In the previous section, we saw the basics about torque. In this section we will see angular momentum
• We are familiar with the ‘role played by force’ in translational motion
♦ In rotational motion, that role is played by ‘moment of force’
♦ We saw this in the previous section
• Now, what about the ‘role played by linear momentum’ in translational motion?
♦ In rotational motion, that role is played by angular momentum
♦ Angular momentum can be considered as the ‘moment of linear momentum’
♦ It is denoted by the letter ‘l’
♦ It is a vector. So it is denoted as: $\mathbf\small{\vec{l}}$
• We know that moment of a force is given by: $\mathbf\small{\vec{\tau}=\vec{r}\times \vec{F}}$
♦ In a similar way, moment of linear momentum is given by:
Eq.7.18: $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
• We are familiar with the ‘role played by force’ in translational motion
♦ In rotational motion, that role is played by ‘moment of force’
♦ We saw this in the previous section
• Now, what about the ‘role played by linear momentum’ in translational motion?
♦ In rotational motion, that role is played by angular momentum
♦ Angular momentum can be considered as the ‘moment of linear momentum’
♦ It is denoted by the letter ‘l’
♦ It is a vector. So it is denoted as: $\mathbf\small{\vec{l}}$
• We know that moment of a force is given by: $\mathbf\small{\vec{\tau}=\vec{r}\times \vec{F}}$
♦ In a similar way, moment of linear momentum is given by:
Eq.7.18: $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
By comparing $\mathbf\small{\vec{l}}$ with $\mathbf\small{\vec{\tau}}$, we can write many useful results:
• $\mathbf\small{|\vec{l}|=|\vec{r}|\times |\vec{p}| \times \sin \theta}$
• $\mathbf\small{|\vec{l}|=|\vec{r}|\times |\vec{p}_\bot|}$
• $\mathbf\small{|\vec{l}|=|\vec{r}_\bot|\times |\vec{p}|}$
• If $\mathbf\small{\vec{p}}$ passes through the origin, $\mathbf\small{|\vec{l}|}$ will become zero
• If $\mathbf\small{\vec{p}}$ is parallel to $\mathbf\small{\vec{r}}$, $\mathbf\small{|\vec{l}|}$ will become zero
• If $\mathbf\small{|\vec{p}| = 0}$, $\mathbf\small{|\vec{l}|}$ will become zero
• We have seen that:
♦ Torque is: Moment of force
♦ Angular momentum is: Moment of Linear momentum
• So now we will see the relation between Torque and Angular momentum
• We will first see it for a single particle in ‘pure rotation’
• We will write it in steps:
1. Consider fig.7.76 that we saw in a previous section
For convenience, it is shown again below:
• Let the mass of the particle be 'm'
2. At the instant when the reading in the stop watch is 't1', the particle is at A
♦ The linear velocity of the particle at that instant is $\mathbf\small{\vec{v}_{(t1)}}$
♦ It is tangential to the circular path
♦ It is shown in magenta color
♦ The linear momentum of the particle at that instant is $\mathbf\small{m\vec{v}_{(t1)}}$
♦ The angular momentum of the particle at that instant is $\mathbf\small{m\vec{v}_{(t1)}r_\bot}$
• At the instant when the reading in the stop watch is 't2', the particle is at B
♦ The linear velocity of the particle at that instant is $\mathbf\small{\vec{v}_{(t2)}}$
♦ It is tangential to the circular path
♦ It is shown in magenta color
♦ The linear momentum of the particle at that instant is $\mathbf\small{m\vec{v}_{(t2)}}$
♦ The angular momentum of the particle at that instant is $\mathbf\small{m\vec{v}_{(t2)}r_\bot}$
