Translational, Rotational and Vibrational Energy: Difference between revisions

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==Main Idea==
==Main Idea==
In many cases, analyzing the kinetic energy of an object is in fact more difficult than just applying the formula <math> K = \cfrac{1}{2}mv^2 </math>. An example of this is when throwing a basketball because not only does it move through the air, but it is also rotating around its own axis. When analyzing more complicated movements like this one, it is necessary to break kinetic energy into different parts, such as rotational, translational, and vibrational, and analyze each one separately to give a more accurate picture.
In many real-world situations, analyzing the kinetic energy of an object is more complex than just applying the formula:


Translational kinetic energy is the kinetic energy associated with the motion of the center of mass of an object. This would be the basketball traveling in the air from one location to another. While relative kinetic energy is the kinetic energy associated to the rotation or vibration of the atoms of the object around its center or axis. Relative kinetic energy would be the rotation of the basketball around it's axis. Later on this page, we go into more depth about the different types of kinetic energy.
<math> K = \cfrac{1}{2}mv^2 </math>


Here is a link to a video which explains kinetic energy in detail: [https://youtu.be/Cobhu3lgeMg]
For example when a basketball is thrown, it is not only moving through space, but also rotating about its own axis. Because of this, the total kinetic energy must be broken into components.  


===Mathematical Model===
The total kinetic energy of a system can be separated into:
=== Total Kinetic Energy ===
* Translational energy (motion of the center of mass)
As we just saw, the total kinetic energy of a multi particle system can be divided into the energy associated with motion of the center of mass and the motion relative to the center of mass.
* Rotational energy (spinning motion)
* Vibrational energy (internal motion of particles)


::<math> K_{total} = K_{translational} + K_{relative} </math>
This breakdown allows us to more accurately analyze motion in physical systems.


The relative kinetic energy is composed of motion due to rotation about the center of mass and vibrations/oscillations of the object. 


::<math> K_{total} = K_{translational} + K_{rotational} + K_{vibrational} </math>
[[File:Rolling Racers - Moment of inertia.gif|Rolling_Racers_-_Moment_of_inertia]]


====Translational Kinetic Energy====
===Mathematical Model===
=== Total Kinetic Energy ===


"Translation" means:
\[
K_{total} = K_{translational} + K_{rotational} + K_{vibrational}
\]


::''To move from one location to another location''
This equation shows that energy must be considered in multiple forms when objects both move and rotate.


By calculating translational kinetic energy, we can track how one object moves from one location to another. Since the translational kinetic energy is associated with the movement of the center of mass of the object, it is important to know how to calculate the location of the center of mass and the velocity of the center of mass which is shown in the two equations below:


::<math>r_{CM} = \cfrac{m_1r_1 + m_2r_2+m_3r_3 + ...}{m_1 + m_2 +m_3}</math>
====Translational Kinetic Energy====


::<math>v_{CM} = \cfrac{m_1v_1 + m_2v_2+m_3v_3 + ...}{m_1+ m_2 +m_3}</math>
\[
K_{trans} = \frac{1}{2}Mv_{CM}^2
\]


:Here is a link to a video if you want to refresh your knowledge on center of mass: [https://youtu.be/5qwW8WI1gkw]
* \(M\): total mass 
* \(v_{CM}\): velocity of the center of mass  


The motion of the center of mass is described by the velocity of the center of mass. Using the total mass and the velocity of the center of mass, we define the translational kinetic energy as:


::<math>K_{translational} = \cfrac{1}{2}M_{total}v_{CM}^2</math>
The center of mass is calculated as:


====Vibrational Kinetic Energy====
\[
The total energy due to vibrations is the sum of the potential energy associated with interactions causing the vibrations and the kinetic energy of the vibrations.
r_{CM} = \frac{\sum m_ir_i}{\sum m_i}
\quad\quad
v_{CM} = \frac{\sum m_iv_i}{\sum m_i}
\]


::<math>E_{vibrational} = K_{vibrational} + U_{s}</math>
'''Key Idea:''' Translational energy depends only on how the object moves.


