Chapter 7.34 - Rolling without Slipping

In the previous section, we saw conservation of angular momentum. In this section, we will see rolling motion

1. Consider a disc rolling on a horizontal surface (Fig.7.137 (a) below)
• Radius of the disc is R

Fig.7.137

■ Since it is rolling, it will have two types of motion:
(i) Translational motion
(ii) Rotational motion
2. Let us write all the available information about the rotational motion:
(i) The disc is a symmetric body
• So the rotation will be about an axis passing through the center C
• Also, this axis will be perpendicular to the plane of the disc (that is., plane of the computer screen)
(ii) The angular velocity of rotation is $\mathbf\small{\vec{\omega}}$
• All particles (except the particle at C) in the disc will be rotating with this same angular velocity
(iii) The particle at C will not have any rotational motion
3. Next we write all the available information about the translational motion:
(i) The particle at C does not have any rotational motion. But it does have translational motion
(ii) Since C is the center of mass (CM) of the disc, we will denote the translational velocity of C as $\mathbf\small{\vec{v}_{CM}}$ 
• This $\mathbf\small{\vec{v}_{CM}}$ is parallel to the surface on which the disc rolls
(iii) Since the disc is rigid, all particles will have the same translational velocity
4. Our next aim is to find this $\mathbf\small{\vec{v}_{CM}}$
• Consider fig.b, An arc AP is marked in magenta color on the periphery of the disc  
• This arc subtends an angle of θ at the center C
5. The fig.b shows the instant at which A is in contact with the ground
6. The disc rolls towards the right
• Point A will lose contact with the surface
• As the rolling proceeds, new points on the arc will come into contact with the surface 
7. Consider the instant at which P comes into contact with the surface   
8. What is the linear distance covered by the disc between the following two instances:
    ♦ Instance mentioned in (5)
    ♦ Instance mentioned in (7)
Answer: Obviously, the linear distance will be equal to the length of the arc AP
(Note that, this will be true only if there is no slipping/skidding between the disc and the surface. For our present discussion, we assume that there is no such slipping/skidding)
9. We know how to find this length of arc
• We have: angle = ^{Arc length}⁄_radius   (Where angle is in radians)
• So Arc length AP = Linear distance traveled by the disc =  Rθ.
10. Let 't' be the time duration between the two instances
• Then we can write two more information:
(i) The disc travels a linear distance of Rθ during a time 't' s
(ii) The disc turns through an angle θ during a time 't' s
11. Let us write the above equation again:
• Linear distance traveled by the disc =  Rθ.
• Dividing both sides by 't', we get:
$\mathbf\small{\frac{\text{Linear distance traveled by the disc}}{t}=\frac{R\theta}{t}}$
• But $\mathbf\small{\frac{\text{Linear distance traveled by the disc}}{t}}$ is the linear velocity with which the disc travels during the time interval 't'
    ♦ 'Linear velocity of the disc' is the 'velocity of the center of mass'
    ♦ We denoted it as $\mathbf\small{\vec{v}_{CM}}$  
• Also, $\mathbf\small{\frac{\theta}{t}}$ is the angular velocity with which the disc turns during the time interval 't'
    ♦ We denoted the angular velocity as $\mathbf\small{\vec{\omega}}$
■ Thus we get: 
Eq.7.32: $\mathbf\small{|\vec{v}_{CM}|=R|\vec{\omega}|}$
• Since it is a rigid body, every particle in the disc will have a translational velocity of $\mathbf\small{|\vec{v}_{CM}|}$ towards the right
12. This equation is similar to: $\mathbf\small{\vec{v}=R\,\vec{\omega}}$ 
• $\mathbf\small{\vec{v}}$ is the 'tangential velocity' of a particle at the periphery of a rotating body 
• But $\mathbf\small{\vec{v}_{CM}}$ is the linear velocity with which the disc rolls
• We must clearly note the difference between the two
■ Eq.7.32 gives the condition for 'no slipping'
• Because if there is slipping, arc length will not be equal to the linear distance. The equation will not be valid any more
13. Let us see another interesting information:
• Fig.7.138(a) below shows 3 points in the disc:
(i) The bottom most point A
(ii) The center of mass C
(iii) The top most point B

Fig.7.138

• A and B pass through a vertical through C
14. First we will consider point B:
• The disc has an angular velocity of $\mathbf\small{|\vec{\omega}|}$ 
• Then the point B will be moving towards the right with a tangential velocity of $\mathbf\small{R\,|\vec{\omega}|}$
• Note that, this velocity will be parallel to the ground. That is., parallel to $\mathbf\small{|\vec{v}_{CM}|}$
• Also, this velocity is towards the right
15. We have seen that, due to translation, already every point in the disc has a velocity of $\mathbf\small{R|\vec{\omega}|}$ towards the right. For point B, this is shown in fig.b
• So for point B, the two velocities will add up
• Thus we can write:
Point B moves towards the right with a net velocity of $\mathbf\small{2R|\vec{\omega}|}$. This is shown in fig.c
16. But at the next instant, a succeeding particle takes over the position of B. This new particle will experience the same effect
■ So in general, we can write:
At any instant, the top most point of the disc will be travelling towards the right with a velocity of $\mathbf\small{2R|\vec{\omega}|}$
17. Next we will consider point C
• The point C is in the axis of rotation. So it will receive no contribution (towards velocity) from the rotation. Because, every point on the axis will be having zero rotation
• But C has translational velocity $\mathbf\small{|\vec{v}_{CM}|}$. It is the only velocity it has. This is shown in fig.b
■ Thus we can write:
Point C moves towards the right with a net velocity of $\mathbf\small{R|\vec{\omega}|}$. This is shown in fig.c        
18. Finally we consider point A
• The disc has an angular velocity of $\mathbf\small{|\vec{\omega}|}$ 
• Then the point A will be moving towards the left with a tangential velocity of $\mathbf\small{R\,|\vec{\omega}|}$. This is shown in fig.a
• Note that, this velocity will be parallel to the ground. That is., parallel to $\mathbf\small{|\vec{v}_{CM}|}$
• Also, this velocity is towards the left
19. We have seen that, due to translation, already every point in the disc has a velocity of $\mathbf\small{R|\vec{\omega}|}$ towards the right. For point A, this is shown in fig.b
• So for point A, the two velocities will cancel each other
• Thus we can write:
Point A has zero net velocity. This is shown in fig.c
20. But at the next instant, the succeeding particle takes over the position of A, This new particle will experience the same effect
■ So in general, we can write:
At any instant, the bottom most point of the disc will be having zero net velocity
• That means, the point of contact (with the ground) of a rolling wheel/disc will be at rest