3. The time elapsed between the two instants = Δt = (t2-t1)
4. During this time interval (Δt), the angular momentum of the particle changes from $\mathbf\small{m\vec{v}_{(t1)}r_\bot}$ to $\mathbf\small{m\vec{v}_{(t2)}r_\bot}$
5. So 'change in the angular momentum' = $\mathbf\small{m\vec{v}_{(t2)}r_\bot\;-\;m\vec{v}_{(t1)}r_\bot}$
⇒ 'Change in the angular momentum' = $\mathbf\small{mr_\bot [\vec{v}_{(t2)}\;-\;\vec{v}_{(t1)}]}$
6. If we divide the above result by the time duration Δt, we will get the 'rate of change of angular momentum'
• So we can write:
Rate of change of angular momentum = $\mathbf\small{\frac{mr_\bot [\vec{v}_{(t2)}\;-\;\vec{v}_{(t1)}]}{\Delta t}}$
7. But $\mathbf\small{\frac{[\vec{v}_{(t2)}\;-\;\vec{v}_{(t1)}]}{\Delta t}=\vec{a}_{\text{average}}}$
• Where $\mathbf\small{\vec{a}_{\text{average}}}$ is the average acceleration with which the particle moves from A to B
8. So we get:
• Rate of change of angular momentum = $\mathbf\small{mr_\bot \vec{a}_{\text{average}}}$
• Rearranging this, we get:
Rate of change of angular momentum = $\mathbf\small{r_\bot (m\vec{a}_{\text{average}})}$
• But '$\mathbf\small{(m\vec{a}_{\text{average}})}$' is the 'average force' ($\mathbf\small{\vec{F}_{\text{average}}}$) experienced by the particle while moving from A to B
9. So we can write:
• Rate of change of angular momentum = $\mathbf\small{r_\bot \vec{F}_{\text{average}}}$
• But $\mathbf\small{r_\bot \vec{F}_{\text{average}}}$ is the average torque ($\mathbf\small{\vec{\tau}_{\text{average}}}$) experienced by the particle while moving from A to B
10. So we can write:
• Rate of change of angular momentum = $\mathbf\small{\vec{\tau}_{\text{average}}}$
11. If the time duration Δt is very small, we can call it an instant
• The 'Rate of change of angular momentum' calculated at an instant is not average torque
• It is instantaneous torque $\mathbf\small{\vec{\tau}}$
12. Thus we get the relation between angular momentum and torque:
• 'Rate of change of angular momentum' calculated over an instant is the torque experienced by the particle at that instant
• $\mathbf\small{|\vec{l}|=|\vec{r}|\times |\vec{p}| \times \sin \theta}$
• $\mathbf\small{|\vec{l}|=|\vec{r}|\times |\vec{p}_\bot|}$
• $\mathbf\small{|\vec{l}|=|\vec{r}_\bot|\times |\vec{p}|}$
• If $\mathbf\small{\vec{p}}$ passes through the origin, $\mathbf\small{|\vec{l}|}$ will become zero
• If $\mathbf\small{\vec{p}}$ is parallel to $\mathbf\small{\vec{r}}$, $\mathbf\small{|\vec{l}|}$ will become zero
• If $\mathbf\small{|\vec{p}| = 0}$, $\mathbf\small{|\vec{l}|}$ will become zero
Relation between torque and angular momentum
♦ Torque is: Moment of force
♦ Angular momentum is: Moment of Linear momentum
• So now we will see the relation between Torque and Angular momentum
• We will first see it for a single particle in ‘pure rotation’
• We will write it in steps:
1. Consider fig.7.76 that we saw in a previous section
For convenience, it is shown again below:
 |
| Fig.7.76 |
2. At the instant when the reading in the stop watch is 't1', the particle is at A
♦ The linear velocity of the particle at that instant is $\mathbf\small{\vec{v}_{(t1)}}$
♦ It is tangential to the circular path
♦ It is shown in magenta color
♦ The linear momentum of the particle at that instant is $\mathbf\small{m\vec{v}_{(t1)}}$
