Chapter 6.13 - Vertical Circular Motion

In the previous section, we saw the potential energy when a spring is stretched/extended vertically. In this section we will see motion of objects in a vertical circle.

1. In fig.6.43(a) below, a small object of mass ‘m’ is moving along a vertical circle

Fig.6.43

• Four points are marked on the circular path.  
    ♦ A and C are the lowest and highest points
    ♦ B and D are the quadrant points
2. Let us analyze the travel of the object along the circular path ABCD
• Wherever be the position of the object along ABCD, it will be experiencing the downward acceleration due to gravity ‘g’
• When the object travels upwards from A to C along B, this g will be negative (opposite to the direction of motion) 
    ♦ So the linear speed ‘v’ of the object goes on decreasing
• When the object travels downwards from C to A along D, this g will be positive
    ♦ So the ‘v’ of the object goes on increasing
■ We can write: 
When an object is in a vertical circular motion, it’s linear speed ‘v’ changes continuously
■ Also we can write: 
When an object is in a vertical circular motion, since it’s linear speed is not uniform, it is not a uniform circular motion (UCM).
• We have seen the details about UCM in a previous chapter (Details here) 
3. When we discussed about UCM, we considered the motion in a horizontal circle. 
• When the circle is horizontal, there is no 'change in height' for the object
    ♦ So the effect due to g will be same at every point along the path
• Thus there is no change in linear speed ‘v’
    ♦ As a consequence, the centripetal acceleration $\mathbf\small{\frac{v^2}{r}}$ will be uniform
    ♦ The centripetal force $\mathbf\small{\frac{mv^2}{r}}$ also will be uniform
4. But in our present case, we saw that the linear speed ‘v’ changes continuously.  
    ♦ So the centripetal acceleration $\mathbf\small{\frac{v^2}{r}}$  changes continuously
    ♦ The centripetal force $\mathbf\small{\frac{mv^2}{r}}$ will also change continuously
5. Another difference between UCM and ‘motion in vertical circle’:
• In UCM, the linear speed ‘v’ remains constant. So there is no linear acceleration
• In the case of ‘motion in vertical circle’ the ‘v’ changes continuously. So there will be linear acceleration also in addition to centripetal acceleration

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We can write a summary:

■ UCM:
• Centripetal acceleration only
    ♦ Magnitude of this centripetal acceleration is a constant
• No linear acceleration
■ Motion in vertical circle:
• Centripetal acceleration is present
    ♦ Magnitude of this centripetal acceleration is not constant
• Linear acceleration is also present
    ♦ If there is no force other than gravity, magnitude of this linear acceleration is constant

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Now we will continue the discussion:

6. If a person ties an object of mass ‘m’ to the end of a string and whirl it in a vertical circle (fig.6.43.b), is he doing any work on the object? Let us analyze:
(i) Fig.6.43(c) above shows the force and velocity when the object is at any point ‘P’
• The only contact that the person have with the object is ‘through the string’
• If he wants to do any work on the object, it will be through the string  
(ii) The tension ‘T’ in the string will always be in a radial direction
• The linear speed ‘v’ will always be tangential to the path
(iii) Any tangent is perpendicular to the radius
• So the force ‘T’ is perpendicular to the direction of motion
(v) That means, ‘T’ cannot do any work on the object. (Details here)
7. The only external force that is capable of doing any work in this situation is the ‘force of gravity’
• ‘Force of gravity’ is a conservative force
    ♦ It does not depend on the path
    ♦ The work done by it can be stored up for later use
• So the law of conservation of energy can be applied to motion in a vertical circle
• That is., E_i = E_f
8. Let us apply it to the points A and B
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write E_A = E_B
9. Let us write the various energies:
(i) Total energy E_A:
• Kinetic energy = $\mathbf\small{\frac{m\,v_A^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_A}$ 
    ♦ Where ‘h_A’ is the height of the point ‘A’ from the datum 
    ♦ In the previous section we saw that, the datum can be drawn through any convenient point
    ♦ Let us assume that the datum passes through ‘A’ itself. Then h_A = 0
    ♦ So potential energy at A = 0
(Remember that, once the datum is fixed, we must follow it for the whole problem)
■ So total energy E_A = $\mathbf\small{\frac{m\,v_A^2}{2}}$
(ii) Total energy E_B
• Kinetic energy = $\mathbf\small{\frac{m\,v_B^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_B}$ 
    ♦ Where ‘h_B’ is the height of the point ‘B’ from the datum 
    ♦ From the fig., we can see that h_B = r, the radius of the circle
■ So total energy E_B = $\mathbf\small{\frac{m\,v_B^2}{2}+mgr}$
10. Equating the two energies we get:
$\mathbf\small{\frac{m\,v_A^2}{2}=\frac{m\,v_B^2}{2}+mgr}$
Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{v_A^2=v_B^2+2gr}$
$\mathbf\small{\Longrightarrow v_B=\sqrt{v_A^2-2gr}}$
■ So, if we know the velocity v_A at the lower most point A, we can easily obtain the velocity v_B at the quadrant point B on the right side