Kinetic energy of a rolling body

We will write this in steps:
1. A rolling body has both rotational and translational motion
2. We know that kinetic energy due to rotation is $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2}$
Where:
• I = Moment of inertia of the body about the axis of rotation
• $\mathbf\small{|\vec{\omega}|}$ = Magnitude of the angular velocity
3. Also we know that kinetic energy due to translation is $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
Where:
• m = mass of the body
• $\mathbf\small{|\vec{v}_{CM}|}$ = magnitude of the linear velocity
4. So the total kinetic energy possessed by a rolling body is given by:
Eq.7.33: $\mathbf\small{K=\frac{1}{2}I\,|\vec{\omega}|^2+\frac{1}{2}m\,|\vec{v}_{CM}|^2}$

Solved example 7.44
A disc of mass 5 kg and radius 50 cm rolls on the ground with out slipping. It's linear velocity is 10 ms^-1. What is the kinetic energy possessed by the disc?
Solution:
1. Kinetic energy due to rotation = $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2}$
(i) Moment of inertia (I) should be selected carefully
• A disc can have a moment of inertia about an axis which can be an extension of one of it's diameters
• It can also have a moment of inertia about an axis perpendicular to the disc. Obviously, a disc can roll only if the axis is perpendicular to the disc. So we must take the I corresponding to this axis
Thus we have: $\mathbf\small{I=\frac{1}{2}m\,R^2}$
(ii) Next we calculate $\mathbf\small{|\vec{\omega}|}$
• Given that the disc rolls without slipping. So the condition $\mathbf\small{|\vec{v}_{CM}|=R|\vec{\omega}|}$ will be satisfied
• So we get: $\mathbf\small{|\vec{\omega}|=\frac{|\vec{v}_{CM}|}{R}}$
(iii) Substituting these expressions in (1), we get:
Kinetic energy due to rotation = $\mathbf\small{\frac{1}{2}\left(\frac{1}{2}m\,R^2\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{4}}$
2. Kinetic energy due to translation = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
3. So total kinetic energy = $\mathbf\small{\frac{m|\vec{v}_{CM}|^2}{4}+\frac{m|\vec{v}_{CM}|^2}{2}=\frac{3m|\vec{v}_{CM}|^2}{4}}$
• Substituting the values, we get:
Total kinetic energy = $\mathbf\small{\frac{3(5\,\rm{kg})(10\,\rm{m\,s^{-1}})^2}{4}=375\;\rm{J}}$
• Recall that dimensions of energy is: ML²T^-2.

Solved example 7.45
Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?
Solution:
• In this problem, we apply the law of conservation of energy
• The body is initially at rest at the top most point of the inclined plane. This is shown in fig.7.139 below:

Fig.7.139

• In that position, it has potential energy equal to mgh
• Since it is initially at rest, there is no initial kinetic energy
• When it reaches at the base, the total energy that it possesses must be equal to the total energy at top
• At the base, there is no potential energy. There is only kinetic energy
■ So we get:
Total kinetic energy of the body at the base = Total potential energy at the top
• Now we consider each body separately
1. Ring:
(i) Rotational kinetic energy 
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(m\,R^2\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{2}}$ 
• Note that for a ring, I = mR²
• Also we have: $\mathbf\small{|\vec{\omega}|=\frac{|\vec{v}_{CM}|}{R}}$
(ii) Translational kinetic energy = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
(iii) So total kinetic energy = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2+\frac{m|\vec{v}_{CM}|^2}{2}=m|\vec{v}_{CM}|^2}$  
(iv) potential energy = mgh
(iii) Equating the two energies, we get: $\mathbf\small{mgh=m|\vec{v}_{CM}|^2}$
$\mathbf\small{\Rightarrow |\vec{v}_{CM}|=\sqrt{gh}}$
2. Solid cylinder
(i) Rotational kinetic energy 
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(\frac{m\,R^2}{2}\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{4}}$ 
• Note that for a solid cylinder, $\mathbf\small{I=\frac{m\,R^2}{2}}$
(ii) Translational kinetic energy = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
(iii) So total kinetic energy = $\mathbf\small{\frac{m|\vec{v}_{CM}|^2}{4}+\frac{m|\vec{v}_{CM}|^2}{2}=\frac{3m|\vec{v}_{CM}|^2}{4}}$  
(iv) potential energy = mgh
(v) Equating the two energies, we get: $\mathbf\small{mgh=\frac{3m|\vec{v}_{CM}|^2}{4}}$
$\mathbf\small{\Rightarrow |\vec{v}_{CM}|=\sqrt{\frac{4gh}{3}}}$
3. Solid sphere
(i) Rotational kinetic energy 
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(\frac{2m\,R^2}{5}\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2=\frac{m|\vec{v}_{CM}|^2}{5}}$ 
• Note that for a solid sphere, $\mathbf\small{I=\frac{2m\,R^2}{5}}$
(ii) Translational kinetic energy = $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
(iii) So total kinetic energy = $\mathbf\small{\frac{m|\vec{v}_{CM}|^2}{5}+\frac{m|\vec{v}_{CM}|^2}{2}=\frac{7m|\vec{v}_{CM}|^2}{10}}$  
(iv) potential energy = mgh
(v) Equating the two energies, we get: $\mathbf\small{mgh=\frac{7m|\vec{v}_{CM}|^2}{10}}$
$\mathbf\small{\Rightarrow |\vec{v}_{CM}|=\sqrt{\frac{10gh}{7}}}$
4. We will now compare the results. We will use the subscripts r, c and s for the ring, solid cylinder and solid sphere respectively:
We have:
(i) $\mathbf\small{|\vec{v}_{CM(r)}|=\sqrt{gh}}$
(ii) $\mathbf\small{|\vec{v}_{CM(c)}|=\sqrt{\frac{4gh}{3}}}$
(iii) $\mathbf\small{|\vec{v}_{CM(s)}|=\sqrt{\frac{10gh}{7}}}$
^• 4⁄₃ = 1.333
^• 10⁄₇ = 1.429
• Thus we see that $\mathbf\small{\Rightarrow |\vec{v}_{CM(s)}|}$ has the greatest value
■ So we can write:
The sphere reaches the ground with the maximum velocity