♦ The angular momentum of the particle at that instant is $\mathbf\small{m\vec{v}_{(t1)}r_\bot}$
• At the instant when the reading in the stop watch is 't2', the particle is at B
♦ The linear velocity of the particle at that instant is $\mathbf\small{\vec{v}_{(t2)}}$
♦ It is tangential to the circular path
♦ It is shown in magenta color
♦ The linear momentum of the particle at that instant is $\mathbf\small{m\vec{v}_{(t2)}}$
♦ The angular momentum of the particle at that instant is $\mathbf\small{m\vec{v}_{(t2)}r_\bot}$
3. The time elapsed between the two instants = Δt = (t2-t1)
4. During this time interval (Δt), the angular momentum of the particle changes from $\mathbf\small{m\vec{v}_{(t1)}r_\bot}$ to $\mathbf\small{m\vec{v}_{(t2)}r_\bot}$
5. So 'change in the angular momentum' = $\mathbf\small{m\vec{v}_{(t2)}r_\bot\;-\;m\vec{v}_{(t1)}r_\bot}$
⇒ 'Change in the angular momentum' = $\mathbf\small{mr_\bot [\vec{v}_{(t2)}\;-\;\vec{v}_{(t1)}]}$
6. If we divide the above result by the time duration Δt, we will get the 'rate of change of angular momentum'
• So we can write:
Rate of change of angular momentum = $\mathbf\small{\frac{mr_\bot [\vec{v}_{(t2)}\;-\;\vec{v}_{(t1)}]}{\Delta t}}$
7. But $\mathbf\small{\frac{[\vec{v}_{(t2)}\;-\;\vec{v}_{(t1)}]}{\Delta t}=\vec{a}_{\text{average}}}$
• Where $\mathbf\small{\vec{a}_{\text{average}}}$ is the average acceleration with which the particle moves from A to B
8. So we get:
• Rate of change of angular momentum = $\mathbf\small{mr_\bot \vec{a}_{\text{average}}}$
• Rearranging this, we get:
Rate of change of angular momentum = $\mathbf\small{r_\bot (m\vec{a}_{\text{average}})}$
• But '$\mathbf\small{(m\vec{a}_{\text{average}})}$' is the 'average force' ($\mathbf\small{\vec{F}_{\text{average}}}$) experienced by the particle while moving from A to B
9. So we can write:
• Rate of change of angular momentum = $\mathbf\small{r_\bot \vec{F}_{\text{average}}}$
• But $\mathbf\small{r_\bot \vec{F}_{\text{average}}}$ is the average torque ($\mathbf\small{\vec{\tau}_{\text{average}}}$) experienced by the particle while moving from A to B
10. So we can write:
• Rate of change of angular momentum = $\mathbf\small{\vec{\tau}_{\text{average}}}$
11. If the time duration Δt is very small, we can call it an instant
• The 'Rate of change of angular momentum' calculated at an instant is not average torque
• It is instantaneous torque $\mathbf\small{\vec{\tau}}$
12. Thus we get the relation between angular momentum and torque:
• 'Rate of change of angular momentum' calculated over an instant is the torque experienced by the particle at that instant
• In the above discussion, we did the calculations based on the fact that $\mathbf\small{r_\bot}$ is constant
• Now we will see the general case where $\mathbf\small{r_\bot}$ is not a constant. We will write it in steps:
1. 'Calculation over an instant' can be easily done using calculus
• In calculus, the calculation of 'Rate of change of angular momentum' over an instant is written as:
$\mathbf\small{\frac{d}{dt} \text{(Angular momentum)}}$
(This is read as: 'd by dt of Angular momentum')
• In short form, we write: $\mathbf\small{\frac{dl}{dt}}$
(This is read as: 'dl by dt')
• This is the differentiation of 'l', with respect to time 't'
• When we do the differentiation, what do we get?