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11. Next, let us apply the law of conservation of energy to the points A and C
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write E_A = E_C
12. Let us write the various energies:
(i) Total energy E_A:
• Kinetic energy = $\mathbf\small{\frac{m\,v_A^2}{2}}$
• Gravitational potential energy = 0
■ So total energy E_A = $\mathbf\small{\frac{m\,v_A^2}{2}}$
(ii) Total energy E_C
• Kinetic energy = $\mathbf\small{\frac{m\,v_C^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_C}$ 
    ♦ Where ‘h_C’ is the height of the point ‘C’ from the datum 
    ♦ From the fig., we can see that h_C = 2r, the diameter of the circle
■ So total energy E_B = $\mathbf\small{\frac{m\,v_C^2}{2}+2mgr}$
13. Equating the two energies we get:
$\mathbf\small{\frac{m\,v_A^2}{2}=\frac{m\,v_B^2}{2}+2mgr}$
Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{v_A^2=v_C^2+4gr}$
$\mathbf\small{\Longrightarrow v_C=\sqrt{v_A^2-4gr}}$
■ So, if we know the velocity v_A at the lower most point A, we can easily obtain the velocity v_C at the top most point C

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14. Next, let us apply the law of conservation of energy to the points A and D
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write E_A = E_D
15. Let us write the various energies:
(i) Total energy E_A:
• Kinetic energy = $\mathbf\small{\frac{m\,v_A^2}{2}}$
• Gravitational potential energy = 0
■ So total energy E_A = $\mathbf\small{\frac{m\,v_A^2}{2}}$
(ii) Total energy E_D
• Kinetic energy = $\mathbf\small{\frac{m\,v_D^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_D}$ 
    ♦ Where ‘h_D’ is the height of the point ‘D’ from the datum 
    ♦ From the fig., we can see that h_D = r, the radius of the circle
■ So total energy E_D = $\mathbf\small{\frac{m\,v_D^2}{2}+mgr}$
16. Equating the two energies we get:
$\mathbf\small{\frac{m\,v_A^2}{2}=\frac{m\,v_D^2}{2}+mgr}$
• Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{v_A^2=v_D^2+2gr}$
$\mathbf\small{\Longrightarrow v_D=\sqrt{v_A^2-2gr}}$
■ So, if we know the velocity v_A at the lower most point A, we can easily obtain the velocity v_D at the quadrant point D on the left side
■ Also it is interesting to note that magnitudes of v_B and v_D are the same
• However, they are opposite in directions
    ♦ v_B is upwards
    ♦ v_D is downwards

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17. So we successfully calculated the velocities at the two quadrant points and the top most point.
• What about the velocities at the intermediate points?
• For that, we will have to use trigonometry. Let us see how it is done:
18. Consider the instant when the object is at a point 'P' shown in fig.6.44(a) below:

Fig.6.44

• At that instant, the string makes an angle θ with the vertical
19. We want the vertical height (h_P) of 'P' above the datum
• The datum passes through 'A' 
• So we want the vertical height of 'P' above 'A'
20. Drop a perpendicular from P onto OA. This is shown in fig.6.44(b)
• Let 'Q' be the foot of this perpendicular
• Then AQ = h_P. In the right triangle OPQ, cos θ = ^OQ⁄_OP
• So OQ = OP cos θ = r cos θ
• Thus we get: h_P = AQ = (OA - OQ) = (r - r cos θ) = r(1- cos θ)

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• Now that we know the height of 'P' above the datum, we can apply the law of conservation of energy to the two points A and P
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write E_A = E_P
22. Let us write the various energies:
(i) Total energy E_A:
We have already calculated it as: $\mathbf\small{\frac{m\,v_A^2}{2}}$
(ii) Total energy E_P
• Kinetic energy = $\mathbf\small{\frac{m\,v_P^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_P}$ 
    ♦ Where ‘h_P’ is the height of the point ‘P’ from the datum 
    ♦ From the fig., we have calculated it as: r(1- cos θ)
■ So total energy E_P = $\mathbf\small{\frac{m\,v_P^2}{2}+mgr(1-\cos \theta)}$
23. Equating the two energies we get:
$\mathbf\small{\frac{m\,v_A^2}{2}=\frac{m\,v_P^2}{2}+mgr(1-\cos \theta)}$
• Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{v_A^2=v_P^2+2gr(1-\cos \theta)}$
$\mathbf\small{\Longrightarrow v_P^2=v_A^2-2gr(1-\cos \theta)}$
$\mathbf\small{\Longrightarrow v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$
■ So, if we know the velocity v_A at the lower most point A, we can easily obtain the velocity v_P at any point P on the circle. The only additional information required is the angle which OP makes with the vertical

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• If we can use the expression derived in (23) above for 'any point', it must be applicable to points B, C and D also. Let us check:
24. First we will check the equation at point 'B' 
• We have: $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$ 
• When the object is at B, θ = 90^o
    ♦ So cos θ = cos 90 = 0
• Substituting the known values, we get:
$\mathbf\small{v_B=\sqrt{v_A^2-2gr(1-0)}}$
$\mathbf\small{\Longrightarrow v_B=\sqrt{v_A^2-2gr}}$
■ This is the same expression that we obtained in (10)
25. Next we will check the equation at point 'C' 
• We have: $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$ 
• When the object is at C, θ = 180^o
    ♦ So cos θ = cos 180 = -1
• Substituting the known values, we get:
$\mathbf\small{v_C=\sqrt{v_A^2-2gr(1-(-1))}}$
$\mathbf\small{\Longrightarrow v_C=\sqrt{v_A^2-2gr \times 2}}$
$\mathbf\small{\Longrightarrow v_C=\sqrt{v_A^2-4gr}}$
■ This is the same expression that we obtained in (14)
26. Finally, we will check the equation at point 'D' 
• We have: $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$ 
• When the object is at C, θ = 270^o
    ♦ So cos θ = cos 270 = 0
• Substituting the known values, we get:
$\mathbf\small{v_D=\sqrt{v_A^2-2gr(1-0)}}$
$\mathbf\small{\Longrightarrow v_D=\sqrt{v_A^2-2gr}}$
■ This is the same expression that we obtained in (16)

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In the next section, we will see tension in the string at various points.

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Chapter 5.22 - The Centripetal Force

In the previous section, we saw rolling friction. In this section, we will see centripetal force.

• Consider fig.5.88(a) below. A small object of mass ‘m’ is tied to one end of an inextensible string.