Another method:
• We can derive a general formula (for the velocity at the base) for any body
1. For any body, we have: I = Mk²
• Where k is the radius of gyration of that body (Details here)
2. So for any body, the rotational kinetic energy
= $\mathbf\small{\frac{1}{2}I\,|\vec{\omega}|^2=\frac{1}{2}\left(m\,k^2\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2}$
3. So total energy
= $\mathbf\small{\frac{1}{2}\left(m\,k^2\right)\left(\frac{|\vec{v}_{CM}|}{R}\right)^2+\frac{1}{2}m\,|\vec{v}_{CM}|^2}$
= $\mathbf\small{\frac{1}{2}m\,|\vec{v}_{CM}|^2\left(1+\frac{k^2}{R^2} \right)}$
4. Equating this total energy to potential energy, we get:
$\mathbf\small{mgh=\frac{1}{2}m\,|\vec{v}_{CM}|^2\left(1+\frac{k^2}{R^2} \right)}$
5. From this, we get:
Eq.7.34: $\mathbf\small{|\vec{v}_{CM}|^2=\frac{2gh}{\left(1+\frac{k^2}{R^2} \right)}}$
• We see that, the velocity at the base is independent of the mass
• In an earlier chapter on pure translational motion also, we have proved that, the velocity of at the base of an inclined plane is independent of the mass

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In the next section, we will see relation between rolling motion and friction

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Chapter 7.33 - Conservation of Angular momentum

In the previous section, we obtained the expression for torque. In this section, we will see conservation of angular momentum

1. We know that, 'rate of change of angular momentum with time' is torque
• Mathematically, we can write this as:
$\mathbf\small{\frac{\vec{L}_2-\vec{L}_1}{\Delta t}= \vec{\tau}}$
• Where:
    ♦ $\mathbf\small{\vec{L}_1}$ is the angular momentum when the reading in the stop watch is t₁.
    ♦ $\mathbf\small{\vec{L}_2}$ is the angular momentum when the reading in the stop watch is t₂.
    ♦ Δ t = (t₂ - t₁)
2. We have seen that, for this calculation of torque, we consider $\mathbf\small{\vec{L}_z}$ only
• We do not have to consider $\mathbf\small{\vec{L}_\bot}$
• So we can write:
$\mathbf\small{\frac{\vec{L}_{z(2)}-\vec{L}_{z(1)}}{\Delta t}= \vec{\tau}}$
3. When the external torque ($\mathbf\small{\vec{\tau}}$) is zero, we get:
$\mathbf\small{\frac{\vec{L}_{z(2)}-\vec{L}_{z(1)}}{\Delta t}=0}$
$\mathbf\small{\Rightarrow (\vec{L}_{z(2)}-\vec{L}_{z(1)})=0}$
$\mathbf\small{\Rightarrow \vec{L}_{z(2)}=\vec{L}_{z(1)}}$
• That means, the angular momentum remains unchanged
• In other words, the angular momentum is a constant
4. Let us analyse this information:
(i) We have seen that $\mathbf\small{\vec{L}_{z}=I|\vec{\omega}|\hat{k}}$
• This quantity is always along the z-axis (the axis of rotation) So we need to consider the magnitudes only
• We can write:
$\mathbf\small{|\vec{L}|=I|\vec{\omega}|}$
• So we get:
    ♦ $\mathbf\small{|\vec{L}_1|=I_1|\vec{\omega}_1|}$
    ♦ $\mathbf\small{|\vec{L}_2|=I_2|\vec{\omega}_2|}$
(ii) If there is no external torque, we will get:
$\mathbf\small{I_1|\vec{\omega}_1|=I_2|\vec{\omega}_2|}$
5. If $\mathbf\small{I_2}$ increases, $\mathbf\small{|\vec{\omega}_2|}$ will decrease so that, the product remains the same
• Similarly, if $\mathbf\small{I_2}$ decreases, $\mathbf\small{|\vec{\omega}_2|}$ will increase so that, the product remains the same
6. Expert classical dancers often perform piroutte
• While performing this act, the axis of rotation passes vertically through the body of the dancer
• When the arms are stretched, I increases and $\mathbf\small{|\vec{\omega}|}$ decreases
• When the arms are brought closer to the body, I decreases and $\mathbf\small{|\vec{\omega}|}$ increases
    ♦ That is., the speed of the spin increases
7. Note that, while performing this act, only the toes are in contact with the floor. So the effect of friction is minimum
• Because of this 'low friction', we can say that no appreciable external torque acts on the spinning performer
8. A circus acrobat and a diver also, while giving the performance, bring their arms close to the body to reduce I

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Now we will see some solved examples

Solved example 7.40
(a) A child stands at the center of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction.
(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
Solution:
Part (a):
1. We have: $\mathbf\small{I_1\,|\vec{\omega}_1|=I_2\,|\vec{\omega}_2|}$
• Given that: $\mathbf\small{I_2=\frac{2}{5}I_1}$
2. Substituting the values, we get: $\mathbf\small{I_1\times40=\frac{2}{5}I_1\,|\vec{\omega}_2|}$
⇒ $\mathbf\small{|\vec{\omega}_2|}$ = 100 rpm
Part (b):
1. We have to find the kinetic energy
• We have: Eq.7.26: $\mathbf\small{K=\frac{1}{2}I\,|\vec{\omega}|^2}$
2. Substituting the values, we get:
• Initial kinetic energy $\mathbf\small{K_1=\frac{1}{2}\times I_1 \times 40^2=800I_1}$  
• Final kinetic energy $\mathbf\small{K_2=\frac{1}{2}\times I_2 \times 100^2=5000I_2}$  
3. Taking ratios, we get: $\mathbf\small{\frac{K_1}{K_2}=\frac{800I_1}{5000I_2}=\frac{4I_1}{25I_2}}$
4. But given that: $\mathbf\small{I_2=\frac{2}{5}I_1}$
• Substituting this in (3), we get: $\mathbf\small{\frac{K_1}{K_2}=\frac{4I_1}{25\times \frac{2I_1}{5}}=\frac{2}{5}}$
$\mathbf\small{\Rightarrow K_2=\frac{5}{2}K_1=2.5K_1}$
5. So it is clear that the kinetic energy increased 2.5 times
The reason for increase can be written as follows:
(i) The angular momentum remains the same
$\mathbf\small{I_1\,|\vec{\omega}_1|=I_2\,|\vec{\omega}_2|}$
(ii) $\mathbf\small{|\vec{L}|=I\,|\vec{\omega}|}$ is a linear relation
• But $\mathbf\small{K=\frac{1}{2}I\,|\vec{\omega}|^2}$ is an exponential relation. Because, $\mathbf\small{|\vec{\omega}|}$ has an exponent '2'
• So mathematically, K₂ will not be equal to K₁ even if I₂ has a lower value
(iii) Considering the physical aspect, we know that energy cannot be created. There must be an input source
• In this problem, the source is the muscular work done by the child while he folds his hands back to his body
6. In this problem, we did not convert the angular speed from rpm to rad s^-1
• This is because, when ratios are taken, the units cancel out