• Let us find out:
2. $\mathbf\small{\frac{dl}{dt}=\frac{d(\vec{r}\times \vec{p})}{dt}}$
(∵ $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$)
• Note that, here $\mathbf\small{\vec{r}}$ is not a constant
• So we have to differentiate the product $\mathbf\small{(\vec{r}\times \vec{p})}$
• We have to apply the product rule:
• We get: $\mathbf\small{\frac{dl}{dt}=\left [\frac{d(\vec{r}\times \vec{p})}{dt} \right ]=\left [\frac{d\vec{r}}{dt}\times \vec{p}+\vec{r} \times \frac{d\vec{p}}{dt}\right ]}$
3. Consider the first term on the right:
$\mathbf\small{\frac{d\vec{r}}{dt}\times \vec{p}}$
• We know that $\mathbf\small{\frac{d\vec{r}}{dt} = \vec{v}}$
• So $\mathbf\small{\frac{d\vec{r}}{dt}\times \vec{p}=\vec{v}\times \vec{p}}$
$\mathbf\small{\Rightarrow \frac{d\vec{r}}{dt}\times \vec{p}=\vec{v}\times m\vec{v}}$
• But $\mathbf\small{\vec{v}}$ and $\mathbf\small{m\vec{v}}$ are parallel vectors
• So their cross product is $\mathbf\small{\vec{0}}$
• Thus the first term vanishes
4. We get:
$\mathbf\small{\frac{dl}{dt}=\left [\frac{d(\vec{r}\times \vec{p})}{dt}\right ]=\left [\vec{r} \times \frac{d\vec{p}}{dt} \right ]}$
• Now, $\mathbf\small{\frac{d\vec{p}}{dt}=\vec{F}}$
• So we get: $\mathbf\small{\frac{dl}{dt}=\left [\frac{d(\vec{r}\times \vec{p})}{dt}\right ]=\left [\vec{r}\times \vec{F} \right ]}$
5. But $\mathbf\small{\vec{r}\times \vec{F}=\vec{\tau}}$
• Thus we get:
Eq.7.19: $\mathbf\small{\frac{dl}{dt}=\vec{r}\times \vec{F}=\vec{\tau}}$
■ Thus we get the same relation as before:
• 'Rate of change of angular momentum' calculated over an instant is the torque experienced by the particle at that instant
1. In translational motion:
• Time rate of change of linear momentum = Force
• This we know as Newton's second law
2. In rotational motion:
• Time rate of change of angular momentum = Torque
• This is analogous to the Newton's second law
(i) Torque experienced by a single particle
(ii) Angular momentum experienced by a single particle
• Next we will see these:
(i) Torque experienced by a system of particles
(ii) Angular momentum experienced by a system of particles
First we will see angular momentum
1. 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
2. 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: $\mathbf\small{\vec{p}_i=m_i \times \vec{v}_i}$
3. We know that, if we differentiate angular momentum ($\mathbf\small{\vec{l}_i}$) of a single particle, we will get the torque ($\mathbf\small{\vec{\tau}_i}$) experienced by that particle (see Eq.7.19)
• Let us now differentiate the total angular momentum ($\mathbf\small{\vec{L}}$) of a system of particles:
• We have: $\mathbf\small{\frac{dL}{dt}=\frac{d}{dt}\left (\sum\limits_{i=1}^{i=n}{\vec{l}_i} \right )}$
4. So we have to differentiate a quantity which is 'a sum of several quantities'
• For that, we have to differentiate each item and then add the results
• That is: $\mathbf\small{\frac{dL}{dt}=\frac{d}{dt}(\vec{l}_1)+\frac{d}{dt}(\vec{l}_2)+\frac{d}{dt}(\vec{l}_3)\;+\; .\; .\; .\;+\frac{d}{dt}(\vec{l}_n)}$
5. But when we differentiate the angular momentum of a particle, we get the torque experienced by that particle (Eq.7.19)
• So the result in (4) can be written as: $\mathbf\small{\frac{dL}{dt}=\vec{\tau}_1+\vec{\tau}_2+\vec{\tau}_3\;+\; .\; .\; .\;+\;\vec{\tau}_n}$
6. But
• $\mathbf\small{\vec{\tau}_1=\vec{r}_1 \times \vec{F}_1}$
• $\mathbf\small{\vec{\tau}_2=\vec{r}_2 \times \vec{F}_2}$
• $\mathbf\small{\vec{\tau}_3=\vec{r}_1 \times \vec{F}_3}$
so on . . .
