Chapter 7.37 - Maximum allowable force on Cart wheels

In the previous section, we saw the maximum acceleration with which a car can be set into motion. In this section we will see the case of a cart. 
• We have seen that, a cart needs external force to move. If this force is too large, the wheels will slip
• So there is a maximum allowable force. Our next aim is to find the magnitude of this allowable force

1. Consider the wheel shown in fig.7.146(a) below
It has a mass of 7 kg. It's radius is 0.35 m

Fig.7.146

𝝁_s = 0.8 and 𝝁_k = 0.6
2. A force is being applied at it's center
• This force is denoted as $\mathbf\small{|\vec{F}_{max}|}$
• This is because, it is the maximum force allowable. If a greater force is applied, the wheel will slide instead of rolling
3. Fig.b shows the remaining forces also which are acting on the wheel
• The weight $\mathbf\small{m|\vec{g}|}$ acts downwards at the center
• The normal force $\mathbf\small{|\vec{N}|}$ acts upwards from the point of contact with the ground
• We have the frictional force $\mathbf\small{|\vec{f_s}|=\mu_s|\vec{N}|=\mu_sm|\vec{g}|}$
• We use the static friction because, we want the wheel to be in 'pure rolling' motion
• Also we know that, for a cart wheel, the frictional force is opposite to the direction of motion
4. This frictional force is the 'cause of the torque' which rotates the wheel
• Because, it is the only force which does not pass through the center of the wheel
5. Now we will do the further calculations as two sets
• Set 1 is related to rotational motion
• Set 2 is related to translational motion
6. Set 1:
(i) The frictional force is the 'cause of the torque'
• So we have: $\mathbf\small{|\vec{\tau}|=|\vec{f_s}|R=\mu_sm|\vec{g}|R}$
• $\mathbf\small{\mu_sm|\vec{g}|R}$ is the maximum frictional force possible
• So we can write: $\mathbf\small{|\vec{\tau}_{max}|=\mu_sm|\vec{g}|R}$
(ii) But $\mathbf\small{|\vec{\tau}|=I\,|\vec{\alpha}|}$
• Where:
    ♦ $\mathbf\small{I}$ is the moment of inertia of the wheel
    ♦ $\mathbf\small{|\vec{\alpha}|}$ is the angular acceleration of the wheel
(iii) For pure rolling, we have eq.7.36 that we derived in the previous section: $\mathbf\small{|\vec{a}|=R\,|\vec{\alpha}|}$
• Rearranging this, we get: $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{a}|}{R}}$
(iv) For our present case, we must consider the maximum allowable angular acceleration
• So we must denote it as $\mathbf\small{|\vec{\alpha}_{max}|}$
• So we have: $\mathbf\small{|\vec{\alpha}_{max}|=\frac{|\vec{a}_{max}|}{R}}$
(v) Thus from (ii), we get: $\mathbf\small{|\vec{\tau}_{max}|=I\,\frac{|\vec{a}_{max}|}{R}}$
• Now, assuming the I of the wheel to be the same as that of the disc, we have: $\mathbf\small{I=\frac{mR^2}{2}}$
• So we get: $\mathbf\small{|\vec{\tau}_{max}|=\left(\frac{mR^2}{2}\right)\,\frac{|\vec{a}_{max}|}{R}=\frac{mR|\vec{a}_{max}|}{2}}$  
(vi) So the result in (i) becomes: $\mathbf\small{\frac{mR|\vec{a}_{max}|}{2}=\mu_sm|\vec{g}|R}$
• From this we get: $\mathbf\small{|\vec{a}_{max}|= 2\mu_s|\vec{g}|}$
7. Set 2:
We have: Net force = mass × acceleration
• In our present case, Net force = $\mathbf\small{|\vec{F}_{max}|-|\vec{f_s}|}$
⇒ Net force = $\mathbf\small{|\vec{F}_{max}|-\mu_s\,m|\vec{g}|}$
8. The two sets are complete
• Now we take the result from set 1 (vi)
• And apply it in the result from set 2
• We get:
Net force = $\mathbf\small{|\vec{F}_{max}|-\mu_s\,m|\vec{g}|=m\times2\mu_s|\vec{g}|}$
• Thus we get:
Eq.7.37: $\mathbf\small{|\vec{F}_{max}|=m\times2\mu_s|\vec{g}|+\mu_s\,m|\vec{g}|=3\mu_sm|\vec{g}|}$ 
Substituting the values, we get: $\mathbf\small{|\vec{F}_{max}|}$ = 3 × 0.8 × 7 × 9.8 = 164.64 N

Now we will see a solved example
Solved example 7.49
The wheel shown in fig.7.146(c) above, has a radius of 0.35 m and a mass of 6 kg. A horizontal force of 85 N is applied at it's center. What is the linear acceleration of the wheel?  𝝁_s = 0.4 and 𝝁_k = 0.3
Take g = 9.8 ms^-2  
Solution: 
1. First of all, we have to determine whether the wheel would slip or not, when the 85 N is applied
• For that, we find the maximum allowable force $\mathbf\small{|\vec{F}_{max}|}$ 
• If 85 N is greater than this maximum allowable force, the wheel will slip
2. We have Eq.7.37: $\mathbf\small{|\vec{F}_{max}|=m\times2\mu_s|\vec{g}|+\mu_s\,m|\vec{g}|=3\mu_sm|\vec{g}|}$
• Substituting the values, we get: $\mathbf\small{|\vec{F}_{max}|}$ = 3 × 0.4 × 6 × 9.8 = 70.56 N
3. So the applied force of 85 N is greater than the maximum allowable force. So the wheel will slip
4. If the wheel slips, static friction is not valid any more. We must consider only kinetic friction
• The kinetic frictional force acting on the wheel = $\mathbf\small{|\vec{f_s}|=\mu_km|\vec{g}|}$ 
• Substituting the values, we get:
Kinetic friction acting on the wheel = 0.3 × 6 × 9.8 = 17.64 N
5. So the net force acting on the wheel = (85 - 17.64) = 67.36 N
6. We have: Net force = mass × acceleration
• Thus we get: Acceleration of the wheel = ^{Net force}⁄_mass = ^67.36⁄₆ = 11.23 ms^-2.