Fig.5.88

• A person holds the other end of the string and whirls the object around in a horizontal plane.
    ♦ Obviously, the path followed by the object will be a circle. 
    ♦ The radius of that circle will be equal to the length of the string. 
• Let the person whirl it in such a way that, the motion is uniform. This is shown in fig.b 
• We have seen 'uniform circular motion' in a previous section. 
• We want to know the forces acting on the object which is under a 'uniform circular motion'. 
• We will write the analysis in steps:
1. When the object moves around the circle, the string will be in tension. 
• Remember that, a string can take only tension. It cannot take compression. We have seen the details in a previous section. 
2. So we can draw two opposite arrows on the string. This is shown in fig.c. 
• The two arrows tell us these:
    ♦ The string pulls at the hand
    ♦ The string pulls at the object
3. Let us choose the object as a sub-system. This is shown by the red rectangle in fig.c
• The FBD will be as shown in fig.d
• We see that the tension in the string is pulling the object towards the center of the circle
4. We know that FBDs show only those forces which are experienced by the sub-systems.
• They do not show forces which are applied by the sub-systems
• So we can write: The object experiences a pull $\mathbf\small{\vec{T}}$ towards the center
5. Let the object be at any position along the circle. We can draw the FBD at that position.
• Wherever (along the circle) be the position of the object, We will see the same $\mathbf\small{\vec{T}}$ in its FBD.
• That means, the object is subjected to a constant pull $\mathbf\small{\vec{T}}$ towards the center.
6. If we cut the string while the circular motion is on, the $\mathbf\small{\vec{T}}$ will disappear. (see fig.5.89.a below)

Fig.5.89

• This is because, when the string is cut, the medium to provide $\mathbf\small{\vec{T}}$ is no longer available
7. The object will then fly off in a straight line.
• This straight line will be tangential to the circle.
• The ‘point of tangency’ is the ‘position of the object at the exact instant when the cut was made’. This is shown in fig.5.89(b) above.
■ So we see that a ‘force towards the center’ is essential to maintain circular motion.
• We want the details of this force.
8. Earlier, we have seen centripetal acceleration $\mathbf\small{\vec{a_c}}$ (Details here)
• Its magnitude is given by $\mathbf\small{|\vec{a_c}|=\frac{v^2}{R}}$
    ♦ Where v is the speed
    ♦ R is the radius of the circle
• Its direction changes continuously, being always directed towards the center
9. By the second law, any object subjected to acceleration will experience a force $\mathbf\small{\vec{F}}$
• The magnitude of this force is given by $\mathbf\small{|\vec{F}|=\text{mass}\,\times\,|\vec{a}|}$
10. Now, our object is also subjected to acceleration $\mathbf\small{\vec{a_c}}$
• So a force must be acting on it. It is known as the centripetal force and is denoted as $\mathbf\small{\vec{f_c}}$
• Obviously, the magnitude of $\mathbf\small{\vec{f_c}}$ will be given by:
Eq.5.5:  $\mathbf\small{|\vec{f_c}|=\frac{mv^2}{R} }$
11. The direction of $\mathbf\small{\vec{f_c}}$ will be the same as the direction of $\mathbf\small{\vec{a_c}}$
• That means the direction of the $\mathbf\small{\vec{f_c}}$ is changing continuously. It is always directed towards the center
12. So in fig.5.88(d) above, $\mathbf\small{|\vec{T}|=|\vec{f_c}|}$
• No object can perform a circular motion without the help of $\mathbf\small{\vec{f_c}}$
• In fact, if an object has to move along even a small part of a curve, the centripetal force is to be provided
■ For planets moving around the sun, the centripetal force is provided by the gravitational force
■ What about a car moving along a curved path? We will see it in detail:

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The motion of a car along a circular road
Two cases come under this topic:
Case 1: The road is level
Case 2: The road  is banked
We will see each case in detail:
Case 1: The motion of a car on a level road
1. Fig.5.90(a) below shows a car moving along a circular road. 
• It is moving along the center-line of the road. The center-line is shown using a dashed curve.