Solved example 7.41
A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20 cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m².
(a) What is his new angular speed? (Neglect friction)
(b) Is kinetic energy conserved in the process? If not, from where does the change come about?
Solution:
Part (a):
1. We have: $\mathbf\small{I=\sum\limits_{i=1}^{i=n}{\left(m_i\;r_{i(\bot)}^2 \right)} }$
(Eq.7.25, Chapter 7.23)
• Let us apply this equation for the present case:
(i) First for the man and platform alone:
• Consider each particle of the man-platform system
(ii) Write the mass (m) of each of those particles
• Write the perpendicular distance ($\mathbf\small{r_\bot}$) of each particle from the axis
• Find the sum $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(m_i\;r_{i(\bot)}^2 \right)} }$
• This sum is given to us as 7.6 kg m². So we do not need to calculate it
(iii) But two more particles are present:
• Two 5 kg weights, one in each hand
• They are initially at a distance of 90 cm from the axis
• So the 'initial I' = 7.6 + (2 × 5 × 0.9²) = 15.7 kg m²
(iv) Similarly, 'final I' = 7.6 + (2 × 5 × 0.2²) = 8.0 kg m².
2. We have: $\mathbf\small{I_1\,|\vec{\omega}_1|=I_2\,|\vec{\omega}_2|}$
• Substituting the values, we get: $\mathbf\small{15.7\times30=8.0\times|\vec{\omega}_2|}$
⇒ $\mathbf\small{|\vec{\omega}_2|}$ = 58.88 rpm
Part (b):
1. We have to find the kinetic energy
• We have: Eq.7.26: $\mathbf\small{K=\frac{1}{2}I\,|\vec{\omega}|^2}$
2. Substituting the values, we get:
• Initial kinetic energy $\mathbf\small{K_1=\frac{1}{2}\times 15.7 \times 30^2=7065\,\rm{J}}$  
• Final kinetic energy $\mathbf\small{K_2=\frac{1}{2}\times 8.0\times 58.88^2=13867.42\,\rm{J}}$  
3. We see that, kinetic energy increases. So it is not conserved
The reason for increase can be written as follows:
(i) The angular momentum remains the same
$\mathbf\small{I_1\,|\vec{\omega}_1|=I_2\,|\vec{\omega}_2|}$
(ii) $\mathbf\small{|\vec{L}|=I\,|\vec{\omega}|}$ is a linear relation
• But $\mathbf\small{K=\frac{1}{2}I\,|\vec{\omega}|^2}$ is an exponential relation. Because, $\mathbf\small{|\vec{\omega}|}$ has an exponent '2'
• So mathematically, K₂ will not be equal to K₁ even if I₂ has a lower value
(iii) Considering the physical aspect, we know that energy cannot be created. There must be an input source
• In this problem, the source is the muscular work done by the man while he brings his hands closer to his body
4. In this problem, we did not convert the angular speed from rpm to rad s^-1
• This is because, when ratios are taken, the units cancel out

Solved example 7.42 
A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it. (Hint: The moment of inertia of the door about the vertical axis at one end is ML2/3)
Solution:
1. Linear momentum of the bullet when it hits the door = mv = 0.01 × 500 = 5 kg ms^-1
2. This linear momentum gets converted to angular momentum because, the door starts to rotate
• Angular momentum = Moment of linear momentum
= Linear momentum × r
= 5 × 0.5 = 2.5 kg m² s^-1    
3. This angular momentum is imparted to the door
• Angular momentum of the door = I𝛚
•  I of the door = $\mathbf\small{\frac{ML^2}{3}=\frac{12 \times 1^2}{3}=4}$ kg m²
4. So we get:
2.5 = 4 𝛚 
⇒ 𝛚 = 0.625 rad s^-1.

