Banking of a Road

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Safe Velocity of a Vehicle on an Unbanked Road:

  • The necessary centripetal force required to negotiate a turn by a vehicle moving along a horizontal unbanked curved road is provided by the force of friction between the wheels (tyres) and the surface of the road.
  • Let us consider a vehicle of a mass ‘m’ is moving along a horizontal curved road of radius ‘r’ with speed ‘v’.

Let μ be the coefficient of friction between the road surface and the wheels then

Banking of Road

  • This expression gives the maximum speed with which a vehicle can be moved safely along a horizontal curved road. If speed is more than this velocity, then there is a danger that the vehicle will get thrown (skid) off the road.

Banking of Road:

  • To make the turning of a vehicle on a curved road safer, the outer edge of the road is raised above the inner edge making some inclination with the horizontal. This is known as banking of road.
  • When the road is banked then, the inclination of the surface of the road with the horizontal is known as the angle of banking.

Necessity of Banking of Road:

  • As the speed of vehicle increases, the centripetal force needed for the circular motion of vehicle also increases. In the case of unbanked road necessary centripetal force is provided by the friction between the tyres and the surface of the road. But there is a maximum limit for frictional force, which depends on the coefficient of friction between the wheels and road.
  • When the centripetal force needed exceeds the maximum limit of frictional force, the vehicle skids and tries to go off the curved path resulting in an accident.
  • Without proper friction, a vehicle will not be able to move on the curved road with large speed. To avoid this we may increase the force of friction making the road rough. However, this results in the wear and tear of the tyres of the vehicle. Also, the force of friction is not always reliable because it changes when roads are oily or wet due to rains etc. To eliminate this difficulty, the curved roads are generally banked.
  • Due to banking of the road, the necessary centripetal force is provided by the component of the normal reaction.



The Expression for the Angle of Banking of Road:

  • Consider a vehicle of mass ‘m’ is moving with speed ‘v’ on a banked road of radius ‘r’ as shown in the figure. Let ‘θ’ be the angle of banking. The weight “mg” of the vehicle acts vertically downwards through its centre of gravity G, and N is the normal reaction exerted on the vehicle by banked road AC. N is perpendicular to road AC. Let ‘f’ be the frictional force between the road and the tyres of the vehicle.
  • Now, the normal reaction is resolved into two components (N cosθ) along the vertical (acting vertically upward) and (N sinθ) along the horizontal (towards left) as shown in the figure. Similarly, the frictional force ‘f’ can be resolved into two components (f sinθ) along the vertical (acting vertically downward) and( f cosθ) along the horizontal (towards left) as shown in the figure.

The free body diagram of a car is as follows

Considering equilibrium

Total upward force = Total downward force

∴   N cosθ  = mg + f sinθ

∴ mg = N cosθ – f sinθ ….. (1)

 The horizontal components N sinθ  and f cosθ  provides necessary centripetal force

mv2/r = N sinθ + f cosθ …… (2)

Dividing equation (2) by (1)

Now, frictional force f  = μsN
Where μs = coefficient of friction between road and tyres

  • This is an expression for the angle of banking for curved road considering friction between the road and tyres of the vehicle.
  • Note:

For the given pair of road and tyre μs = tan λ,  where λ = angle of friction

  • Case – I:

When the road is horizontal θ = 0°

This is an expression for safe velocity on unbanked road

  • Case – II:

When the frictional force between the road and tyres of the vehicle is negligible μs = 0.


This is an expression for the angle of banking of a road.

When friction is not mentioned in the problem, use this expression to solve the problems.

  • The expression for safe velocity on the banked road is

  • The speed will be maximum when tan θ = 1 i.e. θ = 45°. It means the vehicle can be driven with maximum safe speed only when the angle of banking = 45°.



Factors affecting the angle of banking:

  • The angle of banking depends on the speed of the vehicle, the radius of the curved road and the acceleration due to gravity g at that place.
  • The expression does not contain the term m representing mass, thus the angle of banking is independent of mass ‘m’ of the vehicle. Thus the angle of banking is the same for heavy and light vehicles.
  • the angle of banking depends on the radius (r) of the curved road. The angle of banking is inversely proportional to the radius of curvature.
  • For the given radius of the curve, the angle of banking increases with the speed.
  • For the given radius of the curve and speed of the vehicle angle of banking is inversely proportional to the acceleration due to gravity at that place. The acceleration due to gravity is more on the pole than that at the equator. Thus for the given radius of the curve and speed of the vehicle angle of banking is less at the pole than that at the equator.
  • If l is the width of the banked road, then the elevation of an outer edge of the road over the inner edge (provided angle of banking is small) is given by

otherwise h = l sinθ

  • In actual practice, some frictional forces are always present even on the banked road. So that the actual safe velocity is always greater than the calculated safe velocity on the banked road.