The easiest way to find vibrational kinetic energy is by knowing the other energy terms and isolating the vibrational kinetic energy. This is when there is no rotational kinetic energy:
[[File:Center_of_mass_diagram.svg|300px|center|thumb|The center of mass represents the average position of mass in a system.]]
 
::<math>E_{total} = K_{trans} + K_{vibrational} + U_{s} +E_{rest}</math>
---
::<math>K_{vibrational} = E_{total} - (K_{trans} + U_{s} +E_{rest})</math>


====Rotational Kinetic Energy====
====Rotational Kinetic Energy====
[[File:Kinetic_energy.png|300px|right|thumb|Here are links to two videos that cover rotational kinetic energy and moment of interia: [https://youtu.be/craljBk-E5g][https://youtu.be/XlFlZHfAZeE]]]
The total energy due to vibrations is the sum of the potential energy associated with interactions causing the vibrations and the kinetic energy of the vibrations.  
Rotational kinetic energy is the energy due to the rotation about the center of mass. It can be calculated by finding the angular momentum and inertia of the system, which will be discussed in greater detail in the next two sections. The equation used to find kinetic rotational energy is below:
 
::<math>K_{rotational} =\frac{1}{2} I_{cm}{\omega}^2</math>
 
Another important rotational equation is:


::<math></math>
\[
K_{rot} = \frac{1}{2}I\omega^2
\]


=====Moment of Inertia=====
* \(I\): moment of inertia
The moment of inertia of an object shows the difficulty of rotating an object, since the larger the moment of inertia the more energy is required to rotate the object at the same angular velocity as an object with a smaller moment of inertia. The moment of inertia of an object is defined as the sum of the products of the mass of each particle in the object with the square of their distance from the axis of rotation. The general formula for calculating the moment of inertia of an object is:
* \(\omega\): angular velocity


::<math>I = m_1{r_{1}}^2 + m_2{r_{2}}^2 + m_3{r_{3}}^2 +...</math>


:::Here <math> r_1, r_2, r_3 </math> represent the perpendicular distance from the point/axis of rotation.
====Moment of Inertia====
\[
I = \sum m_ir_i^2
\]


:or
This measures how difficult it is to rotate an object.


::<math>I = \sum_{i} m_{i}{r_{i}}^2</math>
'''Important Insight:'''
Mass farther from the axis increases rotational energy because of the \(r^2\) term.


:For a body with a uniform distribution of mass this can be turned into an integral:
[[File:Moment_of_inertia_diagram.svg|300px|right|thumb|Mass farther from the axis increases rotational inertia.]]


::<math>I = \int r^2 \ dm</math>
---


The units of rotational inertia are <math> kg \cdot m^2 </math>
=====Angular Speed and Velocity=====


[[File:4c906c92cebe30d9486deb2a682acf561d23c9c1.png|900px|center]]
\[
\omega = \frac{2\pi}{T}
\]


=====Angular Speed and Acceleration=====
\[
The angular speed is the rate at which the object is rotating. It is given in the following formula:
v = \omega r
[[File:Angularvelocity.png|right|200px]]
\]


::<math>\omega = \cfrac{2\pi}{T}</math>, where
Points farther from the center move faster.


:::<math>T =</math> the period of the rotation
---


The angle in which a disk turns is <math>2 \pi</math> in a time <math>T</math>. It is measured in radians per second. The tangential velocity of an object is related to its radius r at the angular speed because the tangential velocity increases when the distance from the center of an object increases. It is shown in the equation below:
=====Vibrational Kinetic Energy=====


::<math>v(r)= \omega r</math>
Vibrational energy comes from internal motion of particles within an object.