Solved example 7.43
Two discs of moments of inertia I₁ and I₂ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds 𝛚₁ and 𝛚₂ are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take 𝛚₁ ≠ 𝛚₂.
Solution:
Part (a):
1. Initial angular momentum of disc 1 = I₁𝛚₁
• Initial angular momentum of disc 2 = I₁𝛚₂.
• Sum of the angular momenta = I₁𝛚₁ + I₂𝛚₂
2. When the two discs are in contact, the 'moment of inertia of the combination' (I) is given by:
I = (I₁ + I₂) 
3. Let 𝛚 be the angular velocity of the combination
• Then the the angular momentum of the combination = I𝛚 = (I₁ + I₂)𝛚.
4. Applying the law of conservation of angular momentum, we get:
I₁𝛚₁ + I₂𝛚₂ = (I₁ + I₂)𝛚.
• Thus we get: $\mathbf\small{\omega = \frac{I_1\omega_1+I_2\omega_2}{I_1+I_2}}$
Part (b):
1. Total kinetic energy before the combination = $\mathbf\small{K_i=\frac{1}{2}I_1\omega_1^2+\frac{1}{2}I_2\omega_2^2}$
2. Kinetic energy of the combination = $\mathbf\small{K_f=\frac{1}{2}I\omega^2=\frac{1}{2}(I_1+I_2)\left[ \frac{I_1\omega_1+I_2\omega_2}{I_1+I_2}\right]^2}$
$\mathbf\small{\Rightarrow K_f=\frac{1}{2}\left[ \frac{(I_1\omega_1+I_2\omega_2)^2}{I_1+I_2}\right]}$
3. $\mathbf\small{K_i-K_f=\frac{1}{2}I_1\omega_1^2+\frac{1}{2}I_2\omega_2^2-\frac{1}{2}\left[ \frac{(I_1\omega_1+I_2\omega_2)^2}{I_1+I_2}\right]}$
$\mathbf\small{=\left[\frac{I_1\omega_1^2+I_2\omega_2^2}{2}\right]-\left[ \frac{(I_1\omega_1+I_2\omega_2)^2}{2(I_1+I_2)}\right]}$
$\mathbf\small{=\left[\frac{(I_1\omega_1^2+I_2\omega_2^2)(I_1+I_2)}{2(I_1+I_2)}\right]-\left[ \frac{(I_1\omega_1+I_2\omega_2)^2}{2(I_1+I_2)}\right]}$
$\mathbf\small{=\left[\frac{(I_1\omega_1^2+I_2\omega_2^2)(I_1+I_2)-(I_1\omega_1+I_2\omega_2)^2}{2(I_1+I_2)}\right]}$
• Expansion of the numerator is:
$\mathbf\small{I_1^2\omega_1^2+I_1I_2\omega_2^2+I_1I_2\omega_1^2+I_2^2\omega_2^2-I_1^2\omega_1^2-2I_1I_2\omega_1 \omega_2-I_2^2\omega_2^2}$
$\mathbf\small{=I_1I_2\omega_1^2-2I_1I_2\omega_1\omega_2+I_1I_2\omega_2^2}$
$\mathbf\small{=I_1I_2(\omega_1^2-2\omega_1\omega_2+\omega_2^2)}$
$\mathbf\small{=I_1I_2(\omega_1-\omega_2)^2}$
So we get: $\mathbf\small{K_i-K_f=\left[\frac{I_1I_2(\omega_1-\omega_2)^2}{2(I_1+I_2)}\right]}$
4. $\mathbf\small{(\omega_1-\omega_2)}$ may be positive or negative
• But $\mathbf\small{(\omega_1-\omega_2)^2}$ will be surely positive
• So $\mathbf\small{K_i-K_f}$ is positive
5. That means $\mathbf\small{K_i}$ is greater than $\mathbf\small{K_f}$
• That means there is energy loss
• This energy loss is due to the friction between the two discs

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In the next section, we will see rolling motion

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Chapter 7.32 - Torque when rotation is about a fixed axis

In the previous section, we obtained two results for rotation about a fixed axis:
• If the rotation is symmetric (symmetric objects rotating about an axis of symmetry), then:
    ♦ Eq,7.31: $\mathbf\small{\vec{L}=\vec{L}_z}$
• If the rotation is not symmetric, then
    ♦ Eq.7.30: $\mathbf\small{\vec{L}=\vec{L}_z+\vec{L}_\bot}$

• In this section, we will derive the expression for torque acting on such bodies
1. We will consider the general case when the rotation is not symmetric
• That is., we will consider the Eq.7.30
2. $\mathbf\small{\vec{L}}$ is the angular momentum
• We know that, 'change in angular momentum' per unit time will give torque
• So $\mathbf\small{\frac{d \vec{L}}{dt} }$ will give the torque acting on the body
3. When we differentiate the left side, we must differentiate the terms on the right side also
• So we can write: $\mathbf\small{\frac{d \vec{L}}{dt}=\frac{d \vec{L}_z}{dt}+\frac{d \vec{L}_\bot}{dt}}$
4. The term on the left is torque
• So each of the two terms on the right side must be torques
5. The last term is related to torques which are not parallel to the axis of rotation
• We have seen that:
When the axis is fixed, those non-parallel torques have no effect
• So we need not consider the last term
6. We can write:
$\mathbf\small{\frac{d \vec{L}}{dt}=\frac{d \vec{L}_z}{dt}}$
• Let us differentiate the right side:
$\mathbf\small{\frac{d \vec{L}_z}{dt}=\frac{d (I|\vec{\omega}|\hat{k})}{dt}}$
7. In this differentiation, we are taking the ['change in $\mathbf\small{(I|\vec{\omega}|\hat{k})}$' per unit time] at an instant 
• The body is rigid and the axis is fixed. So I does not change
• $\mathbf\small{\hat{k}}$ has a constant magnitude and direction. It also does not change
• So those two items can be taken outside. We get: $\mathbf\small{\frac{d \vec{L}_z}{dt}=I \hat{k}\frac{d (|\vec{\omega}|)}{dt}}$
8. But $\mathbf\small{\frac{d (|\vec{\omega}|)}{dt}}$ is the angular acceleration $\mathbf\small{\alpha}$
So we get: $\mathbf\small{\frac{d \vec{L}_z}{dt}=(I \alpha) \hat{k}}$
$\mathbf\small{\Rightarrow \vec{\tau}=(I \alpha) \hat{k}}$
$\mathbf\small{\Rightarrow |\vec{\tau}|=I \alpha}$
Thus we get the expression for the torque acting on a body rotating about a fixed axis
9. This expression is applicable to all rigid bodies rotating about a fixed axis. So we do not need to consider 'symmetric bodies in symmetric rotation' separately
• That means, we do not need to write the above steps for Eq.7.31

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Now we will see some solved examples

Solved example 7.36
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s^-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Solution:
1. A solid cylinder is a symmetric body. It is rotating about an axis of symmetry
• So we can use Eq.7.31: $\mathbf\small{\vec{L}=\vec{L}_z=I|\vec{\omega}|\,\hat{k}}$
2. This solid cylinder rotates about it's axis
• So $\mathbf\small{I=\frac{MR^2}{2}}$
• Substituting the values, we get: $\mathbf\small{I=\frac{20\times 0.25^2}{2}=0.625\, \rm{kg\;m^2}}$  
3. Given that $\mathbf\small{|\vec{\omega}|}$ = 100 rad s^-1
4. Thus we get:
Angular momentum = $\mathbf\small{I|\vec{\omega}|\hat{k}=0.625\times 100=62.5 \,\hat{k}}$
• Here $\mathbf\small{\hat{k}}$ is the unit vector parallel to the axis of the cylinder
5. Unit of angular momentum:
• Angular momentum is defined as the moment of linear momentum
• That is., we are multiplying the linear momentum by a distance
• The unit of linear momentum is kg ms^-1
• So the unit of angular momentum is kg m²s^-1
• Thus the angular momentum is: 62.5 kg m²s^-1
6. Next we have to find the kinetic energy
• We have: Eq.7.26: $\mathbf\small{K=\frac{1}{2}I\,|\vec{\omega}|^2}$
• Substituting the values, we get: $\mathbf\small{K=\frac{1}{2}\times 0.625 \times 100^2=3125\, \rm{J}}$ 