Solved example 7.50
A horizontal force F acts at the center of a sphere of mass m. The coefficient of static friction between the sphere and the ground is 𝝁. What is the maximum allowable value of F so that, the sphere rolls without slipping ?
Solution:
1. Let f be the force of friction. it acts opposite to the direction of motion
• So the net force acting on the sphere = F-f
2. So the acceleration of the sphere = $\mathbf\small{a=\frac{F-f}{m}}$
3. For pure rolling, we have: $\mathbf\small{a=R\alpha}$
4. But $\mathbf\small{\alpha=\frac{\tau}{I}}$
$\mathbf\small{\Rightarrow\alpha=\frac{fR}{\frac{2mR^2}{5}}=\frac{5f}{2mR}}$
5. So from (3) we get: $\mathbf\small{a=r\alpha=R\times\frac{5f}{2mR}=\frac{5f}{2m}}$
6. So from (2), we get: $\mathbf\small{a=\frac{F-f}{m}=\frac{5f}{2m}}$
$\mathbf\small{\Rightarrow{F-f}=\frac{5f}{2}}$
$\mathbf\small{\Rightarrow F=\frac{7f}{2}}$
7. But f = 𝝁mg
So the result in (6) becomes: $\mathbf\small{F=\frac{7\mu\,mg}{2}}$
8. Thus we can write:
The applied force must be less than or equal to $\mathbf\small{\frac{7\mu\,mg}{2}}$

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So we have completed a discussion on the rotation of cart wheels which needs external force to propagate. Analysis of such wheels are applicable to 'front wheels of cars' also. In the next section, we will see the case of rolling along inclined planes

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Chapter 7.36 - Acceleration of a Car

In the previous section, we saw the case of a car in which the real wheels are self propagating. We also saw the case of a cart whose wheels need external force to rotate

1. Consider an independent circular disc. The word 'independent' is specified to clarify that, the disc has no external connections like to an axle or an engine
• It is kept at a height away from the ground. This is shown in fig.7.145 (a) below:

Fig.7.145

2. Assume that this disc has the ability to self propagate. Like a rear car wheel
• So let it spin. When it has attained considerable  angular speed, say $\mathbf\small{\vec{\omega}_i}$, stop the engine
3. If there is no frictional resistance or air resistance, that disc will continue to spin for ever with angular velocity $\mathbf\small{\vec{\omega}_i}$
• We know the reason:
If there is no external torque, angular momentum $\mathbf\small{I\vec{\omega}_i}$ will be conserved
4. Lower the disc slowly so that the disc gently touches the floor
• Care should be taken to ensure that, while placing the disc, no horizontal force is produced on the disc
5. Assume that the floor is a rough surface
• So there will be friction between the disc and the floor
• In the previous section, we have seen that in the case of a self propagating disc, the frictional force will be in the direction of motion
6. We have to determine whether this friction is static or kinetic
(i) The disc is continuously spinning. There is no instant at which any point on the periphery is at rest
(ii) Let the point of contact be P
(iii) Consider the instant at which the contact with the floor takes place. Even at that instant, the particle P is in motion
(iv) If it is to be at rest, the following two velocities must be equal:
    ♦ The tangential velocity of P, which is towards the left
    ♦ The velocity of the center of mass C, which is towards the right
(v) But at the instant of touch down, the 'velocity of the center of mass towards the right' is zero
(vi) So the point of contact is not at rest at the instant of touch down
(vii) So the frictional force is that of kinetic friction
(viii) This is shown as $\mathbf\small{|\vec{f_k}|}$ in the fig.b
7. The $\mathbf\small{|\vec{f_k}|}$ is acting on the disc. That means an external force is acting on the disc
• So the center of mass of the disc will have linear acceleration $\mathbf\small{|\vec{a}|}$ 
• We have: Force = mass × acceleration
So this acceleration can be obtained using the expression: $\mathbf\small{|\vec{a}|=\frac{|\vec{f_k}|}{m}}$
8. Note: For translational motion, we assume that the external net force acts at the center of mass
• So in our present case, the $\mathbf\small{\vec{f_k}}$ can be assumed to be acting at the center of the disc
• So the expression written in (7) is valid
9. Thus we obtained the linear acceleration $\mathbf\small{|\vec{a}|}$
• The initial linear velocity $\mathbf\small{|\vec{v}_i|}$ is zero
• So the linear velocity $\mathbf\small{|\vec{v}_{(t)}|}$ after a time interval of 't' s after touch down will be given by:
$\mathbf\small{|\vec{v}_{(t)}|=|\vec{v}_i|+|\vec{a}|\,t}$
$\mathbf\small{\Rightarrow|\vec{v}_{(t)}|=0+|\vec{a}|\,t}$
$\mathbf\small{\Rightarrow|\vec{v}_{(t)}|=|\vec{a}|\,t}$
10. Next we consider the angular motion
• The force $\mathbf\small{\vec{f_k}}$ acts on the disc
• But $\mathbf\small{\vec{f_k}}$ it does not pass through the center of the disc
• So the disc will experience a torque
• Obviously, this torque is given by: $\mathbf\small{|\vec{\tau}|=|\vec{f_k}|\,R}$ 
11. If there is a torque, there will be an acceleration $\mathbf\small{|\vec{\alpha}|}$
• This $\mathbf\small{|\vec{\alpha}|}$ can be obtained using the expression: $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}}$
• Recall that $\mathbf\small{\vec{\tau}=I\,\vec{\alpha}}$ corresponds to $\mathbf\small{\vec{F}=m\,\vec{a}}$
12. Thus we obtained the angular acceleration $\mathbf\small{|\vec{\alpha}|}$
• The initial angular speed is $\mathbf\small{\vec{\omega}_i}$
• So the angular speed $\mathbf\small{|\vec{\omega}_{(t)}|}$ after a time interval of 't' s after touch down will be given by:
$\mathbf\small{|\vec{\omega}_{(t)}|=|\vec{\omega}_i|-|\vec{\alpha}|\,t}$
■ Why is a negative sign given to $\mathbf\small{|\vec{\alpha}|}$?
• The answer can be written in 5 steps:
(i) The torque is created by $\mathbf\small{\vec{f_k}}$
(ii) From the direction of $\mathbf\small{\vec{f_k}}$, it is obvious that, the torque opposes the initial angular velocity
(iii) That means, the angular speed will go on decreasing
(iv) That means, there is angular 'deceleration'
(v) So we give a negative sign  
13. Let us write the two available equations together:
• From (9), we have: $\mathbf\small{|\vec{v}_{(t)}|=|\vec{a}|\,t}$ 
• From (12), we have: $\mathbf\small{|\vec{\omega}_{(t)}|=|\vec{\omega}_i|-|\vec{\alpha}|\,t}$
14. The rolling of the disc just after touch down is not 'pure rolling'
    ♦ We proved this by showing that the velocity of the contact point P is not zero
• We also saw in (12) that, the disc suffers angular deceleration
    ♦ So a time will come at which the angular velocity becomes so low that, the condition $\mathbf\small{|\vec{v}_{CM}|=R\,|\vec{\omega}|}$ is satisfied
    ♦ At that time, the rolling changes to pure rolling
Let the pure rolling begin after 't' seconds
• For pure rolling, we have: $\mathbf\small{|\vec{v}_{CM}|=R\,|\vec{\omega}|}$
• All points of the disc will have the same linear velocity $\mathbf\small{|\vec{v}_{CM}|}$
• So we can simply write $\mathbf\small{|\vec{v}|}$ instead of $\mathbf\small{|\vec{v}_{CM}|}$
• So we can write as:
When pure rolling begins: $\mathbf\small{|\vec{v}_{(t)}|=R\,|\vec{\omega}_{(t)}|}$
15. Now we substitute for $\mathbf\small{|\vec{v}_{(t)}|}$ and $\mathbf\small{|\vec{\omega}_{(t)}|}$ 
• We use the equations in (13) for the substitution. We get:
$\mathbf\small{|\vec{a}|\,t=R\,\left(|\vec{\omega}_i|-|\vec{\alpha}|\,t\right)}$
$\mathbf\small{\Rightarrow|\vec{a}|\,t=R|\vec{\omega}_i|-R|\vec{\alpha}|\,t}$
$\mathbf\small{\Rightarrow(|\vec{a}|+R|\vec{\alpha}|)\,t=R|\vec{\omega}_i|}$
• Thus we get:
Eq.7.35: $\mathbf\small{t= \frac{R|\vec{\omega}_i|}{(|\vec{a}|+R|\vec{\alpha}|)}}$
• Where:
    ♦ t is the time after which pure rolling begins
    ♦ R is the radius of the disc
    ♦ $\mathbf\small{|\vec{\omega}_i|}$ is the angular velocity at the time of touch down
    ♦ $\mathbf\small{|\vec{a}|=\frac{|\vec{f_k}|}{m}}$
    ♦ $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$