The angular acceleration a rotating object goes through to change its angular speed is given by:
* Important in molecules and thermal systems 
* Usually not directly calculated in introductory physics problems 


::<math>a(r) = \alpha r</math>
---


===Computational Model===
===Physical Intuition===
Here is a rotating rod computational model example:
Consider a rolling wheel:


https://trinket.io/glowscript/31d0f9ad9e
* Moves forward -> translational energy
* Spins -> rotational energy
* Internal atoms vibrate -> vibrational energy


==Examples==
==Examples==


===Simple===
===Conceptual Example===


A player throws a mid-court pass horizontally with a <math>624g</math> basketball. This pass covers <math>15 \ m</math> in <math>2 \ s</math>.  
A bowling ball rolls without slipping.


:'''a) What is the basketball’s average translational kinetic energy while in flight?  
'''Which energies are present?'''


::Since the ball is moving relative to the gym, we can describe its average velocity, and thus translational kinetic energy as:
* Translational ✔ 
* Rotational  ✔ 
* Vibrational ✖ (ignored at this level)


:::<math>v_{avg} = \frac{d}{t} = \frac{15}{2} = 7.5 \ \frac{m}{s}</math>
---


:::<math>K_{avg_{b}} = \frac{1}{2}m{v_{avg}}^2 = \frac{1}{2} \times 0.624 \times (7.5)^2 = 17.55 \ J</math>
===Calculation Example===


An average molecule of air in the basketball, has a mass of <math>29 \ u</math>, and an average speed of <math>500 \ \frac{m}{s}</math>, relative to the basketball. There are about <math>3 \times 10^{23}</math> molecules inside it, moving in random directions, when the ball is properly inflated.
A solid disk rolls without slipping.


:'''b) What is the average translational kinetic energy of the random motion of all the molecules inside, relative to the basketball?
Given:
* \(m = 2 \, kg\)
* \(v = 3 \, m/s\) 
* \(I = \frac{1}{2}mr^2\)


::Since the average speed of a molecule is <math>500 \ \frac{m}{s}</math>, we can calculate the average translational kinetic energy of a given molecule:
Step 1: Translational Energy


:::<math>K_{avg_{g}} = \frac{1}{2}m_{g}{v_{avg_{g}}}^2</math>
\[
K_{trans} = \frac{1}{2}mv^2 = 9 \, J
\]


::The mass of a molecule in kilograms <math>(m_{g})</math> will be the atomic mass <math>(m_{A})</math> times a conversion factor <math>(A)</math>:
Step 2: Rotational Energy


:::<math>m_{g} = m_{A}A = 29 \times 1.66 \times 10^{-27} = 4.814 \times 10^{-26} \ kg</math>
\[ K_{rot} = \frac{1}{2}I\omega^2
\]


::Thus, the translational kinetic energy of an average molecule relative to the ball is:
Using \( \omega = \frac{v}{r} \): \[ K_{rot} = \frac{1}{4}mv^2 = 4.5 \, J \]


:::<math>K_{avg_{g}} = \frac{1}{2}m_{g}{v_{avg_{g}}}^2 = \frac{1}{2} \times 4.814 \times 10^{-26} \times (500)^2 = 6.0175 \times 10^{-21} \ J</math>
Total Energy:  


::Multiplying this kinetic energy by the number of molecules will give us our answer:
\[ K_{total} = 13.5 \, J
\]


:::<math>K_{avg_{total}} = N K_{avg_{g}} = 3 \times 10^{23} \times 6.0175 \times 10^{-21} = 1,805.25 \ J</math>
---


::The kinetic energy possessed by the gas relative to the ball is <math>1,805.25</math> Joules.
===Common Mistakes===


:'''c) How fast would the basketball have to travel relative to the court to have a kinetic energy equal to the amount in part (b)?
* Forgetting rotational energy in rolling problems 
* Using incorrect relationship between \(v\) and \(\omega\)
* Ignoring moment of inertia differences 
* Assuming only translational motion matters


::The basketball would have to have a translational kinetic energy equal to <math>1,805.25</math> Joules. To do this the ball's speed would have to satisfy:
---


:::<math>K_{ball} = \frac{1}{2}m_{ball}v^2</math>
==Computational Model==


::Solving for <math>v</math> gives:
GlowScript simulation:


:::<math>v = \sqrt{\frac{2K_{ball}}{m_{ball}}} = \sqrt{\frac{2 \times 1,805.25}{0.624}} = 76.07 \ \frac{m}{s}</math>
https://trinket.io/glowscript/31d0f9ad9e


===Middling===
This model helps visualize rotational motion and energy changes.