High Speed on Unbanked Road:

  • During motorcycle race, the riders negotiate the curve on a flat road in that case they have to tilt their motorcycle and themselves to compensate the angle of banking at that place.

  • A cycling race track is called velodrome which has a saucer-shaped track. The rider has to take a position on a track as per his/her speed.



Example – 1:

  • To what angle must a racing track of radius of curvature 600 m be banked so as to be suitable for a maximum speed of 180 km/h?
  • Solution:
  • Given: maximum speed v = 180 km/hr = 180 x 5/18 = 50 m/s ,Radius of curvature = r = 600 m,  g = 9.8 m/s2
  • To Find: Angle of banking = θ = ?

Ans: Angle of banking = 23° 18’

Example – 2:

  • A curve in the road is in the form of an arc of a circle of radius 400 m. At what angle should the surface of the road be laid inclined to the horizontal so that the resultant reaction of the surface acting on a car running at 120 km/h is normal to the surface of the road?
  • Solution:
  • Given: speed of the car = v = 120 km/hr = 120 x 5/18 = 33.33 m/s, radius of curve = r = 400m, g = 9.8 m/s2
  • To Find:  Angle of banking = θ = ?

Ans: Angle of banking = 15° 49’

Example – 3:

  • A vehicle enters a circular bend of radius 200 m at 72 km/h. The road surface at the bend is banked at 10°. Is it safe? At what angle should the road surface be ideally banked for safe driving at this speed ? If the road is 5m wide, what should be the elevation of the outer edge of road surface above the inner edge?
  • Solution:
  • Given: Speed of vehicle = v = 72 km/hr = 72 x 5/18 = 20 m/s , Angle of banking = θ = 10°, radius of curve = r = 200 m, g = 9.8 m/s2
  • To Find:  Part – I: To check safety, part – II: nagle of banking = θ = ? for given speed, elevation = h =?
  • Part – I:

Now, the velocity of the vehicle (20 m/s) is greater than the safe velocity (18.59 m/s).

Hence it is unsafe to drive at a speed 72 km/hr.

  • Part – II:

 

Ans: It is unsafe to drive at the speed 72 km/hr. The angle of banking for speed 72 km/hr = 11°32’,

Elevation of the outer edge over inner edge = 1m

Example – 4:

  • Find the minimum radius of an arc of a circle that can be negotiated by a motorcycle riding at 21 m/s if the coefficient of friction between the tyres and the ground is 0.3. What is the angle made with the vertical by the motorcyclist? g = 9.8 m/s2.
  • Solution:
  • Given: Velocity of motor cycle = v = 21 m/s , Coefficient of friction = μ  = 0.3,  g = 9.8 m/s2
  • To Find: radius of curvature = r = ?,  Angle with vertical = θ= ?

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: Radius of the circular path = 150 m, Angle of banking = 16°42’



Example – 5:

  • What is the angle of banking necessary for a curved road of 50 m radius for safe driving at 54 km/h? If the road is not banked, what is the coefficient of friction necessary between the road surface and tyres for safe driving at this speed?
  • Solution:
  • Given: velocity of vehicle = v = 54 km/hr = 54 x 5/18 = 15 m/s, radius of curve = r = 50m, g = 9.8 m/s2
  • To Find: Angle of banking = θ = ?, coefficient of friction = μ = ?

When the road is not banked, the necessary centripetal force is provided by the friction between the road and tyres.

Ans: The angle of banking = 24°42’, the coefficient of friction = 0.4592

Example – 6:

  • A train of mass 105 kg rounds a curve of radius 150 m at a speed of 20m/s. Find the horizontal thrust on the outer rail if the track is not banked. At what angle must the track be banked in order that there is no thrust on the rail? g= 9.8 m/s2.
  • Solution:
  • Given: mass of train = m = 105 kg, velocity of train = v = 20 m/s, radius of curve = r = 400m,  g = 9.8 m/s2
  • To Find: Horizontal thrust = F = ?,  Angle of banking = θ = ?

The horizontal thrust is equal to the centripetal force in magnitude.