We will see a solved example:
Solved example 7.46
A wheel spinning at an angular velocity of 70 rad s^-1 is slowly lowered on to the ground with out any horizontal force. Radius of the wheel is 0.4 m and it's mass is 15 kg. 
(a) After what time will the pure rolling begin? 
(b) What is the distance traveled by the wheel before the beginning of pure rolling ?
Coefficient of kinetic friction between the wheel and the ground is 0.5
Take g = 9.8 ms^-2  
Solution:
Part (a):
1. We have Eq.7.35: $\mathbf\small{t= \frac{R|\vec{\omega}_i|}{(|\vec{a}|+R|\vec{\alpha}|)}}$
2. Calculation of $\mathbf\small{|\vec{a}|}$:
• $\mathbf\small{|\vec{a}|=\frac{|\vec{F}|}{m}=\frac{|\vec{f_k}|}{m}=\frac{\mu_k\,m\,|\vec{g}|}{m}=\mu_k|\vec{g}|}$
• So we get: $\mathbf\small{|\vec{a}|=\mu_k|\vec{g}|=(0.5 \times 9.8)}$ = 4.9 ms^-2 
3. Calculation of $\mathbf\small{|\vec{\alpha}|}$:
• $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
• For a disc $\mathbf\small{I=\frac{mR^2}{2}}$
• So we get: $\mathbf\small{|\vec{\alpha}|=\frac{2R\,|\vec{f_k}|}{mR^2}=\frac{2\,|\vec{f_k}|}{mR}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{2\,\mu_k\,m\,|\vec{g}|}{mR}=\frac{2\,\mu_k\,|\vec{g}|}{R}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{2(0.5)(9.8)}{0.4}=25.4}$ rad s^-2.
4. Substituting the results from (2) and (3) in (1), we get:
$\mathbf\small{t= \frac{0.4 \times 70}{[(4.9)+0.4(25.4)]}=1.86\, \rm{s}}$

