13.2 - Law of Equipartition of Energy

In the previous section, we explained the gas laws using kinetic theory. In this section, we will see the law of equipartition of energy.

The details can be written in 17 steps:
1. Consider a gas in thermal equilibrium at temperature T.
If 'm' is the mass of each molecule in that gas, then the average value of energy ($\small{\epsilon_t}$) of each molecule can be obtained as:
$\small{\epsilon_t~=~\frac{1}{2}m\,v_x^2~+~\frac{1}{2}m\,v_y^2~+~\frac{1}{2}m\,v_z^2~=~\frac{3}{2}K_B\,T}$

2. $\small{\frac{3}{2}K_B\,T}$ is three times $\small{\frac{1}{2}K_B\,T}$. We know that, there is no preferred direction. So we can write:
    ♦ $\small{\frac{1}{2}m\,v_x^2~=~\frac{1}{2}K_B\,T}$
    ♦ $\small{\frac{1}{2}m\,v_y^2~=~\frac{1}{2}K_B\,T}$
    ♦ $\small{\frac{1}{2}m\,v_z^2~=~\frac{1}{2}K_B\,T}$

3. Now we will write about the number of coordinates required for a molecule.
• If the molecule is free to move in space, we need three coordinates to locate it.
• If the molecule is constrained to move in a plane, we need two coordinates to locate it.
• If the molecule is constrained to move along a line, we need only one coordinate to locate it.

4. The above information in (3) can be written in another form also:
• If the molecule is free to move in space, it has three degrees of freedom.
• If the molecule is constrained to move in a plane, it has two degrees of freedom.
• If the molecule is constrained to move along a line, it has one degree of freedom.

5. The above information in (3) and (4) can be written in yet another form also. Motion of a body as a whole from one point to another is called translation. So we can write:
• If the molecule is free to move in space, it has three translational degrees of freedom.
• If the molecule is constrained to move in a plane, it has two translational degrees of freedom.
• If the molecule is constrained to move along a line, it has one translational degree of freedom.

6. We just saw that, if a molecule is free to move in space, it has three translational degrees of freedom. Each of those three degrees, will contribute a term towards the total kinetic energy of the molecule.
• The degree of freedom in the x-direction will contribute $\small{\frac{1}{2}m\,v_x^2}$
• The degree of freedom in the y-direction will contribute $\small{\frac{1}{2}m\,v_y^2}$
• The degree of freedom in the z-direction will contribute $\small{\frac{1}{2}m\,v_z^2}$

7. Consider a diatomic gas like oxygen ($\small{O_2}$) or nitrogen ($\small{N_2}$). Each molecule of such a gas will have two atoms.
• In fig.13.2 below, the brown spheres represent atoms.
• Two brown spheres are joined together by a yellow line. • This line represents the bond between the two atoms.
    ♦ The green line is perpendicular to the yellow line.
    ♦ The red line is also perpendicular to the yellow line.
    ♦ The green and red lines are perpendicular to each other.

Fig.13.2

8. In the fig. on the left side, the molecule as a whole, is rotating about the green line. So green line is the axis of rotation. It is marked as (1).
• Due to this rotation, the molecule will have a rotational energy equal to $\small{\frac{1}{2}\,I_1\,\omega_1^2}$
    ♦ $\small{I_1}$ is the moment of inertia of the molecule about axis (1)
    ♦ $\small{\omega_1}$ is the angular speed of the molecule about axis (1)

9. In the fig. on the right side, the molecule as a whole, is rotating about the red line. So red line is the axis of rotation. It is marked as (2).
• Due to this rotation, the molecule will have a rotational energy equal to $\small{\frac{1}{2}\,I_2\,\omega_2^2}$
    ♦ $\small{I_2}$ is the moment of inertia of the molecule about axis (2)
    ♦ $\small{\omega_2}$ is the angular speed of the molecule about axis (2)

10. The molecule can rotate about the yellow line also. But the moment of inertia of the molecule about the yellow line is very small. So this rotation will not contribute much to the total rotational energy. We can safely ignore this rotation.

