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Banking of Roads is defined as the process of raising the edges of a curved road above the inner edge so that there is required centripetal force for the vehicles to take a safe turn.
- This is usually done to avoid skidding off of a vehicle.
- The edges are inclined in a horizontal manner by lifting the outer edge of the road.
- Banking angle is the angle at which the road is inclined.
- Banking of Road can be calculated with the help of friction and without friction.
- The maximum speed of the moving vehicle is kept under control.
- The advantage of banking of roads is that tumbles and overturns are avoided.
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Key Terms: Banking of Roads, Banking Angle, Frictional Force, Centripetal Force, Skidding, Radius of Curvature, Friction, Circular Motion, Force, Speed
What Is Banking of Roads?
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Banking of roads is the process of lifting the outer edge of the road in such a way that it is higher than that of its inner edge. The surface of the road seems like some inclined plane.
- The angle at which the road is inclined is known as the banking angle.
- When the vehicle moves through a curved road, the horizontal component force is acting upon the vehicle.
- The component provides the necessary centripetal force, which further avoids skidding of the vehicle.
- The maximum allowed speed on an inclined or banked turn depends on the mass of the vehicle.
- As per the banking angle, it depends upon the coefficient of friction and its radius of curvature.
- The power of friction varies from road to road, as roadways are sometimes moist and greasy due to rain and other factors.
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What Is Centripetal force?
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Centripetal force pushes or pulls an object in an angular or circular motion to the direction of the centre of the circle. It will make the body move along the curved path.
- It depends upon the angle of motion and radius of the circular path.
- The direction of force is orthogonal to the motion of the body.
- It is also known as angular velocity.
- In the case of particle accelerators, angular velocity is equal to the speed of light in a vacuum.
- The centripetal force is obtained from an inertial frame of reference.
- Spinning a ball on a string is another example of centripetal force.
The formula for Centripetal force is given by:
\(F_r = \frac{mv^2}{r}\)
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Derivation of Banking of Road
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Consider a car of mass ‘m’ and moving with a speed ‘v’ across a circular road of radius ‘r’. Let ‘\(\theta\)’ be the banking angle and f be the frictional force acting between the road and the car’s tyres as shown in the figure.
Here,
- Total upwards force = Total downward force
Ncos\(\theta\) = mg + fsin\(\theta\)
Where, Ncos\(\theta\) → Normal reaction along the vertical axis
fsin\(\theta\) → Frictional force along the vertical axis
mg → Weight of the vehicle acting downwards
Thus,
mg = Ncos\(\theta\) - fsin\(\theta\) ……(i)
- By replacing mg, we get
\(\frac{mv^2}{r}\)= Nsin\(\theta\) + fcos\(\theta\) ……(ii)
Where, Nsin\(\theta\) → Normal reaction along the horizontal axis
fcos\(\theta\) → Frictional force along the horizontal axis
Dividing equation (ii) by (i), we get,
where v²rg = Nsin\(\theta\) + fcos\(\theta\)Ncos\(\theta\) - fsin\(\theta\)
- We know that, Frictional Force \(f = \mu_s N \frac{v^2}{rg}\)
- On substitution and further simplification, we get
\(v = \sqrt{\frac{rg tan\theta + \mu_s}{1- \mu_s tan \theta}}\)...(iii)
\(v_{max} = tan^{-1} \frac{v^2}{rg}\)
When there is no friction, the vertical component of the road acts as a normal force due to which the weight of the vehicle is balanced, and the horizontal component gives in to the centripetal force in the direction of the centre of the curvature of the road. The inclination occurs at the horizontal and longitudinal axis.
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Banking of Road Formula
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Banking of the road is calculated in the presence of friction and without friction. The formula required in case of friction and without friction is as follows:
- Banking Road Formula Without Friction: In this case, the vertical component of force balances the weight of the vehicle. The horizontal force is responsible for producing the required centripetal forces.
The required formula for banking of road is as follows: v= grtan
where, r =centre of curvature and g = acceleration due to gravity
- Banking Road Formula With Friction: In this case, friction force will apply vertically downward and horizontally toward the centre.
The required formula for banking of road is as follows: F= μ N
where Velocity is given by: Vmax= grtan +1-tan
Zero banking angle
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If the banking angle is zero, it means the surface of the road is flat. The normal force does not contribute to the centripetal force any more as it becomes vertical now.
- It is responsible for balancing the weight of the vehicle.
- Friction force acts only on a rough finish surface and thus can provide a centripetal force.
- Therefore, the vertical component of the forces creates a balance.
When the banking angle is 0, substitute in eq(iii) of derivation of banking of road, we will get
\(v = \sqrt {\mu rg}\)
Read More: Sliding Friction
Importance of Banking of Roads
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Banking of Roads is mainly done for the following reasons.
- Banking of road provides the vehicles with the required centripetal force to take a safe turn.
- The concept also prevents vehicles from skidding.