Part (b):
1. Consider the two instances:
(i) Instance at which the contact with the floor is made
(ii) Instance at which the pure rolling begins
2. From part (a), we have: Time interval between the two instances = 1.86 s
• Linear acceleration with which the wheel travels during this time interval = (0.5 × 9.8) = 4.9 ms^-2.
• Initial velocity of this linear motion = 0
• So we can find the velocity at the instance when pure rolling begins 
3. We can use the relation $\mathbf\small{|\vec{v}_f|=|\vec{v}_i|+|\vec{a}|\,t}$
• For our present case: $\mathbf\small{|\vec{v}_f|=0+|\vec{a}|\,t}$
$\mathbf\small{\Rightarrow|\vec{v}_f|=|\vec{a}|\,t}$
• Substituting the values, we get: $\mathbf\small{|\vec{v}_f|}$ = (4.9 × 1.86) = 9.114 ms^-1.
4. Let $\mathbf\small{|\vec{x}|}$ be the distance traveled by the wheel during this interval
• We can use the relation: $\mathbf\small{|\vec{v}_f|^2-|\vec{v}_i|^2=2|\vec{a}|\,|\vec{x}|}$
• Substituting the values, we get: $\mathbf\small{9.1^2-0=2\times4.9\times|\vec{x}|}$
• Thus we get: $\mathbf\small{|\vec{x}|}$ = 8.45 m

Solved example 7.47
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 𝝅 rad s^-1. Which of the two will start to roll earlier ? The co-efficient of kinetic friction is 𝝁_k = 0.2
Solution:
• The solid disc and ring are placed simultaneously
• Both have the same initial angular velocity of 10 𝝅 rad s^-1 
• As soon as they are placed, both will move horizontally in the same direction
• But for both of them, this motion will not be 'pure rolling'
• Pure rolling will start only after some time 't' after the 'instant of touch down'
• This 't' will be different for the two objects
• We have to find which object has the smallest 't'
• Given that the radius is the same 10 cm for both the objects. So we will write: 
    ♦ Radius of disc = Radius of ring = R = 10 cm
1. First we will consider the solid disc
We have:
Eq.7.35: $\mathbf\small{t= \frac{R|\vec{\omega}_i|}{(|\vec{a}|+R|\vec{\alpha}|)}}$
• Where:
    ♦ t is the time after which pure rolling begins
    ♦ R is the radius of the disc
    ♦ $\mathbf\small{|\vec{\omega}_i|}$ is the angular velocity at the time of touch down
    ♦ $\mathbf\small{|\vec{a}|=\frac{|\vec{f_k}|}{m}}$
    ♦ $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
2. Calculation of $\mathbf\small{|\vec{a}|}$:
• $\mathbf\small{|\vec{a}|=\frac{|\vec{F}|}{m}=\frac{|\vec{f_k}|}{m}=\frac{\mu_k\,m\,|\vec{g}|}{m}=\mu_k|\vec{g}|}$
• So we get: $\mathbf\small{|\vec{a}|=\mu_k|\vec{g}|=(0.2 \times 9.8)}$ = 1.96 ms^-2 
3. Calculation of $\mathbf\small{|\vec{\alpha}|}$:
• $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
• For a disc $\mathbf\small{I=\frac{mR^2}{2}}$
• So we get: $\mathbf\small{|\vec{\alpha}|=\frac{2R\,|\vec{f_k}|}{mR^2}=\frac{2\,|\vec{f_k}|}{mR}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{2\,\mu_k\,m\,|\vec{g}|}{mR}=\frac{2\,\mu_k\,|\vec{g}|}{R}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{2(0.2)(9.8)}{R}=\frac{3.92}{R}}$ rad s^-2.
4. Substituting the results from (2) and (3) in (1), we get:
$\mathbf\small{t_d= \frac{R \times 10 \pi}{[(1.96)+R(\frac{3.92}{R})]}=1.7\pi R\, \rm{s}}$
5. Now we consider the ring
We have:
Eq.7.35: $\mathbf\small{t= \frac{R|\vec{\omega}_i|}{(|\vec{a}|+R|\vec{\alpha}|)}}$
6. Calculation of $\mathbf\small{|\vec{a}|}$:
• $\mathbf\small{|\vec{a}|=\frac{|\vec{F}|}{m}=\frac{|\vec{f_k}|}{m}=\frac{\mu_k\,m\,|\vec{g}|}{m}=\mu_k|\vec{g}|}$
• So we get: $\mathbf\small{|\vec{a}|=\mu_k|\vec{g}|=(0.2 \times 9.8)}$ = 1.96 ms^-2 
7. Calculation of $\mathbf\small{|\vec{\alpha}|}$:
• $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
• For a ring $\mathbf\small{I=mR^2}$
• So we get: $\mathbf\small{|\vec{\alpha}|=\frac{R\,|\vec{f_k}|}{mR^2}=\frac{|\vec{f_k}|}{mR}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{\mu_k\,m\,|\vec{g}|}{mR}=\frac{\mu_k\,|\vec{g}|}{R}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{(0.2)(9.8)}{R}=\frac{1.96}{R}}$ rad s^-2.
8. Substituting the results from (6) and (7) in (5), we get:
$\mathbf\small{t_r= \frac{R \times 10 \pi}{[(1.96)+R(\frac{1.96}{R})]}=2.55\pi R\, \rm{s}}$
9. Let us write the results together:
• From (4) we have: t_d = 1.7 𝝅R
• From (8) we have: t_d = 2.55 𝝅R
• Obviously, t_d is lesser. That means, the disc will start pure rolling earlier

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A practical application:
• We have seen that, a spinning disc, if brought into contact with the ground will slip
• But when we set a car into motion, the rear wheels are not 'already spinning'
• We slowly transmit the energy from the engine to the rear wheels
• So the rear wheels slowly begins to roll. There will not be any slipping
• If a large energy is transmitted in a small interval of time (that is 'large power'), those wheels will spin
• We want to know the maximum power that can be transmitted so that, spinning does not occur
• The following solved example demonstrates this aspect:

Solved example 7.48
A rear wheel drive car has a total mass of 1200 kg. The wheels have a radius of 0.3 m
(a) What is the maximum torque that can be imparted to the particles on the periphery of the rear wheels?
(b) What is the maximum linear acceleration with which the driver can set the car into motion? 
Coefficient of static friction = 0.6
Solution:
Part (a):
1. Assume that the total mass of the car is distributed equally among the four wheels
• So mass carried by each wheel = ¹²⁰⁰⁄₄ = 300 kg
⇒ Weight acting down on each wheel = (300 × 9.8) = 2940 N
⇒ Force of friction on each wheel = 𝝁_smg = (0.6 × 2940) = 1764 N
2. This is the maximum force which the ground can apply on the particle P
('Particle P' is the particle of the wheel which is in contact with the ground. See fig.7.145.b above)
• Also, this '1764 N frictional force' acts towards the right
3. This frictional force is 'made available' because, particle P applies a force tangentially towards the left
• If this 'force applied by the particle' is greater than the 'force available from the ground', then the wheel will slip
4. So we can write:
The maximum force which the particle P can exert
= 1764 N
5. For the next step, we first write some basics which we have already learned in rotational motion:
(i) The wheel is initially at rest. So the initial angular velocity = 0 
(ii) We gradually apply the energy produced in the engine onto the wheel
• So the wheel gradually begins to rotate. That means it's angular velocity changes from zero
(iii) A change in angular velocity can be brought about only by an application of a torque
• When we transmit the energy from the engine to the wheel, we are indeed applying a torque
(iv) So every particle on the wheel will be experiencing a torque
• The torque experienced by a particle will depend on it's distance 'r' from the center of the wheel
• We have the equation: Torque = Force ×  perpendicular distance from center
6. In the present case:
• Let the torque experience by the particle P be $\mathbf\small{|\vec{\tau}|}$
• Let the tangential force experienced by the particle P be $\mathbf\small{|\vec{F}|}$  
(Remember that the above torque and force are due to the energy transmitted from the engine)
• We can write: $\mathbf\small{|\vec{\tau}|=|\vec{F}|R}$
7. The particle P transmits this $\mathbf\small{|\vec{F}|}$ tangentially to the ground
• When the ground experience this $\mathbf\small{|\vec{F}|}$, it puts up an equal and opposite reaction force. This reaction force is the friction
8. We saw that the maximum friction available is 1764 N
• So $\mathbf\small{|\vec{F}|}$ must not exceed 1764 N
• If $\mathbf\small{|\vec{F}|}$ exceeds this value, the wheel will slip
9. So from (6), we get: $\mathbf\small{|\vec{\tau}|}$ = (1764 × 0.3) = 529.2 N m
• We can write:
The energy transmitted (or the power, which is 'energy transmitted per second') from the engine should be in such a way that, the torque experienced by the particles at the periphery of the wheel do not exceed 529.2 N m

---

• We have been using this sentence many times: 'Energy produced in the engine is transmitted to the rear wheels'

• A detailed description is appropriate here:
(i) Inside the engine, chemical energy is converted to mechanical energy
• This energy is used to rotate the rear wheels of the car. Thus the car moves forward
(ii) Since the 'rotation of the wheels' is what we want, we consider the 'torque' rather than the 'force'
(iii) We know that: Work done by the force = force × linear displacement   
• Similarly, Work done by torque = torque × angular displacement
(We saw this in chapter 7.29)
(iv) 'Work done' is actually energy available from the engine 
(v) We see many advertisements (for engines) specifying the torque which can be produced
• Another way for specifying the capacity of the engine is by giving the value of power
• Power is the energy available in unit time (We saw this in chapter 7.30)
(vi) Whether be it power or energy, for our present discussions, we need to consider only two points:
(i) The car engine produces the energy (or power) to rotate the rear wheels of the car
(Most cars are rear wheel driven. Which means energy is given to the rear wheels only)
(ii) The car engine does not give a push or pull in the linear direction

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Now we take up part (b):
1. Frictional force available at one rear wheel = 1764 N
• So total frictional force available  at the two rear wheels = (1764 × 2) = 3528 N
[Remember that, this forward force of 3528 N is not made available (from the ground) free of cost. It is a reaction force, which can be produced only if the car tire applies the same force in the backward direction. Recall that, if a space astronaut want to move towards the left, he must push onto a massive object on his right side. It is the only option available to him unless he is wearing a jet thruster backpack]
3. Given that, total mass of the car = 1200 kg
• So maximum possible acceleration of the car = ^Force⁄_mass = ³⁵²⁸⁄₁₂₀₀ = 2.94 m s^-2
4. So the driver must set the car into motion in such a way that, the acceleration of the car does not exceed 2.94 m s^-2
• What does that mean?
Let us analyze:
(i) The car is initially at rest
• When it is set in motion, it attains a velocity v
(ii) So there is a 'change in velocity' because, velocity changes from 0 to v
• If there is a change in velocity, obviously, there must be an acceleration
• So without an acceleration, no vehicle can attain a required velocity
(iii) We know that $\mathbf\small{a=\frac{v_2-v_1}{t}}$
• 't' is in the denominator
• So a large acceleration implies that, a large 'change in velocity' is brought about in a small time
• Such a large acceleration will cause the car to jerk and the rear wheels will spin
(iv) So the driver must not give a large acceleration
• In our present case, the acceleration must not exceed 2.94 ms^-2