Fig.5.90

• The radius of the center-line of the road is 'R'.
2. Now, centripetal force must be provided for the car so that it can perform the circular motion.
■ Where does this force come from?
• The forces acting on the car are shown in fig.b
• The vertical forces will cancel each other. 
• There will be a frictional force, which is horizontal. 
• This force acts at the interface between the tire and the surface of the road.
■ It is this frictional force that provides the necessary centripetal force.
3. We know that there are two types of friction: static and kinetic
• It is the static friction that provides the centripetal force. 
• The reason can be written in the following steps (i) to (iv) 
(i) The car will be always trying to move off in a linear direction
(ii) But friction keeps the car in the circular path
(iii) That means friction opposes the impending linear motion. 
(iv) So it is the static friction which is playing the role here.  
4. We know that 'external force opposing the static frictional force' must be always less than or equal to $\mathbf\small{\mu_s |\vec{F_N}|}$
Otherwise, the static frictional force will no longer be able to hold (Details here)
5. In our present case, the force which opposes the static frictional force is the centripetal force $\mathbf\small{\vec{f_c}}$
So we can write: $\mathbf\small{|\vec{f_c}|\leq \mu_s |\vec{F_N}|}$
$\mathbf\small{\Longrightarrow \frac{mv^2}{R}\leq \mu_s |\vec{F_N}|}$
$\mathbf\small{\Longrightarrow \frac{mv^2}{R}\leq \mu_s \times mg}$
$\mathbf\small{\Longrightarrow \frac{v^2}{R}\leq \mu_s \times g}$
$\mathbf\small{\Longrightarrow {v^2}\leq \mu_s \times Rg}$
$\mathbf\small{\Longrightarrow {v}\leq \sqrt{\mu_s  Rg}}$
6. In the final expression that we obtained in the step (5) above, μ_s, R and g are constants.
• 'v' is the only variable. This is obvious because a car can move with variable speeds. The radius of the road will remain the same
• So it is important to keep the speed less than or equal to $\mathbf\small{\sqrt{\mu_s  Rg}}$
• Otherwise, the friction capacity will be exceeded. The tire will not be able to provide the necessary centripetal force. And the car will skid off from the circular path causing an accident.
7. So drivers must be warned about the maximum speed they can attain on a curved road
• From the result in step (5), we see that the maximum allowable speed is $\mathbf\small{\sqrt{\mu_s  Rg}}$
• We can write:
Eq.5.6: $\mathbf\small{v_{max}=\sqrt{\mu_s  Rg}}$
• This maximum allowable speed is displayed on the roadsides in such a way that drivers from both directions see them much before they reach the curve. Some examples can be seen here.    
• From Eq.5.6, we see that $\mathbf\small{v_{max}}$ is independent of the mass of the car. That means the $\mathbf\small{v_{max}}$ obtained is applicable to heavy vehicles and light vehicles alike
• The factors on which $\mathbf\small{v_{max}}$ depends on are: μ_s, R and g 
• There are, however, many more factors to be considered before finalizing the value of $\mathbf\small{v_{max}}$. Those factors are detailed in Highway Engineering textbooks.

Case 2: The motion of a car on a banked road
1. In this case, the outer edge of the road is raised to a higher level than the inner edge. This is shown in fig.5.91(a) below:

Fig.5.91

2. We see that the force of friction $\mathbf\small{\vec{f_s}}$ is now inclined. 
• This is obvious because frictional force will always be parallel to the surface of contact. 
3. The other two forces are: 
• $\mathbf\small{\vec{W_{car}}}$ acting vertically downwards
• $\mathbf\small{\vec{F_N}}$ acting perpendicular to the surface of the road.
4. For our present analysis, we do not want any inclined forces. All forces that we bring into calculations should be either vertical or horizontal.
• $\mathbf\small{\vec{W_{car}}}$ is already vertical
• So we will resolve the inclined forces $\mathbf\small{\vec{F_N}}$ and $\mathbf\small{\vec{f_s}}$ into their rectangular components.
5. First, we take $\mathbf\small{\vec{F_N}}$. See fig.5.91(b)
(i) Draw horizontal and vertical dashed lines through the tail end of $\mathbf\small{\vec{F_N}}$
(ii) Identify the 'position of θ'. 
• By the properties of right triangles, we will see that θ is between $\mathbf\small{\vec{F_N}}$ and the vertical
(iii) Once θ is fixed, we can easily write the components (Details here):
    ♦ The component which is adjacent to the angle gets the cosine
    ♦ The other component gets the sine
■ Thus we get:
• The horizontal component as: $\mathbf\small{(|\vec{F_N}|\sin \theta)\hat{i}}$
• The vertical component as: $\mathbf\small{(|\vec{F_N}|\cos \theta)\hat{j}}$
• They are shown in fig.c
6. Next, we take $\mathbf\small{\vec{f_s}}$. See fig.5.91(d)
(i) Draw horizontal and vertical dashed lines through the tail end of $\mathbf\small{\vec{f_s}}$
(ii) Identify the 'position of θ'. 
• By the properties of right triangles, we will see that θ is between $\mathbf\small{\vec{f_s}}$ and the horizontal
(iii) Once θ is fixed, we can easily write the components:
    ♦ The component which is adjacent to the angle gets the cosine
    ♦ The other component gets the sine
■ Thus we get:
• The horizontal component as: $\mathbf\small{(|\vec{f_s}|\cos \theta)\hat{i}}$
• The vertical component as: $\mathbf\small{(|\vec{f_s}|\sin \theta)\hat{j}}$
• They are shown in fig.e
7. So both the inclined forces are now resolved into their respective components.
• In fig.f, those components are arranged in order. So now we can see their effects:  
(i) There is no motion in the vertical direction. So the vertical components can be related as:
$\mathbf\small{|\vec{F_N}|\cos \theta=|\vec{f_s}|\sin \theta+mg}$
(ii) The two horizontal components together provide the necessary centripetal force. So we can write:
$\mathbf\small{|\vec{F_N}|\sin \theta+|\vec{f_s}|\cos \theta=\frac{mv^2}{R}}$
8. So we see that, both $\mathbf\small{\vec{F_N}}$ and $\mathbf\small{\vec{f_s}}$ contributes towards providing the centripetal force
• Now, the maximum contribution that can be expected from $\mathbf\small{\vec{f_s}}$ is $\mathbf\small{\mu_s |\vec{F_N}|}$
• When this maximum possible contribution is obtained, the car will be able to move with the maximum allowable velocity $\mathbf\small{v_{max}}$
• So the result in 7(ii) can be written as: $\mathbf\small{|\vec{F_N}|\sin \theta+\mu_s|\vec{F_N}|\cos \theta=\frac{mv_{max}^2}{R}}$
$\mathbf\small{\Longrightarrow|\vec{F_N}|(\sin \theta+\mu_s\cos \theta)=\frac{mv_{max}^2}{R}}$
9. The previous result in 7(i) can also be rewritten in terms of μ_s
$\mathbf\small{|\vec{F_N}|\cos \theta=\mu_s|\vec{F_N}|\sin \theta+mg}$
$\mathbf\small{\Longrightarrow|\vec{F_N}|(\cos \theta-\mu_s\sin \theta)=mg}$