7. So we want the net force $\mathbf\small{\vec{F}}$ on each particle
• The 'net force on a particle' is the sum of all external and internal forces
That is:
• $\mathbf\small{\vec{F}_1=\vec{F}_{1(external)}+\vec{F}_{1(internal)}}$
• $\mathbf\small{\vec{F}_2=\vec{F}_{2(external)}+\vec{F}_{2(internal)}}$
• $\mathbf\small{\vec{F}_3=\vec{F}_{3(external)}+\vec{F}_{3(internal)}}$
so on . . .
8. So the result in (6) becomes:
• $\mathbf\small{\vec{\tau}_1=\vec{r}_1 \times \vec{F}_{1(external)}+\vec{r}_1 \times \vec{F}_{1(internal)}}$
$\mathbf\small{\Rightarrow \vec{\tau}_1=\vec{\tau}_{1(external)}+\vec{\tau}_{1(internal)}}$
• $\mathbf\small{\vec{\tau}_2=\vec{r}_2 \times \vec{F}_{2(external)}+\vec{r}_2 \times \vec{F}_{2(internal)}}$
$\mathbf\small{\Rightarrow \vec{\tau}_2=\vec{\tau}_{2(external)}+\vec{\tau}_{2(internal)}}$
(i) Newtons third law applies to the 'mutual interaction between any two particles'
(ii) The 'mutual interaction between any two particles' act along the line joining those two particles
■ Thus the vector sum of all internal forces is a null vector
■ So the vector sum: $\mathbf\small{[\vec{\tau}_{1(internal)}]+[\vec{\tau}_{2(internal)}]+[\vec{\tau}_{3(internal)}]\;+\; .\; .\; .\;+\;[\vec{\tau}_{n(internal)}]}$
will be a null vector
11. So the result in (9) becomes:
$\mathbf\small{\frac{dL}{dt}=[\vec{\tau}_{1(external)}]+[\vec{\tau}_{2(external)}]+[\vec{\tau}_{3(external)}]\;+\; .\; .\; .\;+\;[\vec{\tau}_{n(external)}]}$
• The right side is the sum of 'external torques on all the particles'. We can denote it as $\mathbf\small{\vec{\tau}_{ext}}$
• So we get:
Eq.7.21: $\mathbf\small{\frac{dL}{dt}=\vec{\tau}_{ext}}$
12. 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'
• The analogous of this in translational motion is already known to us:
The 'rate of change of linear momentum of a system of particles' at any instant is equal to the 'sum of external forces on all the particles'
$\mathbf\small{\frac{dp}{dt}=\vec{F}_{ext}}$
1. Consider the situation where there is no net external torque
• This situation can be denoted as: $\mathbf\small{\vec{\tau}_{ext}=0}$
2. Based on Eq.7.21 above, we can write:
If $\mathbf\small{\vec{\tau}_{ext}=0}$, then $\mathbf\small{\frac{dL}{dt}=0}$
3. $\mathbf\small{\frac{dL}{dt}=0}$ means that, 'rate of change of angular momentum' with respect to time is zero
• That means, angular momentum does not change with time
• That means, angular momentum remains a constant
4. So we can write:
• If there is no net external torque, angular momentum remains constant
• Mathematically, this can be written as:
♦ $\mathbf\small{|\vec{L}|=\;\text{a constant}\;\;k}$
♦ Direction of $\mathbf\small{\vec{L}}$ remains the same
5. But since $\mathbf\small{\vec{L}}$ is a vector, it will have components: $\mathbf\small{\vec{L}_x,\;\vec{L}_y,\;\vec{L}_z}$
• If $\mathbf\small{\vec{L}}$ is to remain a constant, there must not be any change in any of the components