---

Now we will derive an interesting relation
1. In the above solved example 7.48, we saw that: 
• The car tires are rolling without slipping
• At the same time, the car is moving with acceleration
2. The ‘acceleration of the car’ is ‘linear acceleration’
• Since the car is a rigid body, 'every non-rotating particle' in it will be having the same linear acceleration
• So the center of the wheel will also be having the same linear acceleration
3. Now consider a disc 
• It is rolling with out slipping
• At the same time, it’s center is experiencing linear acceleration
4. We have:
Eq.7.32: $\mathbf\small{|\vec{v}_{CM}|=R|\vec{\omega}|}$ 
5. Since our present disc is moving with linear acceleration, the linear velocity will change
• So the right side of Eq.7.32 will also change
• But R is a constant. So we can say, if the disc is in pure rolling and is moving with linear acceleration, the angular $\mathbf\small{|\vec{\omega}|}$  also changes
• If the angular velocity changes, it means that there is angular acceleration
6. Consider the instant at which the reading in the stop watch  is 0
• Let the magnitude of linear velocity be $\mathbf\small{|\vec{v}_0|}$
• Let the magnitude of angular velocity be $\mathbf\small{|\vec{\omega}_0|}$ 
7. Consider the instant at which the reading in the stop watch  is t
• Let the magnitude of linear velocity be $\mathbf\small{|\vec{v}_t|}$
• Let the magnitude of angular velocity be $\mathbf\small{|\vec{\omega}_t|}$
8. For linear motion, we have: $\mathbf\small{|\vec{v}_t|=|\vec{v}_0|+|\vec{a}|\,t}$
• So we can write: $\mathbf\small{t=\frac{|\vec{v}_t|-|\vec{v}_0|}{|\vec{a}|}}$
9. For angular motion, we have: $\mathbf\small{|\vec{\omega}_t|=|\vec{\omega}_0|+|\vec{\alpha}|\,t}$
• So we can write: $\mathbf\small{t=\frac{|\vec{\omega}_t|-|\vec{\omega}_0|}{|\vec{\alpha}|}}$
10. The 't' is same in both (7) and (8)
• So we can equate them. We get: $\mathbf\small{\frac{|\vec{v}_t|-|\vec{v}_0|}{|\vec{a}|}=\frac{|\vec{\omega}_t|-|\vec{\omega}_0|}{|\vec{\alpha}|}}$
11. $\mathbf\small{|\vec{\omega}_t|}$ is the angular velocity at the instant when the reading in the stopwatch is 't'
• At that instant, the linear velocity of CM is $\mathbf\small{|\vec{v}_t|}$
• Since there is no slipping, we have: $\mathbf\small{|\vec{v}_t|=R|\vec{\omega}_t|}$
12. $\mathbf\small{|\vec{\omega}_0|}$ is the angular velocity at the instant when the reading in the stopwatch is 0
• At that instant, the linear velocity of CM is $\mathbf\small{|\vec{v}_0|}$
• Since there is no slipping, we have: $\mathbf\small{|\vec{v}_0|=R|\vec{\omega}_0|}$
13. So the result in (9) will become:
$\mathbf\small{\frac{R|\vec{\omega}_t|-R|\vec{\omega}_0|}{|\vec{a}|}=\frac{|\vec{\omega}_t|-|\vec{\omega}_0|}{|\vec{\alpha}|}}$
$\mathbf\small{\Rightarrow\frac{R(|\vec{\omega}_t|-|\vec{\omega}_0|)}{|\vec{a}|}=\frac{|\vec{\omega}_t|-|\vec{\omega}_0|}{|\vec{\alpha}|}}$
$\mathbf\small{\Rightarrow\frac{R}{|\vec{a}|}=\frac{1}{|\vec{\alpha}|}}$
Thus we get:
Eq.7.36: $\mathbf\small{|\vec{a}|=R\,|\vec{\alpha}|}$
14. In the above solved example 7.48, we have: $\mathbf\small{|\vec{a}_{max}|}$ = 2.94 ms^-2
• That means, the maximum allowable linear acceleration for the center of the wheel is 2.94 ms^-2
• The wheel is to roll with out slipping. So the relation $\mathbf\small{|\vec{a}|=R\,|\vec{\alpha}|}$ is valid
• We can write:$\mathbf\small{2.94=0.3\,|\vec{\alpha}|}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=9.8}$ rad s^-2
• That means, the driver must accelerate the car in such a way that, the angular acceleration of the rear wheels do not exceed 9.8 rad s^-2

---

So we have completed a discussion on the rotation of car wheels which are able to self propagate. In the next section, we will see the case of carts which need external force to propagate

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Chapter 7.35 - Friction while Rolling

In the previous section, we saw the condition for rolling without slipping. We also saw kinetic energy of a rolling body. In this section, we will see the 'effect of friction' in rolling motion

For that, first we must understand how friction helps us to walk on the ground. We will write it in steps:
1. We know that, it is difficult (and dangerous) to walk on an icy floor
■ Why is that so?
2. Let us say, a person take the first forward step with his right foot
• The blue rectangle in fig.7.140(a) represents his right shoe
• He presses the right shoe onto the floor to get a grip

Fig.7.140

3. The floor must hold the shoe in position
• If proper grip is available, the man can pull himself forward
4. 'Pulling oneself forward' is a horizontal motion
• The man will be exerting a horizontal force $\mathbf\small{\vec{F}}$ through his right foot. This is shown in fig.a 
• So the floor must hold the shoe in such a way that, the shoe does not move horizontally in the direction of $\mathbf\small{\vec{F}}$ 
5. But the icy floor will not be able to hold the shoe in such a manner
• This is because, there is no friction
6. So when he tries to pull himself forward, the shoe will slip backwards
• He will not be able to move forward
• In fact, if he does not hold on to some proper support with his hands, he will slip and fall, causing serious injuries
7. Things are different if he wears spiked shoes which are manufactured specially for icy conditions
• The spikes will penetrate into the ice. This is shown in fig.b
• So the ice will be able to keep his feet in position
8. Let us see what really happens when spikes are used:
(i) The spikes first penetrate into the ice
(ii) When the man pulls himself forward, the spikes 'presses horizontally' against the ice
(iii) So a horizontal force is applied by the spikes on the ice. This force is in the backward direction
(iv) The ice in turn, applies an equal forward horizontal force (in accordance with Newton's third law) on the spikes 
(v) This forward force is the thrust that helps the man to move forward
9. Let us see how this forward thrust is related to friction:
(i) We saw that, the forward thrust is produced due to the interlocking between the spikes and the ice
(ii) On ordinary floors, spikes are not required because there are already numerous ridges and valleys
(iii) These ridges and valleys will do the interlocking
(iv) These ridges and valleys is the cause of friction. We saw this in an earlier chapter (Fig.5.76 in chapter 5.19). There we also saw that, frictional force will always be parallel to the surfaces in contact 
(v) So the 'forward thrust that helps in walking', is in fact the frictional force. This frictional force is denoted as $\mathbf\small{\vec{f_s}}$ in fig.b
10. Next we want to determine whether this frictional force is static or kinetic
(i) Consider the following instant:
The instant at which the man pulls himself forward
(ii) At that instant, the frictional force acts
• But at that instant, actual motion has not begun. In other words, at that instant, there is no relative motion between the shoe and the ice
(iii) So the friction which helps the forward motion is static friction
11. Also recall that static friction is a self adjusting force
    ♦ It increases when the applied force increases
    ♦ It decreases when the applied force decreases
• The maximum possible friction that can be expected is $\mathbf\small{\vec{f}_{s,max}}$
• This maximum force depends on the nature of the ice and the spikes
• When the man moves forward, the full $\mathbf\small{\vec{f}_{s,max}}$ need not be necessarily mobilized
• A force lesser than $\mathbf\small{\vec{f}_{s,max}}$ may be sufficient
12 Another interesting aspect:
(i) We saw that, the spikes on the right shoe helps the man to pull himself forward
(ii) Once he has pulled himself forward, he puts the left foot forward and lifts the right foot
(iii) Now, it is the duty of the spikes on the left shoe to provide the necessary forward force
(iv) Thus we see that, the left and right spikes 'alternately and continuously' provide the necessary forward force. So the man is able to walk properly