$\mathbf\small{\Longrightarrow|\vec{F_N}|=\frac{mg}{(\cos \theta-\mu_s\sin \theta)}}$
10. In (9) above, we get an expression for $\mathbf\small{|\vec{F_N}|}$. We can use this expression in the place of $\mathbf\small{|\vec{F_N}|}$ in (8). We get:
$\mathbf\small{\frac{mg}{(\cos \theta-\mu_s\sin \theta)}\times(\sin \theta+\mu_s\cos \theta)=\frac{mv_{max}^2}{R}}$
$\mathbf\small{\Longrightarrow\frac{g(\sin \theta+\mu_s\cos \theta)}{(\cos \theta-\mu_s\sin \theta)}=\frac{v_{max}^2}{R}}$
• Dividing numerator and denominator by cos θ, we get:
$\mathbf\small{\frac{g(\mu_s+\tan \theta)}{(1-\mu_s\tan \theta)}=\frac{v_{max}^2}{R}}$
$\mathbf\small{\Longrightarrow {v_{max}^2}=\frac{Rg(\mu_s+\tan \theta)}{(1-\mu_s\tan \theta)}}$
• Thus we get Eq.5.7: $\mathbf\small{ {v_{max}}=\left[\frac{Rg(\mu_s+\tan \theta)}{1-\mu_s\tan \theta}\right]^{\frac{1}{2}}}$
11. If the banked road is friction less, the required centripetal force will have to be provided by the 'banking effect' alone. There will not be any contribution from the friction
Let us write the steps briefly:
(i) Fig.5.91(f) will have only two vertical forces:
• $\mathbf\small{(|\vec{F_N}|\cos \theta)\hat{j}}$ (upwards)
• $\mathbf\small{\vec{W_{car}}}$ (downwards)
(ii) There is no motion in the vertical direction. So we can write:
$\mathbf\small{|\vec{F_N}|\cos \theta=mg}$
(iii) Also, fig.5.91(f) will have only one horizontal force:
• $\mathbf\small{|\vec{F_N}|\sin \theta}$
(iv) This horizontal component provide the necessary centripetal force. So we can write:
$\mathbf\small{|\vec{F_N}|\sin \theta=\frac{mv_0^2}{R}}$
• Where v₀ is the maximum velocity of the car when no contribution from friction is received
(v) From (ii) we have: $\mathbf\small{|\vec{F_N}|=\frac{mg}{\cos \theta}}$
We can use this in the place of $\mathbf\small{|\vec{F_N}|}$ in (iv). We get:
$\mathbf\small{\frac{mg}{\cos \theta} \times \sin \theta=\frac{mv_0^2}{R}}$
$\mathbf\small{\Longrightarrow g \tan \theta=\frac{v_0^2}{R}}$
$\mathbf\small{\Longrightarrow {v_0^2}=Rg \tan \theta}$
Thus we get Eq.5.8:
$\mathbf\small{{v_0}=(Rg \tan \theta})^{\frac{1}{2}}$
• Note that, this result can be easily obtained by putting μ_s = 0 in Eq.5.7
• If a car move with this velocity on a banked road, friction at the contact surface will not be utilized. Thus wear and tear of the tires will be kept to a minimum.  

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Let us write a comparison of the two important results that we saw in the above discussion  
■ When the road is level, we have Eq.5.6: $\mathbf\small{v_{max}=\sqrt{\mu_s  Rg}}$ 
■ When the road is banked, we have Eq.5.7: $\mathbf\small{ {v_{max}}=\left[\frac{Rg(\mu_s+\tan \theta)}{1-\mu_s\tan \theta}\right]^{\frac{1}{2}}}$

• In both the equations there are three items on the right side

• In both the equations, all the three items come under 'square root'

• In both the equations, R and g are present. The difference is in the third item only

    ♦ In Eq.5.6, the third item is μ_s.

    ♦ In Eq.5.7, the third item is $\mathbf\small{\frac{(\mu_s+\tan \theta)}{1-\mu_s\tan \theta}}$

■ $\mathbf\small{\frac{(\mu_s+\tan \theta)}{1-\mu_s\tan \theta}}$ will be always greater than μ_s.

■ So we can write: When banking is provided, the car will be able to move with a greater speed

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In the next section, we will see some solved examples

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