A wheel is mounted on a stationary axel, which is nearly frictionless so that the wheel turns freely. The wheel has an inner ring with a mass of <math>5 \ kg</math> and a radius of <math>10 \ cm</math>, and an outer ring with a mass of <math>2 \ kg</math> and a radius of <math>25 \ cm</math>; the spokes have negligible mass. A string with negligible mass is wrapped around the outer ring and you pull on it, increasing the rotational speed of the wheel.
---


:'''a) During the time that the wheel's rotation changes from 4 revolutions per second to 7 revolutions per second, how much work do you do?'''
==Connectedness==
 
::To find the work done, the best course of action will be to find the change in kinetic energy of the wheel. To do this, we will need to find the change in angular speed and the moment of inertia of the wheel.
 
::We are given the initial and final frequencies of rotation for the wheel:


:::<math>f_{0} = 4 \ s^{-1} \quad \And \quad f_{f} = 7 \ s^{-1}</math>
'''Personal Connection:'''
Dance and sports like tennis involve rotation and motion, similar to energy concepts discussed here.


::The frequency of a rotation is related to its period by:
'''Academic Connection:'''
Important in physics, engineering, and chemistry for analyzing motion and energy systems.


:::<math>T = \frac{1}{f}</math>
'''Industrial Applications:'''
 
* Flywheels for energy storage
::Therefore, we have the initial and final period of rotation:
* Rotating machinery
 
* Engines and turbines
:::<math>T_{0} = \frac{1}{f_{0}} = \frac{1}{4} = 0.25 \ s \quad \And \quad T_{f} = \frac{1}{f_{f}} = \frac{1}{7} = 0.1429 \ s</math>
 
::The period of a rotation is related to the body's angular speed by:
 
:::<math>T = \frac{2\pi}{\omega} \quad \therefore \quad \omega = \frac{2\pi}{T}</math>
 
::Therefore, we can calculate the initial and final angular speeds of the wheel:
 
:::<math>\omega_{0} = \frac{2\pi}{T_{0}} = \frac{2\pi}{0.25} = 25.133 \ \left(\frac{rads}{s}\right) \quad \And \quad \omega_{f} = \frac{2\pi}{T_{f}} = \frac{2\pi}{0.1429} = 43.969 \ \left(\frac{rads}{s}\right)</math>
 
::Now, to find the moment of inertia, we assume the separate rings are thin enough to use the moment of inertia for a thin ring, from the table above:
 
:::<math>I = mr^2</math>
 
::However, since there are two concentric rings that make up the wheel (ignoring the spokes), we must add their moments of inertia to calculate the total moment of inertia for the wheel:
 
:::<math>I_{wheel} = I_{inner} + I_{outer} = m_{inner} r^2_{inner} + m_{outer} {r}^2_{outer} = 5 \times (0.1)^2 + 2 \times (0.25)^2 = 0.175 \ (kg \cdot m^2)</math>
 
::Using the Energy Principle, and assuming the only work done on the wheel is due to the string being pulled, we can say:
 
:::<math>\Delta E_{wheel} = W_{on wheel}</math>
 
:::<math>\Delta U_{wheel} + \Delta K_{transalational_{wheel}} + \Delta K_{rotational_{wheel}} = W_{on wheel} \quad \And \quad \Delta U_{wheel} = 0 \quad \And \quad \Delta K_{translational_{wheel}} = 0</math>
 
:::<math>\Delta K_{rotational_{wheel}} = W_{wheel}</math>
 
::The rotational kinetic energy of the wheel is given by:
 