---

13. Now, instead of walking motion, let us see another type of motion:

• In fig.140 (c), the shoe is being pulled by an external force $\mathbf\small{\vec{F}}$
    ♦ This external pulling force can be provided by a snowmobile for example
• If the man can safely provide proper support to the upper part of his body (like by holding on to a rope attached to the snowmobile), he can slide over the snow
• In this case, the man does not have to put up any effort for the forward movement
    ♦ The effort is done by the snow mobile
14. But if he uses spiked shoes, things will be different
• The spikes will penetrate into the ice. This is shown in fig.d
• So the ice will try to keep his feet in position
15. Let us see what really happens when spikes are used:
(i) The spikes first penetrate into the ice
(ii) When the shoe is pulled forward, the spikes 'presses horizontally' against the ice
(iii) So a horizontal force is applied by the spikes on the ice. This force is in the forward direction
(iv) The ice in turn, applies an equal backward horizontal force (in accordance with Newton's third law) on the spikes 
(v) This backward force tries to prevent the shoe from moving forward
• And we know that, this force on the spikes is the frictional force
16. Here, the force which tries to prevent motion is the static friction
• But once the motion begins, kinetic friction takes over
17. Now we will write a summary:
• When the man propels himself, the friction acts in the forward direction
• When an external force propels him, the friction acts in the backward direction

---

■ The above situations arise in the case of vehicle tires also

• While walking, the left and right feet are in contact with the ground alternately and continuously
• For tires, succeeding portions of the tire comes into contact with the ground continuously
• Each portion of the tire must receive proper grip from the ground
• Icy floor cannot provide this grip
• So the 'portion of the tire' in contact with ice will slip
• Succeeding portions of the tire will also slip
• This causes the tire to spin. It will not roll

We will consider two cases where vehicle tires spin:
A. The case of a car
B. The case of a cart

A. The case of a car
1. Consider fig.7.141 below:

Fig.7.141

• It is a schematic arrangement of the rear wheels of a car
    ♦ The two rear wheels are shown in yellow color
    ♦ The axle connecting the two wheels is shown in cyan color    
    ♦ The engine is shown as a block in grey color
2. Remember that, this is only a schematic arrangement. The actual arrangements in a car will be different
• But all the information that is relevant for our present discussion are the following four:
(i) The engine rotates the axle
    ♦ The rotation of the axle is indicated by the white curved arrow
(ii) Since the wheels are connected to the axle, the wheels also will rotate
    ♦ The rotation of the wheel is indicated by the red curved arrow
(iii) If there is no slipping between the wheels and the ground, the wheels will begin to roll forward
(iv) Thus the car as a whole moves forward
3. So we see that, there is no 'push' or 'pull' on the wheels of a car
• The movement of the wheels is achieved by providing a rotational motion to the wheels

B. The case of a cart
1. Consider fig.7.142 below:

Fig.7.142

• It is a schematic arrangement of the rear wheels of a cart
    ♦ The two rear wheels are shown in yellow color
    ♦ The axle connecting the two wheels is shown in cyan color    
    ♦ There is no engine. Instead, a cyan rod is connected to the axle
2. Remember that, this is only a schematic arrangement. The actual arrangements in a cart will be different
• But all the information that is relevant for our present discussion are the following four:
(i) A pulling force is applied on the rod connected to the axle. As a result, the axle moves forward
(ii) Since the wheels are connected to the axle, the wheels also move forward
(iii) If there is no slipping between the wheels and the ground, the wheels will begin to roll forward
(iv) Thus the cart as a whole moves forward
3. So we see that, there is indeed a 'push' or 'pull' on the wheels of a cart
• The push or pull is applied at the center of the wheels

■ We can write a summary:
• In the case of a car, the wheels are made to roll by applying a rotational motion to the wheels
• In the case of a cart, the wheels are made to roll by applying a force at the center of the wheels

---

■ We will analyse the role of friction in each case

First we will consider the wheel of a car. We will write the analysis in steps:
1. Fig.7.143 (a) below shows one of the rear wheels of a car 
• It is a side view. So the engine is not visible
• In the fig., the engine is transmitting power to the wheel. The wheel is spinning
• But the car is resting on icy floor
• The icy floor is so smooth that, there is no friction

