Chapter 8.6 - Cavendish's Experiment

In the previous section, we saw the forces exerted by a spherical shell on a point mass inside or outside it. In the discussions so far in this chapter, we came across the universal gravitational constant (G) many times. We will now see how scientists determined the actual value of G

In 1798, the British scientist Henry Cavendish set up an apparatus called the torsion balance. His aim was to determine the density of the Earth. Though he succeeded in finding the density, he did not find it necessary to determine the value of G. However, the same setup was used by later scientists to calculate G accurately. We will write the procedure in steps:

1. Fig.8.22 below shows the schematic arrangement of the setup

Fig.8.22

• There are four balls
    ♦ The two small green balls are identical. They have the same diameter
    ♦ The two large magenta balls are identical. They have the same diameter
• All the four balls are made of lead
2. The green balls are attached to the ends of a rod
• The rod is suspended using a thin wire. This thin wire is shown in blue color
    ♦ The top end of this wire is attached to a rigid support
    ♦ The bottom end of this wire is attached at point O
3. The following 3 conditions should be satisfied:
(i) Consider an imaginary line joining the centers of the green balls
• Let the length of this line be l
• Point O must be the exact midpoint of this imaginary line
(ii) Consider an imaginary line joining the centers of the magenta balls
• Length of this imaginary line should be the same l mentioned in (i)
• Point O must be the exact midpoint of this imaginary line also
(iii) The green and magenta balls should be placed on alternate sides. This can be explained using the fig.8.22
• On the left side, the green ball is placed at the rear of the magenta ball
• On the right side, the green ball is placed at the front of the magenta ball
4. When the conditions 3(i) and 3(ii) are satisfied, the centers of all the four spheres will lie on an imaginary circle with center at O
• This imaginary circle is indicated by the dashed yellow curve
5. The setup is complete. Now we can begin the experiment
• Let r be the initial distance between any one magenta ball and it's corresponding green ball. This is shown in fig.8.22
• Due to the gravitational force of attraction, the green balls will move towards the magenta balls
6. The magnitude of the force acting on any one green ball is given by:
$\mathbf\small{|\vec{F}|=\frac{GMm}{r^2}}$
• Where:
    ♦ M is the mass of the larger magenta ball
    ♦ m is the mass of the smaller green ball
• This same magnitude acts on both the green balls. But in opposite directions
7. Since the forces are equal in magnitude but opposite in directions, a torque is created
• Magnitude of this torque = Magnitude of any one force × Perpendicular Distance between the forces
= $\mathbf\small{|\vec{F}|l=\frac{GMml}{r^2}}$
8. The green balls will not touch the magenta balls
• But that is unexpected. The torque must rotate the two green balls and bring them into contact with with the magenta balls
• So what is stopping the green balls ?
The answer can be written in 8 steps:
(i) When the green balls rotate, the rod attached to them also rotates
(ii) When the rod rotates, the blue wire also rotates
(iii) But the blue wire is fixed firmly at it's top end
So the wire is not 'free to rotate'
(iv) The bottom portions of the wire rotate. But the top portions resist rotation
• As a result, the wire is twisted
(v) When the wire is twisted, it tries to resist 'being twisted'
• This is called torsional resistance
• In our present case, 'torsional resistance' means, the torque which is resisting the rotation of the green balls   
• This torsional resistance is indicated by the white curved arrow in fig.8.23 below:

Fig.8.23

(vi) In the fig.8.23, note the two directions:
    ♦ Direction of rotation of the green balls
    ♦ Direction of the white curved arrow
• The above two directions are opposite to each other. This is indeed so because, the torsional resistance will be opposing the rotation of the green balls   
(vii) We calculated the torque created by $\mathbf\small{|\vec{F}|}$ in (7)
• At the initial stages, this torque is able to over come the torsional resistance offered by the wire
• So the green balls move towards the magenta balls
(viii) But as the rotation continues, the torsional resistance increases
• It increases to such a level that, it becomes equal to the value calculated in (7)
• At that stage, the rotation stops
• So at that stage, we can write:
Torsional resistance offered by the wire = $\mathbf\small{\frac{GMml}{r^2}}$
9. Consider the equation written in 8(viii) above
• If we can find the 'torsional resistance offered by the wire' at the final stage, we will be able to calculate G
• This is because, all others are known quantities
• So our next aim is to find this 'torsional resistance'
• For that, we have to perform a separate experiment. The following steps from (10) to (14) give a basic idea about that experiment
10. In fig.8.24 below, a blue cylinder is fixed at one of it's ends. The other end is free

Fig.2.24

11. In the fig.8.25 below, a torsion is applied at the free end
• This torsion is indicated by the cyan curved arrow

Fig.8.25

• Consider any fibre on the surface, say PQ 
• When torsion is applied, the end Q will move to a new position Q'
• But since the other end of the cylinder is fixed, P remains at the same position
12. So we get an angle Q'PQ
• This angle is called the angle of twist (θ)
• It is measured in radians
    ♦ If the cylinder is strong, we will get only a small θ even if we apply a large torsion
    ♦ If the cylinder is weak, we will get a large θ even if we apply a small torsion
• Thus, θ is a property of the object
13. A number of trials are done on the cylinder
• In each trial, a known torsion is applied and the corresponding θ is noted
■ From that data, we get an important information:
The exact torsion ($\mathbf\small{\tau}$) required to obtain a θ of 1 radian
■ Each object has it's own unique value of $\mathbf\small{\tau}$  
14. In our present case, the 'object of interest' is the blue wire used for suspending the green balls
• So the experiment is performed on the blue wire and it's $\mathbf\small{\tau}$ is determined
15. Now we get back to our main experiment
• We have the initial and final positions of any one of the two green balls
• From those positions, we can determine the angle (θ') through which the blue wire is twisted
• Also we have $\mathbf\small{\tau}$, which is the unique property of the wire
16. So, if we multiply θ' by $\mathbf\small{\tau}$, we will get the exact torsional resistance
• That means: Torsional resistance = $\mathbf\small{\tau \theta'}$
17. So the equation in 8(v) becomes: $\mathbf\small{\tau \theta'=\frac{GMml}{r^2}}$
• In this equation, G is the only unknown. So it can be easily calculated

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• So we have seen how scientists determined the value of G
• In the next section, we will see the acceleration due to gravity of the Earth

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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

---

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

---

• 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

---

• 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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