:::<math>K_{rotational} = \frac{1}{2} I_{wheel} \omega_{wheel}</math>
 
::Thus, the change in rotational kinetic energy of the wheel will be:
 
:::<math>\Delta K_{r} = \frac{1}{2} I_{wheel} (\omega^2_{f} - \omega^2_0) = \frac{1}{2} \times 0.175 \times ((43.969)^2 - (25.133)^2) = 113.89 \ J</math>
 
::Since we are assuming there is no friction in this process and the center of mass of the wheel does not move, all the work done is due to you pulling the string:
 
:::<math>W_{you} = \Delta K_{r} = 113.89 \ J</math>
 
 
===Difficult===
Image: https://drive.google.com/file/d/185xWvF0iMpX-dUNc91ZOhCxQIq7-hHM3/preview
 
 
 
A common flywheel design is a flattened disk (cylinder) rotating about an axis perpendicular to its center, as shown in the figure. Let’s assume our cylinder is about a meter across ( R = 0.5 m ) and has a mass of 240 kg (i.e. has a weight of about 530 pounds). The moment of inertia for such a shape is <math> I = (1/2)MR^2 </math>. What velocity do I need the disk to rotate in order to power a house? A typical home uses energy at a rate of roughly 1000 W or 1000 J/s.   
 
 
 
<math> I = (0.5)*(240)*(0.5)^2 = 30 Kg*m^2 </math>
 
Energy consumption in a day:
 
<math> E{k} = 1000 * 24* 3600 = 8.64 * 10^7 J </math>
 
<math> E = (1 / 2)Iw^2 </math>
 
Energy = angular times inertia:
 
<math> w = √(2E/I) </math>
<math> w = √(2*8.64*10^7/30) = 2400 rad/sec </math>
<math> v = r*w = 0.5 * 2400 = 1200 m/sec </math>
 
==Connectedness==
'''1. How is this topic connected to something that you are interested in?'''


*This topic connected to me because I used to dance when I was younger. This section focused on kinetic energy and the different parts of kinetic energy. You could break up different parts of dance and compare it to kinetic energy.
---


*This topic resonated with my tennis experience. If you play tennis there are certain moves that generate specific rotations patterns on the ball and can either increase or decrease the length of its trajectory. Top spin, for example, involves spinning the ball forward and this leads to a positive change in K{rot} and, if we assume no change in K{trans}, an increase in K{total} that extends the ball's trajectory.
===History===


'''2. How is it connected to your major?'''
The concept of kinetic energy developed over time through contributions from scientists such as Aristotle, Leibniz, Bernoulli, and Gaspard-Gustave Coriolis. The term “kinetic energy” was later coined by Lord Kelvin.


*In Chemical Engineering, we will focus on the kinetic energy on the microscopic level and determining the energy of the particles by looking at the translational, rotational, and vibrational energies of the atom, and how they allow chemical reactions to precess.
===Why This Matters for Exams===


'''3. Is there an interesting industrial application?'''
Most physics problems:
* Combine translation and rotation
* Require identifying ALL forms of energy


*There are many machines that use kinetic energy for power, and we will probably see in a few years from now the use of rotational, translational, and vibrational energy to power anything from phones to computers.
Missing one energy component often leads to incorrect answers.


==History==
---
Kinetic energy was first set apart from potential energy by Aristotle. Later, in the 1600's, Leibniz and Bernoulli developed the idea that <math>E \propto mv^2</math>, and they called it the 'living force.' However, it wasn't until 1829 that Gaspard-Gustave Coriolis showed the first signs of understanding kinetic energy the way that we do today by focusing on the transfer on energy in rotating water wheels. Finally, in 1849, Lord Kelvin is said to have coined the term 'kinetic energy.'


