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
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
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
Now we will see some solved examples
(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
In the next section, we will see rolling motion
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