- It also prevents vehicles from overturning or toppling.
Things To Remember
- Banking of road provides the centripetal force that is required for a vehicle to take a safe turn on a curved road.
- It also avoids skidding and prevents overturning or toppling of the vehicle.
- The angle at which the banked road is inclined is called the Banking angle.
- Banking of Road is also present on railway tracks and on the airplane runway so that the trains can take a safer turn and the aircraft wings can get in a proper horizontal position.
- Banking is impossible on a perfectly flat, smooth road because no turning is involved.
- The race tracks have large angles to allow great speeds.
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Previous Year Questions
Sample Questions
Ques. A curve has a radius of 50 meters and a banking angle of 15º. What is the ideal, or critical, speed (the speed for which no friction is required between the car's tires and the surface) for a car on this curve? (3 marks)
Ans. Here, radius of curve, r = 50 m
banking angle, θ = 15º
free-fall acceleration, g = 9.8 m/s2
We have to find out the ideal speed v (the speed for which no friction is required between the car's tires and the surface)
From the free-body diagram for the car:-
Fnet = Fcentripital
mg tanθ = mv2/r
v2 = rg tanθ
v = √rg tanθ
= √(50 m) (9.8 m/s2) (tan 15º) = 11 m/s
If the car has a speed of about 11 m/s, it can negotiate the curve without any friction.
Ques. What happens when the banking angle is zero? (1 mark)
Ans. When the banking angle is zero degree, the road tends to be flat.
Ques. A 1200 kg automobile rounds a level curve of radius 200 m, on an unbanked road with a velocity of 72 km/hr. What is the minimum coefficient of friction between the tyres and road in order that the automobile may not skid. (g = 10 m/s2) (3 marks)
Ans. In an unbanked road, the centripetal force is provided by the frictional force.
So, ffriction = mv2/r But flimiting friction > ffriction
or, μmg = ffriction or μmg = mv2/r
So, μmin = v2/gr = (20×20)/(10×200) = 0.2.
The bank angle is the angle at which the vehicle is inclined about its longitudinal axis with respect to the plane of its curved path.
If the force of friction is not strong enough, the vehicle will skid.
Even if there is very little force of friction the vehicle can still go round the curve with no tendency to skid.
Roads are banked because of the inertia of vehicles driving on the road.
Ques. What do you mean by banking of roads? (2 marks)
Ans. The process of raising the outer edge of a road over its inner edge through a certain angle is known as banking of road. The angle at which the road is banked is called the banking angle.
Ques. A turn of radius 100 m is being designed for a speed of 25 m/s. At what angle should the turn be banked? (3 marks)
Ans. Here, radius of turn, r = 100 m
speed of the car, v = 25 m/s
free-fall acceleration, g = 9.8 m/s2
We have to find out the bank angle, θ.
From the free body diagram of the car,
Fnet = Fcentripetal
mg tanθ = mv2/r
tanθ = \(\frac{v^2}{rg}\)
θ = tan-1(v2/rg)
= tan-1 [(25 m/s)2/ (100 m) (9.8 m/s2)] = 33º
So, the banking angle should be about 33º.
Ques. Consider a circular road of radius 20 meter banked at an angle of 15 degree. With what speed a car has to move on the turn so that it will have a safe turn? (2 marks)
Ans.
The safe speed for the car on this road is 7.1 m per second
Ques. Is Banking of roads possible on a smooth road? (1 mark)
Ans. No. Banking of roads is not possible on a smooth road as it does not have friction.
Ques. The radius of a circular road course track is 500 m and its angle of banking is 10°. Take the coefficient of friction between the road and tyre as 0.25. Calculate the speed, upto which slipping can be avoided. (3 marks)
Ans. The following information is given:
Radius = 500 m
Angle of banking = 10o
Coefficient of friction = 0.25
From the formula derived above,
\(v =\sqrt{ \frac{rg(tan\theta +\mu_s)}{1-\mu_stan\theta}}\)
= \(\sqrt{\frac{500\times9.8\times(tan10^o +0.25)}{(1-0.25tan10^o)}}\)
tan10o = 0.1763
Computing the above value,
v = 46.74 m/s
Ques. Civil engineers generally bank curves on roads in such a manner that a car going around the curve at the recommended speed does not have to rely on friction between its tires and the road surface in order to round the curve. Suppose that the radius of curvature of a given curve is r=60m , and that the recommended speed is v = 40kmph. At what angle should the curve be banked? (2 marks)
Ans. We know that,
\(\theta = \tan^-1 (\frac{v^2}{rg})\)
Hence,
\(\theta = \tan^{-1} (\frac{(\frac{40*1000}{3600})^2}{60*9.81}) =11.8^o\)
Ques. Why is it necessary to bank the roads? (2 marks)
Ans. When a bike or any vehicle moves in a banking road, necessary centripetal force is supplied by the frictional force between wheels and the surface of the road, this way the vehicle safely moves in the curve path, it doesn't skid. So, in a simple way, Due to the lack of centripetal force, vehicles tend to skid. To safely drive on a circular road, banking of the road is necessary .