Solved example 7.37

A 40 kg flywheel in the form of a uniform circular disc of 1 m radius is making 120 rpm. Determine the angular momentum
Solution:
1. A circular disc is a symmetric body. It is rotating about an axis of symmetry
• So we can use Eq.7.31: $\mathbf\small{\vec{L}=\vec{L}_z=I|\vec{\omega}|\,\hat{k}}$
2. The fly wheel rotates about a perpendicular axis passing through the center
• So $\mathbf\small{I=\frac{MR^2}{2}}$
• Substituting the values, we get: $\mathbf\small{I=\frac{40\times 1^2}{2}=20\, \rm{kg\;m^2}}$  
3. Given that $\mathbf\small{|\vec{\omega}|}$ = 120 rpm
• We have to convert it into rad s^-1.
• 1 rotation = 2𝝅 rad
⇒ 120 rotations = 240𝝅 rad
⇒ 240𝝅 rad is covered in 60 s
• So angle covered in 1 s = $\mathbf\small{|\vec{\omega}|=\frac{240 \pi}{60}=4\pi\,\,\text{rad s}^{-1}}$
4. Thus we get:
Angular momentum = $\mathbf\small{I|\vec{\omega}|\hat{k}=20\times 4 \pi=251.2 \,\hat{k}}$
• Here $\mathbf\small{\hat{k}}$ is the unit vector parallel to the axis of the fly wheel
5. Unit of angular momentum:
• Angular momentum is defined as the moment of linear momentum
• That is., we are multiplying the linear momentum by a distance
• The unit of linear momentum is kg ms^-1
• So the unit of angular momentum is kg m²s^-1
• Thus the required answer is: 251.2 kg m²s^-1

Solved example 7.38
A wheel is rotating with an angular momentum of 3 kg m²s^-1. A torque of 12 Nm is applied on the wheel for 4 s. What is the final angular momentum of the wheel?
Solution:
We have: Torque = Time rate of change of angular momentum 
⇒ Torque = ^{(Final angular momentum - Initial angular momentum)}⁄_time
⇒ 12 = ^{(Final angular momentum - 3)}⁄₄
⇒ Final angular momentum = (48+3) = 51 kg m²s^-1

Solved example 7.39
The diameter of a circular disc is 0.5 m and it's mass is 16 kg. What torque will increase it's angular velocity from zero to 120 rpm in 8 s?
Solution:
1. We have: $\mathbf\small{I=\frac{MR^2}{2}}$
• Substituting the values, we get: $\mathbf\small{I=\frac{16\times 0.25^2}{2}=0.5\, \rm{kg\;m^2}}$  
2. Initial angular velocity = 0
• So initial angular momentum = 0
3. Final angular velocity = 120 rpm = 4𝝅 rad s^-1.
• So final angular momentum = $\mathbf\small{I|\vec{\omega}|\hat{k}=0.5\times 4 \pi=6.28 \,\hat{k}}$
• Here $\mathbf\small{\hat{k}}$ is the unit vector parallel to the axis of the disc
4. We have: Torque = Time rate of change of angular momentum 
⇒ Torque = ^{(Final angular momentum - Initial angular momentum)}⁄_time
⇒ Torque = ^{(6.28 - 0)}⁄₈ = 0.785 N m

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In the next section, we will see conservation of angular momentum

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Chapter 7.31 - Angular Momentum about a Fixed axis

In the previous section, we saw power derived from a torque. In this section, we will see .angular momentum in case of rotation about a fixed axis'

In chap 7.17, we saw the basics about angular momentum of a system of particles
We saw the following information:
Information 1:
Eq.7.19: $\mathbf\small{\frac{dl}{dt}=\vec{r}\times \vec{F}=\vec{\tau}}$
• That is: 'Rate of change of angular momentum' calculated over an instant is the torque experienced by the particle at that instant

Information 2:
• To get the total angular momentum $\mathbf\small{\vec{L}}$ of a system of particles, we add the angular momenta of individual particles
• Thus for a system of n particles, we have:
$\mathbf\small{\vec{L}=\vec{l}_1+\vec{l}_1+\vec{l}_1\;+\; .\; .\; .\;+\vec{l}_n=\sum\limits_{i=1}^{i=n}{\vec{l}_i}}$
• Note that, this is an addition of vectors. That is., we have to add the momenta vectorially
• In the above equation, we know that: $\mathbf\small{\vec{l}_i=\vec{r}_i \times \vec{p}_i}$
• So the equation in (1) becomes:
Eq.7.20: $\mathbf\small{\vec{L}=\sum\limits_{i=1}^{i=n}{(\vec{r}_i\times \vec{p}_i)}}$
• If required, the $\mathbf\small{\vec{p}_i}$ can be further split up as: $\mathbf\small{\vec{p}_i=m_i \times \vec{v}_i}$

Information 3:
Eq.7.21: $\mathbf\small{\frac{dL}{dt}=\vec{\tau}_{ext}}$
• Based on this equation, we can write:
The 'rate of change of angular momentum of a system of particles' at any instant is equal to the 'sum of external torques on all the particles'

---

We now want to see a special case:
The angular momentum about a fixed axis
We will write it in steps:
1. From information 2 above, we have:
$\mathbf\small{\vec{L}=\sum\limits_{i=1}^{i=n}{(\vec{r}_i\times \vec{p}_i)}}$
2. In this equation, $\mathbf\small{(\vec{r}_i\times \vec{p}_i)}$ is the angular momentum of a single particle (the i^th particle)
• Also recall that the angular momentum of a single particle is denoted by '$\mathbf\small{\vec{l}}$'
• So the angular momentum of the i^th particle is denoted by '$\mathbf\small{\vec{l}}$'
• That is., $\mathbf\small{\vec{l}_i=(\vec{r}_i\times \vec{p}_i)}$  
3. We will consider that typical particle first.
• That is., we will find the '$\mathbf\small{\vec{l}_i}$ of a typical particle about a fixed axis' first
• Then we will sum up the angular momenta of all the particles to get the $\mathbf\small{\vec{L}}$ of the whole body
4. Consider again fig.7.79 that we saw in a previous section. It is shown again below:

Fig.7.79

• For the single particle (indicated by the red sphere), we have: $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
5. $\mathbf\small{\vec{r}}$ can be split up as: $\mathbf\small{\vec{r}=\vec{OC}+\vec{CP}}$
• So we get: $\mathbf\small{\vec{l}=(\vec{OC}+\vec{CP})\times \vec{p}}$
$\mathbf\small{\Rightarrow \vec{l}=\left(\vec{OC}\times \vec{p}+\vec{CP}\times \vec{p}\right)}$
• But $\mathbf\small{\vec{p}=m \times \vec{v}}$
• So we get: $\mathbf\small{\vec{l}=\left[\vec{OC}\times (m\,\vec{v})+\vec{CP}\times (m\,\vec{v})\right]}$
6. There are two terms on the right side
• Consider the first term: $\mathbf\small{\vec{OC}\times (m\,\vec{v})}$
• It is the cross product of two vectors: $\mathbf\small{\vec{OC}}$ and $\mathbf\small{(m\,\vec{v})}$
• Obviously the resulting vector will be perpendicular to $\mathbf\small{\vec{OC}}$
• The vector $\mathbf\small{\vec{OC}}$ is along the axis of rotation
• So the resulting vector from the first term is perpendicular to the axis of rotation
7. Consider the second term: $\mathbf\small{\vec{CP}\times (m\,\vec{v})}$
• $\mathbf\small{\vec{CP}}$ lies on the plane of the red circle 
• $\mathbf\small{(m\,\vec{v})}$ also lies on the plane of the red circle
• So the resulting vector will be perpendicular to the plane of the red circle
• That means, the resulting vector will be parallel to the axis of rotation
8. Let us calculate this 'parallel vector' obtained  from the second term: $\mathbf\small{\vec{CP}\times (m\,\vec{v})}$
• We will apply the cross product rule: $\mathbf\small{|(\vec{A}\times \vec{B})|=|\vec{A}|\times |\vec{B}|\times \sin \theta}$
• In the present case, we have:
    ♦ $\mathbf\small{|\vec{CP}|=r_\bot}$
    ♦ $\mathbf\small{|\vec{v}|=r_\bot\,|\vec{\omega}|}$
    ♦ Angle between the two vectors = 90^o. So sin θ = sin 90 = 1
• Thus we get: $\mathbf\small{\left|\left(\vec{CP}\times (m\,\vec{v})\right)\right|=r_\bot \times (m\,r_\bot\,|\vec{\omega}|)\times 1}$
$\mathbf\small{\Rightarrow \left|\left(\vec{CP}\times (m\,\vec{v})\right)\right|=(m\,r_\bot^2\,|\vec{\omega}|)}$
9. This is the magnitude of the resulting vector of the second term
• We want the direction also
• We saw that, the resulting vector is parallel to the axis of rotation
    ♦ The axis of rotation is the z-axis
    ♦ The unit vector along the z-axis is $\mathbf\small{\hat{k}}$
• So we get:
The resulting vector from the second term is: $\mathbf\small{(m\,r_\bot^2\,|\vec{\omega}|)\hat{k}}$
10. Since this vector is parallel to the axis of rotation (the z-axis), we will denote it as $\mathbf\small{\vec{l}_z}$ 
• So we can write: $\mathbf\small{\vec{l}_z=(m\,r_\bot^2\,|\vec{\omega}|)\hat{k}}$
• So the result in (5) becomes: $\mathbf\small{\vec{l}=\vec{l}_z+\left(\vec{OC}\times (m\,\vec{v})\right)}$
11. This is a vector sum
• The first component $\mathbf\small{\vec{l}_z}$ is parallel to the axis of rotation
• The second component $\mathbf\small{\left(\vec{OC}\times (m\,\vec{v})\right)}$ is not parallel to the axis of rotation
• So the vector sum $\mathbf\small{\vec{l}=\vec{l}_z+\left(\vec{OC}\times (m\,\vec{v})\right)}$ is not parallel to the axis of rotation
12. We know that the angular velocity $\mathbf\small{\vec{\omega}}$ is parallel to the axis of rotation (See fig.7.78 in chapter 7.15)
• So we get an important result:
The angular velocity $\mathbf\small{\vec{\omega}}$ and angular momentum $\mathbf\small{\vec{l}}$ of a particle are not necessarily parallel
• A comparison with translational motion:
In translational motion we have:
Linear velocity $\mathbf\small{\vec{v}}$ and linear momentum $\mathbf\small{\vec{p}}$ are always parallel to each other

---

13. So we calculated the angular momentum of a single particle
• It is given by the expression in (10)

• Now we add the angular momenta of all the particles
• The sum will give the angular momentum $\mathbf\small{\vec{L}}$ of the whole body
14. So we can write:
$\mathbf\small{L=\sum\limits_{i=1}^{i=n}{\left(l_i \right)}}$
$\mathbf\small{\Rightarrow L=\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}+\sum\limits_{i=1}^{i=n}{\left(\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)}) \right)}}$
Where:
• $\mathbf\small{\vec{l}_{z(i)}}$ is the 'angular momentum component' of the i^th  parallel to the axis of rotation
• $\mathbf\small{\vec{OC}_{(i)}}$ is the vector between the two points:
    ♦ Origin O of the reference system
    ♦ Center C of the circle described by the i^th particle
• $\mathbf\small{m_{(i)}}$ is the mass of the i^th particle
• $\mathbf\small{\vec{v}_{(i)}}$ is the linear velocity of the i^th particle
15. Thus we see that, L also has two components
Consider the first component: $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}}$
• All vectors involved in this summation are parallel to the z-axis (axis of rotation)
• So the result of the summation will be parallel to the z-axis
• So we will denote this component as $\mathbf\small{\vec{L}_z}$
16. Consider the second component: $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)}) \right)}}$
• All vectors involved in this summation are perpendicular to the z-axis. We know the reason:
    ♦ The vector product $\mathbf\small{\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)})}$ will be perpendicular to $\mathbf\small{\vec{OC}_{(i)}}$  
    ♦ $\mathbf\small{\vec{OC}_{(i)}}$ lies along the z-axis
• So the result of the summation will be perpendicular to the z-axis
• So we will denote this component as $\mathbf\small{\vec{L}_\bot}$
17. Thus we can write:
Eq.7.30:
$\mathbf\small{\vec{L}=\vec{L}_z+\vec{L}_\bot}$
• Let us consider each component separately:
We have: $\mathbf\small{\vec{L}_z=\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}}$
• But from (10), we have: $\mathbf\small{\vec{l}_{z(i)}=(m_{(i)}\,r_{\bot (i)}^2\,|\vec{\omega}|)\hat{k}}$
• So, when we take the summation $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}}$,
We are actually taking the summation: $\mathbf\small{\sum\limits_{i=1}^{i=n}{(m_{(i)}\,r_{\bot (i)}^2\,|\vec{\omega}|)\hat{k}}}$
• $\mathbf\small{|\vec{\omega}|\hat{k}}$ is the same for all particles. So it can be taken outside
• So we get: $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}=|\vec{\omega}|\,\hat{k}\sum\limits_{i=1}^{i=n}{(m_{(i)}\,r_{\bot (i)}^2)}}$
18. But $\mathbf\small{\sum\limits_{i=1}^{i=n}{(m_{(i)}\,r_{\bot (i)}^2)}}$ = I, the moment of inertia of the whole body (Eq.7.25 in chapter 7.23)
So we get: $\mathbf\small{\vec{L}_z=\sum\limits_{i=1}^{i=n}{\left(\vec{l}_{z(i)} \right)}=I|\vec{\omega}|\,\hat{k}}$
19. Now consider the second component: $\mathbf\small{\vec{L}_\bot=\sum\limits_{i=1}^{i=n}{\left(\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)}) \right)}}$
• This is a summation of vectors
• Each of those vectors results from the 'vector multiplication' of two vectors: $\mathbf\small{\vec{OC}_{(i)}\;\;\rm{and}\;\; (m_{(i)}\,\vec{v}_{(i)})}$
Let us see an example:
(i) Consider the square rod shown in fig.7.136 (a) below:

Fig.7.136

• It is rotating about the axis shown in blue color
• The direction of rotation is indicated by the yellow curved arrow
• It is a uniform square rod. Also, the blue axis passes through the exact center of the rod
(ii) A particle is isolated at position 'P' along the rod
• It is shown in red color
• Let us write the properties of that particle:
    ♦ $\mathbf\small{\vec{OC}_{(i)}=\vec{OC}}$
    ♦ $\mathbf\small{m_{(i)}=m_P}$
    ♦ $\mathbf\small{\vec{v}_{(i)}=\vec{v}_{(P)}}$
• So for the particle at P, we can easily calculate the cross product of $\mathbf\small{\vec{OC}_{(i)}\;\;\rm{and}\;\; (m_{(i)}\,\vec{v}_{(i)})}$
(iii) For the particle at P, there will be an exact replica on the other side of the axis
• This particle is at Q
(An exact replica is obtained because, the square rod is uniform and also, the axis passes through the exact center)
• The position of Q will be such that, CQ = CP 
• Let us write the properties of the particle at Q:
    ♦ $\mathbf\small{\vec{OC}_{(i)}=\vec{OC}}$
    ♦ $\mathbf\small{m_{(i)}=m_Q}$
    ♦ $\mathbf\small{\vec{v}_{(i)}=\vec{v}_{(Q)}}$
• So for the particle at Q also, we can easily calculate the cross product of $\mathbf\small{\vec{OC}_{(i)}\;\;\rm{and}\;\; (m_{(i)}\,\vec{v}_{(i)})}$
(iv) Let us write a comparison of properties:
• For both the particles, $\mathbf\small{\vec{OC}_{(i)}=\vec{OC}}$
• Because of the symmetry, $\mathbf\small{m_{(P)}=m_Q}$  
• Because of the symmetry, the magnitudes of $\mathbf\small{\vec{v}_{(P)}\;\;\rm{and}\;\; \vec{v}_{(Q)}}$ will be the same
    ♦ But their directions will be exactly opposite to each other
(v) Consider the two cross products:
(a) Cross product of $\mathbf\small{\vec{OC}_{(P)}\;\;\rm{and}\;\; (m_{(P)}\,\vec{v}_{(P)})}$ 
(b) Cross product of $\mathbf\small{\vec{OC}_{(Q)}\;\;\rm{and}\;\; (m_{(Q)}\,\vec{v}_{(Q)})}$ 
• The 'vector obtained as the cross product in (a)' will be equal and opposite to the 'vector obtained as the cross product in (b)'
• So in the summation, the two vectors will cancel each other
(vi) For each particle in the rod, there will be an exact replica on the other side of the axis
• So all particles in the rod, can be grouped into pairs such that:
• In any pair, one member is the exact replica of the other
(vii) If we draw a circle with center at C and passing through P,
    ♦ The point Q will also lie on the circle
    ♦ The point Q qill be diametrically opposite to P
• This is true for all symmetric bodies
• The members of the pairs will lie diametrically opposite on the circle with center at C
(vii) So the net result is that, the summation $\mathbf\small{\vec{L}_\bot=\sum\limits_{i=1}^{i=n}{\left(\vec{OC}_{(i)}\times (m_{(i)}\,\vec{v}_{(i)}) \right)}}$ will become zero 
• That means $\mathbf\small{\vec{L}_\bot=0}$
20. So the result in (1) becomes: $\mathbf\small{\vec{L}=\vec{L}_z}$
• But for this simplification, two conditions should be satisfied:
(i) The object must be uniform and symmetric
(ii) The rotation must be about the axis of symmetry
• For our present discussion we consider only those rotations which satisfy the two conditions
• So we can confidently use the simplified form
We can write:
Eq.7.31:
For symmetric rotation of symmetric bodies, $\mathbf\small{\vec{L}=I|\vec{\omega}|\,\hat{k}}$

---

Let us write a summary about the above discussion:
■ In chap 7.17, we obtained the general equation for the angular momentum of any object:
Eq.7.20: $\mathbf\small{\vec{L}=\sum\limits_{i=1}^{i=n}{(\vec{r}_i\times \vec{p}_i)}}$

■ In the present section, we applied it to objects which rotate about a fixed axis. We obtained:
Eq.7.30:

$\mathbf\small{\vec{L}=\vec{L}_z+\vec{L}_\bot}$
■ Also, we applied it to symmetric objects which rotate symmetrically about a fixed axis. We obtained:
Eq.7.31:
For symmetric rotation of symmetric bodies, $\mathbf\small{\vec{L}=I|\vec{\omega}|\,\hat{k}}$

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In the next section, we will see applications of Eq.7.30

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