• That means, each of the components must remain constant
■ So we can write:
Eq.7.22:
$\mathbf\small{\vec{\tau}_{ext}=0}$
$\mathbf\small{\Rightarrow |\vec{L}|=\;\text{a constant}\;\;k}$
♦ Direction of $\mathbf\small{\vec{L}}$ remains the same
$\mathbf\small{\Rightarrow |\vec{L}_x|=k_x,\;|\vec{L}_y|=k_y,\;|\vec{L}_z|=k_z}$
♦ Direction of $\mathbf\small{\vec{L}_x}$ always remains along the x-axis
♦ Direction of $\mathbf\small{\vec{L}_y}$ always remains along the y-axis
♦ Direction of $\mathbf\small{\vec{L}_z}$ always remains along the z-axis
• We will see the applications of the above 'conservation of angular momentum' in later sections of this chapter
• Now we will see the general case where $\mathbf\small{r_\bot}$ is not a constant. We will write it in steps:
1. 'Calculation over an instant' can be easily done using calculus
• In calculus, the calculation of 'Rate of change of angular momentum' over an instant is written as:
$\mathbf\small{\frac{d}{dt} \text{(Angular momentum)}}$
(This is read as: 'd by dt of Angular momentum')
• In short form, we write: $\mathbf\small{\frac{dl}{dt}}$
(This is read as: 'dl by dt')
• This is the differentiation of 'l', with respect to time 't'
• When we do the differentiation, what do we get?
• Let us find out:
2. $\mathbf\small{\frac{dl}{dt}=\frac{d(\vec{r}\times \vec{p})}{dt}}$
(∵ $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$)
• Note that, here $\mathbf\small{\vec{r}}$ is not a constant
• So we have to differentiate the product $\mathbf\small{(\vec{r}\times \vec{p})}$
• We have to apply the product rule:
• We get: $\mathbf\small{\frac{dl}{dt}=\left [\frac{d(\vec{r}\times \vec{p})}{dt} \right ]=\left [\frac{d\vec{r}}{dt}\times \vec{p}+\vec{r} \times \frac{d\vec{p}}{dt}\right ]}$
3. Consider the first term on the right:
$\mathbf\small{\frac{d\vec{r}}{dt}\times \vec{p}}$
• We know that $\mathbf\small{\frac{d\vec{r}}{dt} = \vec{v}}$
• So $\mathbf\small{\frac{d\vec{r}}{dt}\times \vec{p}=\vec{v}\times \vec{p}}$
$\mathbf\small{\Rightarrow \frac{d\vec{r}}{dt}\times \vec{p}=\vec{v}\times m\vec{v}}$
• But $\mathbf\small{\vec{v}}$ and $\mathbf\small{m\vec{v}}$ are parallel vectors
• So their cross product is $\mathbf\small{\vec{0}}$
• Thus the first term vanishes
4. We get:
$\mathbf\small{\frac{dl}{dt}=\left [\frac{d(\vec{r}\times \vec{p})}{dt}\right ]=\left [\vec{r} \times \frac{d\vec{p}}{dt} \right ]}$
• Now, $\mathbf\small{\frac{d\vec{p}}{dt}=\vec{F}}$
• So we get: $\mathbf\small{\frac{dl}{dt}=\left [\frac{d(\vec{r}\times \vec{p})}{dt}\right ]=\left [\vec{r}\times \vec{F} \right ]}$
5. But $\mathbf\small{\vec{r}\times \vec{F}=\vec{\tau}}$
• Thus we get:
Eq.7.19: $\mathbf\small{\frac{dl}{dt}=\vec{r}\times \vec{F}=\vec{\tau}}$
■ Thus we get the same relation as before:
• 'Rate of change of angular momentum' calculated over an instant is the torque experienced by the particle at that instant