Fig.7.143

2. consider the instant at which any random point on the periphery of the wheel is in contact with the icy floor
3. The floor must hold that point in position
• If proper grip is available, the 'wheel as a whole' can pull itself forward
4. 'Pulling oneself forward' is a horizontal motion
• The wheel will be exerting a horizontal force $\mathbf\small{\vec{F}}$ through the point of contact. This is shown in fig.a 
• So the floor must hold the point in such a way that, it does not move horizontally in the direction of $\mathbf\small{\vec{F}}$ 
5. But the icy floor will not be able to hold the point in such a manner
• This is because, there is no friction
6. So when the wheel tries to pull itself forward, the point of contact on the wheel will slip backwards
• The succeeding point will come into contact with the ice. It also will experience the same 'no grip condition' and will slip
• This process continues and so the wheel will just spin. It will not be able to roll
7. Things are different if the wheels are spiked
• The spikes will penetrate into the ice. This is shown in fig.b
• So the ice will be able to keep the point in position
8. Let us see what really happens when spikes are used:
(i) The spikes first penetrate into the ice
(ii) When the wheel spin further, the spikes 'presses horizontally' against the ice
(iii) So a horizontal force is applied by the spikes on the ice. This force is in the backward direction
(iv) The ice in turn, applies an equal forward horizontal force (in accordance with Newton's third law) on the spikes 
(v) This forward force is the thrust that helps the wheel to roll forward
9. Let us see how this forward thrust is related to friction:
• We saw that, the forward thrust is produced due to the interlocking between the spikes and the ice
• On ordinary floors, spikes are not required because there are already numerous ridges and valleys
• These ridges and valleys is the cause of friction 
• We saw this in an earlier chapter (Fig.5.76 in chapter 5.19)
• Also we saw that, frictional force will always be parallel to the surfaces in contact 
• So the 'forward thrust that helps in rolling', is in fact the frictional force
• This frictional force is denoted as $\mathbf\small{\vec{f_s}}$ in fig.b
10. Next we want to determine whether this frictional force is static or kinetic
• Consider the following instant:
The instant at which the wheel pulls itself forward
• At that instant, the frictional force acts
• But at that instant, actual motion has not begun. In other words, at that instant, there is no relative motion between the point and the ice
• So the friction which helps the forward motion is static friction
11. Also recall that static friction is a self adjusting force
    ♦ It increases when the applied force increases
    ♦ It decreases when the applied force decreases
• The maximum possible friction that can be expected is $\mathbf\small{\vec{f}_{s,max}}$
• This maximum force depends on the nature of the ice and the spikes
• When the wheel moves forward, the full $\mathbf\small{\vec{f}_{s,max}}$ need not be necessarily mobilized
• A force lesser than $\mathbf\small{\vec{f}_{s,max}}$ may be sufficient
12 Another interesting aspect:
• We saw that, the spikes on the wheel helps the wheel to pull itself forward
• Once the wheel has pulled itself forward, the succeeding point comes into contact with the ice
• Now, it is the duty of the 'spikes at the new point of contact' to provide the necessary forward force
• This process continues and so the wheel is able to roll forward

---

Now we will consider the cart

1. Fig.7.144 (a) below shows one of the rear wheels of a cart 
• It is a side view.
• In the fig., a horizontal force is being applied at the center of the wheel
• But the cart is resting on icy floor
• The icy floor is so smooth that, there is no friction

Fig.7.144

2. We see the weight vector $\mathbf\small{m\vec{g}}$ in the down ward direction
    ♦ This vector passes through the center 'C' of the wheel
• We see the normal reaction vector $\mathbf\small{\vec{N}}$
    ♦ This vector also passes through the center 'C' of the wheel
• We see no other forces
• Forces passing through the center of mass of a body cannot cause it to rotate. So the wheel will not rotate
• When the cart is pulled. the wheel will just slide
3. But if the spiked wheels are used, things will be different

• The spikes will penetrate into the ice. This is shown in fig.b
• So the ice will try to keep the point of contact in position
4. Let us see what really happens when spikes are used:
(i) The spikes first penetrate into the ice
(ii) When the wheel is pulled forward, the spikes 'presses horizontally' against the ice
(iii) So a horizontal force is applied by the spikes on the ice. This force is in the forward direction
(iv) The ice in turn, applies an equal backward horizontal force (in accordance with Newton's third law) on the spikes 
(v) Note that, this backward horizontal force is applied by the ice on the spikes
• That means, the spikes experience this force
• Since the spikes are attached rigidly at the periphery of the wheel, the wheel will experience this force
• The wheel will experience this force as:
A force acting tangentially at the periphery
• So this is the force that we were waiting for. It does not pass through the center 'C'
• The wheel will begin to rotate
5. We know that, this force on the spikes is the frictional force
■ So we can write:
Friction helps the wheel of a cart to roll. If there is no friction, the wheel will only slide

---

Let us write a comparison between the two types of wheels:

    ♦ The rear wheels of a car
    ♦ The wheels of a cart
• It can be written in two steps: 
1. For the rear wheels of a car, rotation is already present. Because, the engine provides rotation to the axle, and so the wheels which are connected to the axle naturally rotates. If there is no friction, these rotating tires will just spin. The car will not move
• For the rear wheels of a cart, rotation is not readily present. Even if the cart is pulled forward, the wheels will rotate only if friction is present. In the absence of friction, the rear wheels of a cart can only slide
2. Frictional force, if present, acts on the rear wheel of a car
    ♦ The direction of this frictional force is same as the direction of motion of the car
• Frictional force, if present, acts on the wheels of a cart
    ♦ The direction of this frictional force is opposite to the direction of motion of the cart

---

Also we see the following similarities:

1. A car with an engine can self propagate
• A man is also self propagating when he is walking
• So the two sets of figs. are comparable:
    ♦ Set 1: Figs.7.143 (a) and (b)
    ♦ Set 2: Figs.7.140 (a) and (b)
2. A cart without an engine cannot self propagate. It needs an external pulling force
• A man trying to slide behind a snowmobile also needs the external pulling force  
• So the two sets of figs. are comparable:
    ♦ Set 1: Figs.7.144 (a) and (b)
    ♦ Set 2: Figs.7.140 (c) and (d)

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In the next section, we will see situations where the wheels will spin/slide even when friction is present

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