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.
• 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)
• 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:
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.
• 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, μs, 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: μs, 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:
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 μs
$\mathbf\small{|\vec{F_N}|\cos \theta=\mu_s|\vec{F_N}|\sin \theta+mg}$
$\mathbf\small{\Longrightarrow|\vec{F_N}|(\cos \theta-\mu_s\sin \theta)=mg}$
$\mathbf\small{\Longrightarrow|\vec{F_N}|=\frac{mg}{(\cos \theta-\mu_s\sin \theta)}}$
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 μs = 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.
• Consider fig.5.88(a) below. A small object of mass ‘m’ is tied to one end of an inextensible string.
 |
| Fig.5.88 |
♦ 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 |
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:
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 |
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, μs, 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: μs, 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 |
• 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 μs
$\mathbf\small{|\vec{F_N}|\cos \theta=\mu_s|\vec{F_N}|\sin \theta+mg}$
$\mathbf\small{\Longrightarrow|\vec{F_N}|(\cos \theta-\mu_s\sin \theta)=mg}$
$\mathbf\small{\Longrightarrow|\vec{F_N}|=\frac{mg}{(\cos \theta-\mu_s\sin \theta)}}$
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 μs = 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.
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: $\mathbf\small{ {v_{max}}=\left[\frac{Rg(\mu_s+\tan \theta)}{1-\mu_s\tan \theta}\right]^{\frac{1}{2}}}$
■ 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 μs.
♦ 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 μs.
■ So we can write: When banking is provided, the car will be able to move with a greater speed
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