Now we can make a comparison:
• Time rate of change of linear momentum = Force
• This we know as Newton's second law
2. In rotational motion:
• Time rate of change of angular momentum = Torque
• This is analogous to the Newton's second law
• Thus we have completed the discussion on two items:
(ii) Angular momentum experienced by a single particle
• Next we will see these:
(i) Torque experienced by a system of particles
(ii) Angular momentum experienced by a system of particles
First we will see angular momentum
1. 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
2. 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: $\mathbf\small{\vec{p}_i=m_i \times \vec{v}_i}$
3. We know that, if we differentiate angular momentum ($\mathbf\small{\vec{l}_i}$) of a single particle, we will get the torque ($\mathbf\small{\vec{\tau}_i}$) experienced by that particle (see Eq.7.19)
• Let us now differentiate the total angular momentum ($\mathbf\small{\vec{L}}$) of a system of particles:
• We have: $\mathbf\small{\frac{dL}{dt}=\frac{d}{dt}\left (\sum\limits_{i=1}^{i=n}{\vec{l}_i} \right )}$
4. So we have to differentiate a quantity which is 'a sum of several quantities'
• For that, we have to differentiate each item and then add the results
• That is: $\mathbf\small{\frac{dL}{dt}=\frac{d}{dt}(\vec{l}_1)+\frac{d}{dt}(\vec{l}_2)+\frac{d}{dt}(\vec{l}_3)\;+\; .\; .\; .\;+\frac{d}{dt}(\vec{l}_n)}$
5. But when we differentiate the angular momentum of a particle, we get the torque experienced by that particle (Eq.7.19)
• So the result in (4) can be written as: $\mathbf\small{\frac{dL}{dt}=\vec{\tau}_1+\vec{\tau}_2+\vec{\tau}_3\;+\; .\; .\; .\;+\;\vec{\tau}_n}$
6. But
• $\mathbf\small{\vec{\tau}_1=\vec{r}_1 \times \vec{F}_1}$
• $\mathbf\small{\vec{\tau}_2=\vec{r}_2 \times \vec{F}_2}$
• $\mathbf\small{\vec{\tau}_3=\vec{r}_1 \times \vec{F}_3}$
so on . . .
7. So we want the net force $\mathbf\small{\vec{F}}$ on each particle
• The 'net force on a particle' is the sum of all external and internal forces
That is:
• $\mathbf\small{\vec{F}_1=\vec{F}_{1(external)}+\vec{F}_{1(internal)}}$
• $\mathbf\small{\vec{F}_2=\vec{F}_{2(external)}+\vec{F}_{2(internal)}}$
• $\mathbf\small{\vec{F}_3=\vec{F}_{3(external)}+\vec{F}_{3(internal)}}$
so on . . .
8. So the result in (6) becomes:
• $\mathbf\small{\vec{\tau}_1=\vec{r}_1 \times \vec{F}_{1(external)}+\vec{r}_1 \times \vec{F}_{1(internal)}}$
$\mathbf\small{\Rightarrow \vec{\tau}_1=\vec{\tau}_{1(external)}+\vec{\tau}_{1(internal)}}$
• $\mathbf\small{\vec{\tau}_2=\vec{r}_2 \times \vec{F}_{2(external)}+\vec{r}_2 \times \vec{F}_{2(internal)}}$
$\mathbf\small{\Rightarrow \vec{\tau}_2=\vec{\tau}_{2(external)}+\vec{\tau}_{2(internal)}}$
• $\mathbf\small{\vec{\tau}_3=\vec{r}_3 \times \vec{F}_{3(external)}+\vec{r}_3 \times \vec{F}_{3(internal)}}$
$\mathbf\small{\Rightarrow \vec{\tau}_3=\vec{\tau}_{3(external)}+\vec{\tau}_{3(internal)}}$
so on . . .
