicle are as follows:

1. Vertically downward weight mg

2. Normal response N moving upward through the center of gravity and

adjacent to the banked road/surface

The frictional force between the vehicle's tires and the road surface can be subdivided into,

1. f cosθ - along the horizontal axis
2. f sinθ - in a downward vertical direction

The normal reaction can be resolved into two components,

1. N cosθ – Vertical component of normal reaction

2. N sinθ – Horizontal component of normal reaction

• The component N cosθ of Normal Reaction is balanced by the vehicle weight mg and the frictional force component f sin.

N cosθ = mg + f sinθ

∴ mg = N cosθ – f sinθ ⇢ (1)

• The centripetal force is provided by the horizontal component N sinθ and the component f cos of frictional force.

N sinθ + f cosθ = $mv^{2}$ / R

Or, $mv^{2}$ / R = N sinθ + f cosθ ⇢ (2)

Dividing Eq (2) by Eq (1),

$v^{2}$ / Rg = N sinθ + f cosθ / N cosθ – f sinθ ⇢ (3)

• The magnitude of the frictional force is determined by the vehicle's speed on a certain road surface and the vehicle's tires. Let v represent the vehicle's maximum speed,

the frictional force f produced at this speed should be

$f_{m}$ = μ$_{s}$ N ⇢ (4)

$v^{2} max$ / Rg = N sinθ + f cosθ / N cosθ – f sinθ ⇢ (5)

From Eq (4) and Eq (5), we get $v_{max}$

∴ $v_{max}$ = ⇢ (6)

For a curved Horizontal road, θ = 0°, hence Eq (6) becomes,

$v_{max}$ = √{μ Rg} ⇢ (7)

• When Eqs (6) and (7) are compared, it is clear that the maximum safe speed of cars on a banked road is greater than that of a curved horizontal road/level road.

If μ$_{s}$ = 0, then Eq (6) becomes

v$_{max}$ = v$_{0}$ = $\sqrt{rg(0+tan)} / 1-0tan$

v$_{0}$ = $\sqrt{Rgtan}$ ⇢ (8)

• At this speed, the frictional force is insufficient to create the required centripetal force. If the vehicle is driven at this speed on the banked road, there will be some water and tire wear. v is known as optimum speed.

From Eq (8), i.e.

v$_{0}$ = $\sqrt{Rgtan}$

tanθ = v$_{0}$ $^{2}$/ Rg

θ = tan [ v$_{0}$ $^{2}$/ Rg] ⇢ (9)

• The mass of the vehicle m is not included in this formula [Eq (9)] for the angle of banking. As a result, the angle of banking is independent of the vehicle's mass.

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