Ans: Horizontal thrust = 2.67 x 105 N, the angle of banking = 15° 13’

Example – 7:

  • The radius of curvature of a metre gauge railway line at a place where the train is moving at 36 km/h is 50m. If there is no side thrust on the rails find the elevation of the outer rail above the inner rail.
  • Solution:
  • Given: speed of train = v = 36 km/hr = 36 x 5/18 = 10 m/s, radius of curve = r = 50m,   g = 9.8 m/s2 , For metre gauge, distance between rail = l   = 1 m.
  • To Find: elevation = h = ?,

Ans: The elevation of outer rail over inner rail = 0.2 m

Example – 8:

  • Find the angle of banking of the railway track of radius of curvature 3200 m if there is no side thrust on the rails for a train running at 144 km/h. Find the elevation of the outer rail above the inner one if the distance between the rails is 1.6 m.
  • Solution:
  • Given: Speed of train = v = 144 km/hr = 144 x 5/18 = 40 m/s, radius of curve = r = 3200m, g = 9.8 m/s2 , distance between rail = l = 1.6 m.
  • To find: Angle of banking = θ =? elevation = h = ?,

Ans: The angle of banking = 2°55’, the elevation of outer rail over inner rail = 81 mm

Example – 9:

  • A motorcyclist at a speed of 5 m/s is describing a circle of radius 25 m. Find his inclination with the vertical. What is the value of the coefficient of friction between the road and tyres?
  • Solution:
  • Given: Speed of motor cycle = v = 5 m/s, radius of circle = r = 25 m,  g = 9.8 m/s2 ,
  • To find:  Angle of inclination = θ= ?,  Coefficient of friction = μ = ?

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: Angle made with the vertical = 5°49’, The coefficient of friction = 0.1020

Example – 10:

  • A motor van weighing 4400 kg rounds a level curve of radius 200 m on the unbanked road at 60 km/hr. What should be the minimum value of the coefficient of friction to prevent skidding? At what angle the road should be banked for this velocity?
  • Solution:
  • Given: Mass of vehicle = m = 4400 kg, velocity of vehicle = v = 60 km/hr = 60 x 5/18 = 16.67 m/s , r = 200m,  g = 9.8 m/s2
  • To Find:  Coefficient of friction = μ = ?, Angle of banking = θ = ?,

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: The coefficient of friction = 0.1418, The angle of banking = 8°4’



Example – 11:

  • A circular road course track has a radius of 500 m and is banked to 10°. If the coefficient of friction between the road and tyre is 0.25. Compute (i) the maximum speed to avoid slipping (ii) optimum speed to avoid wear and tear of the tyres.
  • Solution:
  • Given: radius of curve = r = 500m,  Angle of banking = θ = 10°, coefficient of friction = μ = 0.25,  g = 9.8 m/s2 ,
  • To find: safe velocity = v = ?, to avoid wear and tear v = ?

Maximum sped to avoid slipping.

Maximum speed to avoid wear and tear

Ans: The maximum velocity to avoid skidding = 46.74 m/s,

The maximum velocity to avoid wear and tear of tyres. = 29.39 m/s.

Example – 12:

  • A rotor has a diameter of 4.0 m. The rotor is rotated about the central vertical axis. The occupant remains pinned against the wall. When the linear velocity of the drum is 8 m/s. Compute the coefficient of static friction between the wall of the rotor and the clothing of occupant. Also, calculate the angular velocity of the drum. How many revolutions will it make in a minute?
  • Solution:
  • Given: diameter = d = 4 m, radius = r = 4/2 = 2 m, linear speed = v =.8 m/s , g = 9.8 m/s2 ,
  • To find:  Coefficient of friction = μ = ?, Angular velocity = ω = ?, rpm =  N = ?

 

As the occupant remains pinned against the wall, his weight is equal to the frictional force.

Frictional force = Weight of a body

 

Ans: Coefficient of friction = 0.3063, Angular velocity = 4 rad/s, Number of revolutions per minute = 38.21.

Overturning of Vehicle:

Unbanked Road:

  • When a vehicle moves along a curved path with very high speed, then there is a chance of overturning of the vehicle. Inner wheel leaves ground first. Let a vehicle of a  mass ‘m’ is negotiating a turn along a curved path of radius ‘r’ with speed ‘v’. Let ‘2a’ be the length of the axle i.e. the distance between the two wheels. Let h be the height of the centre of gravity of the vehicle above the ground. Let R1 and R2 be the reactions on the inner wheel and outer wheel of the vehicle exerted by the road surface.

The frictional force ‘F’ provides the necessary centripetal force.

Taking the moment of forces about the centre of gravity ‘G’

Fh  + R1a = R2a  ………… (2)

Reactions are balanced by the weight of the vehicle

R1 + R2 = mg …….. (3)

Solving equations |(1), (2) and (3)

From these equations, we can see that as the speed increases reaction R1 decreases while reaction R2 increases.

When reaction R1 is zero overturning of the vehicle takes place.

This is an expression for maximum velocity of the vehicle by which it can be driven beyond which overturning of the vehicle takes place.

Banked Road:

If μ < d/h, vehicle skids and if μ >d/h, vehicle overturns



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Science > Physics > Circular MotionYou are Here
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