11. Based on (8) and (9), we can write:
Rotational energy of the molecule is given by:
$\small{\epsilon_r~=~\frac{1}{2}\,I_1\,\omega_1^2~+~\frac{1}{2}\,I_2\,\omega_2^2}$

• It is clear that, a diatomic molecule has two rotational degrees of freedom.
    ♦ One along axis (1)
    ♦ The other along axis (2)

12. So a diatomic molecule has a total energy given by:
$\small{\epsilon_t\,+\,\epsilon_r\,=\,\frac{1}{2}m\,v_x^2\,+\,\frac{1}{2}m\,v_y^2\,+\,\frac{1}{2}m\,v_z^2\,+\,\frac{1}{2}\,I_1\,\omega_1^2\,+\,\frac{1}{2}\,I_2\,\omega_2^2}$

• Note that, for a monatomic molecule, the last two terms will be absent.

13. We have seen the energy contributions from translation and rotation. If a molecule experience vibration, that molecule will posses vibrational energy also.
• Consider the two spheres in fig.13.2 above. If the spheres oscillate along the yellow line, then that molecule as a whole, will posses vibrational energy. It is like a spring with the two spheres at it's ends. We will see more details in later chapters.
• The $\small{O_2}$ molecule is rigid. So this energy does not come into effect at moderate temperatures. But a $\small{CO}$ molecule will posses this energy even at moderate temperatures.

14. For a molecule like $\small{CO}$, the vibrational energy is given by:
$\small{\epsilon_v\,=\,\frac{1}{2} m \left(\frac{dy}{dt} \right)^2\,+\,\frac{1}{2} k y^2}$
    ♦ $\small{k}$ is the force constant of the oscillator
    ♦ $\small{y}$ is the vibrational co-ordinate

15. Now we can write the total energy:
$\small{\epsilon\,=\,\epsilon_t\,+\,\epsilon_r\,+\,\epsilon_v}$

• Where,

$\small{\epsilon_t\,=\,\frac{1}{2}m\,v_x^2\,+\,\frac{1}{2}m\,v_y^2\,+\,\frac{1}{2}m\,v_z^2}$

$\small{\epsilon_r\,=\,\frac{1}{2}\,I_1\,\omega_1^2\,+\,\frac{1}{2}\,I_2\,\omega_2^2}$

$\small{\epsilon_v\,=\,\frac{1}{2} m \left(\frac{dy}{dt} \right)^2\,+\,\frac{1}{2} k y^2}$

16. Now we can analyze the contribution from each type of energy.
(i) Each of the translational degree of freedom, contributes one square term.
(ii) Each of the rotational degree of freedom, contributes one square term.
(iii) Each vibrational frequency, contributes two square terms.

17. Now we can write about the general form of contribution:
• We have seen that, each square term in the translational energy is equal to $\small{\frac{1}{2}K_B\,T}$
• The Scottish physicist James Clerk Maxwell proved that, each square term has an average energy equal to $\small{\frac{1}{2}K_B\,T}$. This is known as the law of equipartition of energy.
• So we can write:
    ♦ Each translational degree of freedom contributes $\small{\frac{1}{2}K_B\,T}$ to the total energy.
    ♦ Each rotational degree of freedom contributes $\small{\frac{1}{2}K_B\,T}$ to the total energy.
    ♦ Each vibrational frequency contributes two times $\small{\frac{1}{2}K_B\,T}$ = $\small{K_B\,T}$ to the total energy.

• We will see the detailed proof of the law in higher classes.

---

In the next section, we will apply the law to predict specific heats of gases theoretically.

Previous

Contents

Next

Copyright©2026 Higher secondary physics.blogspot.com

Chapter 7.34 - Rolling without Slipping

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

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

Fig.7.137

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

Fig.7.138

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

Kinetic energy of a rolling body

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

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

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

Fig.7.139

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

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

---

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

PREVIOUS           CONTENTS          NEXT

Copyright©2019 Higher Secondary Physics. blogspot.in - All Rights Reserved

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

---

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

---

In the next section, we will see rolling motion

PREVIOUS           CONTENTS          NEXT

Copyright©2019 Higher Secondary Physics. blogspot.in - All Rights Reserved