Chapter 8 - Gravitation

In the previous section, we completed a discussion on rotational motion. In this chapter we will see gravitation.

• In our day to day life, we continuously feel the effects of gravitational force. Let us see some examples:
    ♦ Objects let go from a height always fall downwards
    ♦ It is more difficult to lift a heavier mass than a lighter mass
    ♦ Going uphill is more difficult than going downhill
• Through ages, scientists tried to find proper explanations for such phenomena
• The sixteenth century scientist Galileo Galilei, through his experiments, found out that, all falling objects are subjected to acceleration. This can be explained in 5 steps as follows:
1. We know that, a falling object is moving towards the surface of the earth
• So that falling object is in motion
2. Since it is in motion, we can say: It has a velocity
3. But that velocity is continuously increasing
• That means, the velocity is continuously changing
4. Since there is ‘change in velocity’, we can say: The body is subjected to acceleration
5. Galilio found out that, all bodies are subjected to the same acceleration
• That means:
    ♦ Some bodies may be having a greater mass (we call them heavy bodies)
    ♦ Some bodies may be having a lesser mass (we call them light bodies)
    ♦ Whatever be the mass, the acceleration experienced by all bodies will be the same

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• Next we will see another notable discovery made by scientists of the seventeenth century. We will write it in 5 steps:

1. Observe a planet
2. Record it’s position
3. Note down the exact date and time at which the observation is made
4. Wait for one year
5. At the exact date and time, observe the planet
■ It will be located at the same exact position recorded in (2)

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• Danish scientist Tycho Brahe received funds from various Kings and rulers. Those Funds were granted to him, for carrying out extensive scientific research works. So he was able to spend a major part of his life time recording such observations of the planets and stars

• The data recorded by Brahe was later analysed by his assistant Johannes Kepler
• Kepler formulated three laws, which are now known as Kepler’s laws
• Kepler’s laws was the starting point of the works done by Sir Isaac Newton
    ♦ Those works led to the discovery of Universal law of Gravitation
• So let us discuss about the Kepler’s laws in some detail

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Kepler’s first law
All planets move in elliptical orbits with the sun situated at one of the foci of the ellipse
• This law is also known as Law of Orbits

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• To fully understand this law, we must know some of the 'salient features of ellipse'

• Learning a simple method to 'draw an ellipse' will help us to appreciate those salient features
• The method of drawing can be written in 3 steps:
1. Fix two pins at any ‘convenient distance apart’ on a drawing board 
• This is shown in fig.8.1(a) below:

Fig.8.1

• Name the positions of the pins as F₁ and F₂
2. Tie one end of an inextensible string to the pin at F₁
• Tie the other end to the pin at F₂
3. Using the tip of a pencil, stretch the string taut and draw a curve
• The string must be taut all the while
• The closed curve thus obtained is called the ellipse

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Let us see the important features of an ellipse:

1. F₁ and F₂ are called the foci
2. Consider any point T on the ellipse. There will be two distances F₁T and F₂T
• Point T was obtained while keeping the string taut
• So (F₁T + F₂T) = Length of the string
• This is true for any point on the string
■ So we can write:
For any point on the string, the sum of the distances from the foci is the same
3. Draw a line connecting F₁ and F₂
• Extend this line towards either sides so that, it meets the ellipse at P and A (fig.b)
• Mark the midpoint of PA as O
• This O is the center of the ellipse
4. PA is called the major axis of the ellipse
• OP or OA, which is half of the major axis is called semi major axis
5. An interesting case:
• Move F₂ towards the left
• Let it coincide with F₁
• Draw the curve as before
• We will get a circle This is shown in fig.8.2(a) below

Fig.8.2

Now we will see how the 'ellipse and it's foci' are related to 'planetary motion'

1. The sun is present at the focus F₁
• It is marked as S in the above fig.8.2(b)
2. The planet orbits around the sun
• The motion of the planet is along the ellipse
3. Consider the position P
• PS is the smallest distance possible between the planet and the sun
■ The position P is called Perihelion
4. Consider the position A
• AS is the largest distance possible between the planet and the sun
■ The position A is called Aphelion