9. So now the result in (5) becomes:
$\mathbf\small{\frac{dL}{dt}=[\vec{\tau}_{1(external)}+\vec{\tau}_{1(internal)}]+[\vec{\tau}_{2(external)}+\vec{\tau}_{2(internal)}]+[\vec{\tau}_{3(external)}+\vec{\tau}_{3(internal)}]\;+\; .\; .\; .\;+\;[\vec{\tau}_{n(external)}+\vec{\tau}_{n(internal)}]}$
10. The internal forces mentioned above arise due to the 'mutual interaction between the particles'
• We will assume two items here:$\mathbf\small{\frac{dL}{dt}=[\vec{\tau}_{1(external)}+\vec{\tau}_{1(internal)}]+[\vec{\tau}_{2(external)}+\vec{\tau}_{2(internal)}]+[\vec{\tau}_{3(external)}+\vec{\tau}_{3(internal)}]\;+\; .\; .\; .\;+\;[\vec{\tau}_{n(external)}+\vec{\tau}_{n(internal)}]}$
10. The internal forces mentioned above arise due to the 'mutual interaction between the particles'
(i) Newtons third law applies to the 'mutual interaction between any two particles'
(ii) The 'mutual interaction between any two particles' act along the line joining those two particles
■ Thus the vector sum of all internal forces is a null vector
■ So the vector sum: $\mathbf\small{[\vec{\tau}_{1(internal)}]+[\vec{\tau}_{2(internal)}]+[\vec{\tau}_{3(internal)}]\;+\; .\; .\; .\;+\;[\vec{\tau}_{n(internal)}]}$
will be a null vector
11. So the result in (9) becomes:
$\mathbf\small{\frac{dL}{dt}=[\vec{\tau}_{1(external)}]+[\vec{\tau}_{2(external)}]+[\vec{\tau}_{3(external)}]\;+\; .\; .\; .\;+\;[\vec{\tau}_{n(external)}]}$
• The right side is the sum of 'external torques on all the particles'. We can denote it as $\mathbf\small{\vec{\tau}_{ext}}$
• So we get:
Eq.7.21: $\mathbf\small{\frac{dL}{dt}=\vec{\tau}_{ext}}$
12. 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'
• The analogous of this in translational motion is already known to us:
The 'rate of change of linear momentum of a system of particles' at any instant is equal to the 'sum of external forces on all the particles'
$\mathbf\small{\frac{dp}{dt}=\vec{F}_{ext}}$
Conservation of angular momentum
• This situation can be denoted as: $\mathbf\small{\vec{\tau}_{ext}=0}$
2. Based on Eq.7.21 above, we can write:
If $\mathbf\small{\vec{\tau}_{ext}=0}$, then $\mathbf\small{\frac{dL}{dt}=0}$
3. $\mathbf\small{\frac{dL}{dt}=0}$ means that, 'rate of change of angular momentum' with respect to time is zero
• That means, angular momentum does not change with time
• That means, angular momentum remains a constant
4. So we can write:
• If there is no net external torque, angular momentum remains constant
• Mathematically, this can be written as:
♦ $\mathbf\small{|\vec{L}|=\;\text{a constant}\;\;k}$
♦ Direction of $\mathbf\small{\vec{L}}$ remains the same
5. But since $\mathbf\small{\vec{L}}$ is a vector, it will have components: $\mathbf\small{\vec{L}_x,\;\vec{L}_y,\;\vec{L}_z}$
• If $\mathbf\small{\vec{L}}$ is to remain a constant, there must not be any change in any of the components
• That means, each of the components must remain constant
■ So we can write:
Eq.7.22:
$\mathbf\small{\vec{\tau}_{ext}=0}$
$\mathbf\small{\Rightarrow |\vec{L}|=\;\text{a constant}\;\;k}$
♦ Direction of $\mathbf\small{\vec{L}}$ remains the same
$\mathbf\small{\Rightarrow |\vec{L}_x|=k_x,\;|\vec{L}_y|=k_y,\;|\vec{L}_z|=k_z}$
♦ Direction of $\mathbf\small{\vec{L}_x}$ always remains along the x-axis
♦ Direction of $\mathbf\small{\vec{L}_y}$ always remains along the y-axis
♦ Direction of $\mathbf\small{\vec{L}_z}$ always remains along the z-axis
• We will see the applications of the above 'conservation of angular momentum' in later sections of this chapter
In the next section, we will see some solved examples
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