Ques. How is banking of roads used in trains and aircrafts? (2 marks)
Ans. Banking is also present on the railway tracks and on the aircrafts so that the trains can take a safer turn while the aircraft wings get in a proper horizontal position.
Ques. Two turns of a smooth road are banked with the same angle. What is the ratio of maximum velocities for the two turns if the ratio of radii of curvature is 1:3? (3 marks)
Ans. The expression of maximum velocity v,
- v=√grtanθ
Here, g is the gravitational acceleration, ris the radius of curvature, and the banking angle. Since the two turns have the same banking angle, the ratio of maximum velocities
- v1 / v2 = √r1 / r2
where r1 and r2 are the radii of curvature with their ratio being
- r1 / r2 = 1/3
Hence the ratio of velocities is,
- v1 / v2 = √1/3
Ques. A vehicle can have a maximum velocity of 72 km/hr on a smooth turn of radius 100 m. What is the banking angle? (3 marks)
Ans. Maximum velocity, Vm = 54 km/hr = 15 m/s
The radius of banking r = 100 m
Gravitational acceleration g = 10 m/s2
From the expression of maximum velocity: Vm = √grtanθ
- tanθ = v2m/gr
- tanθ = 15 x 15 / 10 x 1000
- tanθ = 0.0225
θ = 1.28893756 degree which is our required banking angle
Ques. A circular race track of radius 100 m is banked at an angle of 45°. What is the (3 marks)
(a) Optimum speed of a race car to avoid wear and tear on its tyres?
(b) Maximum permissible speed to avoid slipping if the coefficient of friction is 0.2?
Ans. The maximum permissible speed on a banked road is given by
\(v_{max} = (Rg \frac{\mu _s + tan \theta}{1- \mu_s tan \theta})^{1/2}\)
Where
- R is the radius of the circular track
- g is the acceleration due to gravity
- θ is the angle of the banked
- µs is the coefficient of friction
- To avoid wear and tear and get optimum speed friction force should be zero i.e. µs = 0.
Therefore the above equation becomes
v0 = (Rg tanθ)1/2
On substituting the value, we get
v0 = (100 x 9.8 x tan 45)1/2 = 31.3 m/s
- Maximum permissible speed to avoid slipping is given by
\(v_{max} = (Rg \frac{\mu _s + tan \theta}{1- \mu_s tan \theta})^{1/2}\)
On substituting the values, we get
vmax = (100 x 9.8 x \(\frac{0.2 + tan45}{1- 0.2 \times tan45}\))1/2
vmax = 38.34 m/s
Ques. A cyclist speeding at 4.5 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The coefficient of static friction between the road and the tyres is 0.1. (3 marks)
(A) Will he slip while taking the turn?
(B) Will he slip if his speed is 9 km/h?
Ans. Friction force provides the necessary centripetal force. He will slip if the turn is too sharp i.e. if the radius of the turn is too small or if his speed is too large.
Maximum speed for not slipping is given by
vmax = √(µsgR)
On substituting the given values, we get
vmax = √(0.1 x 9.8 x 3) = 1.71 m/s
- For radius, R = 3 m, maximum speed should be 1.71 m/s. But his speed is 4.5 km/h or 1.25 m/s. Therefore, he will not slip.
- Since the maximum permissible speed is 1.71 m/s, if his speed is 9 km/h i.e. 2.5 m/s, then he will slip.
Ques. The angle of banking for a cyclist taking a turn at a curve is given by tanθ = vn/rg where symbols have their usual meaning. The value of n is (2 marks)
(a) 1
(b) 2
(c) 3
(d) 4
Ans. The correct answer is b. 2
Explanation: For a perfectly smooth road (i.e. µs = 0), the optimum speed is given by
v = (rg tanθ)1/2
On squaring both sides, we get
v2 = rg tanθ
⇒ tanθ = v2/rg
Ques. Find the maximum speed with which a car can turn on a bend without skidding. If the radius of the bend is 20 m and the coefficient of friction between the road and tyres is 0.4. (2 marks)
Ans. Given
- Coefficient of friction between the tyres and road, µs = 0.4
- The radius of the circular bend, R = 20 m
The maximum speed with which a car can turn on a bend without skidding is given by
vmax = √(µsgR)
On substituting the given values, we get
vmax = √(0.4 x 9.8 x 20) = 8.85 m/s
Ques. When a car moves on a banked road, the centripetal force is due to (2 marks)
(a) Frictional force
(b) Vertical component of the normal force and the frictional force
(c) The horizontal component of the normal force
(d) The horizontal component of the normal force and the frictional force
Ans. The correct answer is d. The horizontal component of the normal force and the frictional force
Explanation: When a car moves on a banked road, the horizontal component of normal reaction force and frictional force provides the necessary centripetal force.
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