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Kepler’s second law
The line that joins any planet to the sun sweeps equal areas in equal intervals of time
• This law is also known as Law of Areas

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• To fully understand this law, we can make use of fig.8.3 below:

Fig.8.3

• A planet is moving around the sun S
• We must complete 4 steps:
Step 1:
1. Note down the time t₁ at which the planet is at any convenient point U
2. Note down the time t₂ at which the planet is at any other convenient point V
3. Calculate Δt = (t₂-t₁)
4. Calculate the area A₁ enclosed between the three items:
(i) Line US
(ii) Line VS
(iii) Arc UV
■ A₁ is the area swept by the line US during a time interval of Δt
Step 2:
1. Note down the time t₃ at which the planet is at any other convenient point W
2. Note down the point X at which the planet reaches exactly at time (t₃+Δt)
• Here Δt must be the same Δt calculated in step 1
• This ensures that, the two items are equal:
(i) Time taken to travel from U to V
(ii) Time taken to travel from W to X
3. Calculate the area A₂ enclosed between the three items:
(i) Line WS
(ii) Line XS
(iii) Arc WX
■ A₂ is the area swept by the line WS during a time interval of Δt
Step 3:
1. Note down the time t₄ at which the planet is at any other convenient point Y
2. Note down the point Z at which the planet reaches exactly at time (t₄+Δt)
• Here Δt must be the same Δt calculated in step 1
• This ensures that, the three items are equal:
(i) Time taken to travel from U to V
(ii) Time taken to travel from W to X
(ii) Time taken to travel from Y to Z
3. Calculate the area A₃ enclosed between the three items:
(i) Line YS
(ii) Line ZS
(iii) Arc YZ
■ A₃ is the area swept by the line YS during a time interval of Δt
Step 4:
■ Compare the areas
We will find that: A₁ = A₂ = A₃

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• The above four steps were carried out several times by Kepler
• Thus he arrived at the second law
• If we have precision instruments to observe the planets, we too can perform the 4 steps and verify the law

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• The second law is based on 'analysis of observations'
    ♦ Observations were made by Tycho Brahe
    ♦ Analysis of those observations were done by Kepler

• While doing the analysis, Kepler discovered that, there is an ‘equality in areas’
• But we want to know the ‘cause of such an equality’
• Indeed, scientists were able to find the ‘cause’
• Before discussing the 'explanation given by scientists', let us see an 'immediate inference'. It can be written in 3 steps:
1. Let us assume that, the three areas in fig.8.3 above, are triangles
• For the △SUV near the aphelion, the base is small
• For the △SWX and △SYZ near the perihelion, the bases are larger
2. In spite of the differences in bases, the areas are the same
• This is due to the difference in heights
    ♦ For the △SUV near the aphelion, the height is large
    ♦ For the △SWX and △SYZ near the perihelion, the heights are smaller
3. Remember that, Δt is same in all the three cases
■ So from (1), it is clear that:
• The planet travel with greater speeds when it is near the perihelion
• That is how it is able to cover greater ‘base distance’ in the same interval of time
■ So we can write the ‘immediate inference’:
The ‘equality in areas’ is due to:
    ♦ Greater speeds near perihelion
    ♦ Lesser speeds near aphelion

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But this leads to more questions:

    ♦ Why do planets move faster when they are near the perihelion ?
    ♦ Why do planets move slower when they are near the aphelion ?
The answers can be written in steps:
1. In fig.8.4 below, the elliptical orbit is shown:

Fig.8.4

• The position of the sun is denoted by S
• The position of the planet is denoted by Q
2. Let S be the origin of the coordinate system
• Then the vector joining S and Q can be considered as the position vector $\mathbf\small{\vec{r}}$ of the planet    
3. After a time duration of Δt, the planet reaches Q’
4. So the vector joining Q and Q' is the displacement vector
• Magnitude of the displacement is given by: (velocity × time)
• So the vector from Q to Q' is: $\mathbf\small{\vec{v}(\Delta t)}$
• '$\mathbf\small{\vec{v}(\Delta t)}$' is a vector
    ♦ It's magnitude is: $\mathbf\small{|\vec{v}|\times\Delta t}$
    ♦ It's direction is same as that of $\mathbf\small{\vec{v}}$
    ♦ In effect, it is the displacement vector from Q to Q' 
5. Consider the △SQQ'
• One side of this triangle is $\mathbf\small{\vec{r}}$
• The other side is $\mathbf\small{\vec{v}(\Delta t)}$
■ Then we have:
• Area of △SQQ' = ¹⁄₂ times the magnitude of ($\mathbf\small{\vec{r}\times[\vec{v}(\Delta t)]}$)
See the solved example 7.11 at the beginning of section 7.14
6. We have: Linear momentum (p) = mass × Linear velocity
• So we get: $\mathbf\small{\vec{p}=\vec{v}(m)}$
$\mathbf\small{\Rightarrow \vec{v}=\frac{\vec{p}}{m}}$
7. So the result in (5) becomes:
Area of △SQQ' = $\mathbf\small{\frac{1}{2}\times\vec{r}\times \left [\frac{\vec{p}}{m}(\Delta t)\right ]}$
8. Dividing both sides by Δt, we get:
$\mathbf\small{\frac{\text{Area of △SQQ'}}{\Delta t}=\frac{1}{2}\times\vec{r}\times \left [\frac{\vec{p}}{m}\right ]}$
$\mathbf\small{\Rightarrow \frac{\text{Area of △SQQ'}}{\Delta t}=\frac{1}{2m}\times \left [\vec{r}\times{\vec{p}}\right ]}$
9. But $\mathbf\small{\left [\vec{r}\times{\vec{p}}\right ]=\vec{L}}$
Where $\mathbf\small{\vec{L}}$ is the angular momentum of the planet
• So the result in (8) becomes:
$\mathbf\small{\frac{\text{Area of △SQQ'}}{\Delta t}=\frac{\vec{L}}{2m}}$
10. If Δt is very small, the distance traveled by the planet in that time duration will be very small
• Then Q' will be very close to Q
• In such a situation, the chord QQ' will nearly coincide with the arc length from Q to Q'
• So the area of △SQQ' will be same as the area of the sector SQQ'
11. So, if Δt is very small, we can modify the result in (9):
$\mathbf\small{\frac{\text{Area of sector SQQ'}}{\Delta t}=\frac{\vec{L}}{2m}}$
• Let us denote the area of sector SQQ' by ΔA
• Then the result becomes:
$\mathbf\small{\frac{\Delta A}{\Delta t}=\frac{\vec{L}}{2m}}$
12. But the observations made by Kepler gives us the following information:
• If we consider the same 'intervals of time' the 'areas swept' will also be the same
    ♦ Same 'intervals of time' means: Δt is a constant
    ♦ 'Areas swept' are same means: ΔA is a constant
13. So in the result in (11), the left side is a constant
• Then right side must also be a constant
• On the right side, '2' and 'm' are already constants
• So $\mathbf\small{\vec{L}}$ must be a constant
14. Thus from the Kepler's second law, we get an important information:
■ The angular momentum of a planet always remains the same

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Let us see the implication of 'constant angular momentum'. It can be written in 6 steps:

1. In fig.8.5(a), the planet is at Q

Fig.8.5

• At that instant, it's position vector is $\mathbf\small{\vec{r}}$
• At that instant, it's velocity vector is $\mathbf\small{\vec{v}}$  
2. Resolve $\mathbf\small{\vec{v}}$ into two components as shown in fig.b
• One component is parallel to $\mathbf\small{\vec{r}}$. It is denoted as $\mathbf\small{\vec{v}_{\shortparallel}}$
• The other component is perpendicular to $\mathbf\small{\vec{r}}$. It is denoted as $\mathbf\small{\vec{v}_{\bot }}$
3. Let the angular momentum of the planet be $\mathbf\small{\vec{L}}$
We have: $\mathbf\small{\vec{L}=\vec{r}\;\times\;\vec{p}}$
$\mathbf\small{\Rightarrow \vec{L}=\vec{r}\;\times\;m\vec{v}}$
$\mathbf\small{\Rightarrow \vec{L}=\vec{r}\;\times\;m(\vec{v}_{\shortparallel}+\vec{v}_{\bot })}$ (∵ $\mathbf\small{\vec{v}}$ will be the vector sum of it's components)
$\mathbf\small{\Rightarrow \vec{L}=\vec{r}\;\times\;(m\vec{v}_{\shortparallel}+m\vec{v}_{\bot })}$
$\mathbf\small{\Rightarrow \vec{L}=(\vec{r}\;\times\;m\vec{v}_{\shortparallel})+(\vec{r}\;\times\;m\vec{v}_{\bot })}$
$\mathbf\small{\Rightarrow \vec{L}=(\vec{0})+(\vec{r}\;\times\;m\vec{v}_{\bot })}$ (∵ cross product of two parallel vectors is a null vector)
$\mathbf\small{\Rightarrow \vec{L}=\vec{r}\;\times\;m\vec{v}_{\bot }}$
$\mathbf\small{\Rightarrow |\vec{L}|=m\;\times\;|\vec{r}|\;\times\;|\vec{v}_{\bot}|\;\times\;\sin 90}$ (∵ angle between $\mathbf\small{\vec{r}}$ and $\mathbf\small{\vec{v}_{\bot }}$ is 90^o)
$\mathbf\small{\Rightarrow |\vec{L}|=m\;\times\;|\vec{r}|\;\times\;|\vec{v}_{\bot}|}$ (∵ sin 90 is 1)
• Thus we get the magnitude of the angular momentum of the planet
4. We see that, the magnitude depends upon 3 items:
(i) The mass m 
(ii) The magnitude of the position vector
(iii) The magnitude of the velocity
5. Mass m is a constant
• Magnitude of the position vector is same as the 'distance of the planet from the sun'
• So we can write:
If the angular momentum is to remain constant, either one of the two changes must take place:
(i) When the 'distance from sun' decrease, the velocity must increase
(ii) When the 'distance from sun' increase, the velocity must decrease
6. Thus we find that:
• When the planet is near the perihelion, it's velocity increases
• When the planet is near the aphelion, it's velocity decreases

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• But this is not all. Kepler's second law, and it's explanation, leads to more questions and discoveries. This can be elaborated in 7 steps as follows:

1. Consider a particle in rotational motion
• We know that, the angular momentum of that particle remains constant if there is no net external torque acting on it
• That is: Angular momentum of that particle remains constant if $\mathbf\small{\vec{\tau}}$ = 0
2. The external torque is given by: $\mathbf\small{\vec{\tau}=\vec{r}\times \vec{F}}$
• We have:$\mathbf\small{|\vec{\tau}|=|\vec{r}|\times |\vec{F}|\times \sin \theta}$
Where θ is the angle between $\mathbf\small{\vec{r}}$ and $\mathbf\small{\vec{F}}$
3. So, if $\mathbf\small{\vec{\tau}}$ is to become zero, one of the two conditions must be satisfied:
(i) $\mathbf\small{|\vec{F}|}$ must be zero OR
(ii) θ must be zero or 180^o
4. We cannot consider (i) because, a force must be present. Otherwise the planet cannot move
• So we conclude that: The external torque is zero because, θ is zero or 180^o
5. θ is zero or 180^o implies that: The $\mathbf\small{\vec{F}}$ is acting along the same direction as $\mathbf\small{\vec{r}}$
• So it was concluded that, the 'force in action' is a central force
6. What is a central force ?
• The answer can be written in 3 steps:
(i) Suppose a force is acting at a point
(ii) Consider the line joining that point and the 'origin of the system'
(iii) If the force in (i) is acting along the line in (ii), then that force is a central force
We will learn more about central force in higher classes
7. In our present case, sun is the origin (That is why we draw the position vector $\mathbf\small{\vec{r}}$ from the sun) 
• So we can conclude that, a force is acting along the line joining the sun and the planet
■ Let us write a summary of the above 7 steps:
• An external force must act on a planet in order to keep that planet in it's orbit
• But this external force must not cause any change to the angular momentum of that planet
• This can be achieved only if the external force is a 'central force'

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• Thus Kepler's second law enabled scientists to reach an important milestone
• An 'important milestone' because, the presence of a central force was detected for the first time
• It was Sir Isaac Newton who finally discovered what this central force really is
• In the next section, we will see Kepler's third law

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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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