==See also==
===Summary===


===Further Reading===
Kinetic energy in real systems consists of multiple components. By separating it into translational, rotational, and vibrational parts, we can more accurately understand and analyze motion.
*[[Point Particle Systems]]<br>
*[[Real Systems]]<br>
*[[Conservation of Energy]]<br>
*[[Potential Energy]]<br>
*[[Thermal Energy]]<br>
*[[Internal Energy]]<br>
*[[Center of Mass]]<br>


===External Links===
===External Links===

Revision as of 23:56, 28 April 2026

SHREYA LAKSHMISHA SPRING 2026

Main Idea

In many real-world situations, analyzing the kinetic energy of an object is more complex than just applying the formula:

[math]\displaystyle{ K = \cfrac{1}{2}mv^2 }[/math]

For example when a basketball is thrown, it is not only moving through space, but also rotating about its own axis. Because of this, the total kinetic energy must be broken into components.

The total kinetic energy of a system can be separated into:

  • Translational energy (motion of the center of mass)
  • Rotational energy (spinning motion)
  • Vibrational energy (internal motion of particles)

This breakdown allows us to more accurately analyze motion in physical systems.


Rolling_Racers_-_Moment_of_inertia

Mathematical Model

Total Kinetic Energy

\[ K_{total} = K_{translational} + K_{rotational} + K_{vibrational} \]

This equation shows that energy must be considered in multiple forms when objects both move and rotate.


Translational Kinetic Energy

\[ K_{trans} = \frac{1}{2}Mv_{CM}^2 \]

  • \(M\): total mass
  • \(v_{CM}\): velocity of the center of mass


The center of mass is calculated as:

\[ r_{CM} = \frac{\sum m_ir_i}{\sum m_i} \quad\quad v_{CM} = \frac{\sum m_iv_i}{\sum m_i} \]

Key Idea: Translational energy depends only on how the object moves.

File:Center of mass diagram.svg
The center of mass represents the average position of mass in a system.

---

Rotational Kinetic Energy

The total energy due to vibrations is the sum of the potential energy associated with interactions causing the vibrations and the kinetic energy of the vibrations.

\[ K_{rot} = \frac{1}{2}I\omega^2 \]

  • \(I\): moment of inertia
  • \(\omega\): angular velocity


Moment of Inertia

\[ I = \sum m_ir_i^2 \]

This measures how difficult it is to rotate an object.

Important Insight: Mass farther from the axis increases rotational energy because of the \(r^2\) term.

File:Moment of inertia diagram.svg
Mass farther from the axis increases rotational inertia.

---

Angular Speed and Velocity

\[ \omega = \frac{2\pi}{T} \]

\[ v = \omega r \]

Points farther from the center move faster.

---

Vibrational Kinetic Energy

Vibrational energy comes from internal motion of particles within an object.

  • Important in molecules and thermal systems
  • Usually not directly calculated in introductory physics problems

---

Physical Intuition

Consider a rolling wheel:

  • Moves forward -> translational energy
  • Spins -> rotational energy
  • Internal atoms vibrate -> vibrational energy

Examples

Conceptual Example

A bowling ball rolls without slipping.

Which energies are present?

  • Translational ✔
  • Rotational ✔
  • Vibrational ✖ (ignored at this level)

---

Calculation Example

A solid disk rolls without slipping.

Given:

  • \(m = 2 \, kg\)
  • \(v = 3 \, m/s\)
  • \(I = \frac{1}{2}mr^2\)

Step 1: Translational Energy

\[ K_{trans} = \frac{1}{2}mv^2 = 9 \, J \]

Step 2: Rotational Energy

\[ K_{rot} = \frac{1}{2}I\omega^2 \]

Using \( \omega = \frac{v}{r} \): \[ K_{rot} = \frac{1}{4}mv^2 = 4.5 \, J \]

Total Energy:

\[ K_{total} = 13.5 \, J \]

---

Common Mistakes

  • Forgetting rotational energy in rolling problems
  • Using incorrect relationship between \(v\) and \(\omega\)
  • Ignoring moment of inertia differences
  • Assuming only translational motion matters

---

Computational Model

GlowScript simulation:

https://trinket.io/glowscript/31d0f9ad9e

This model helps visualize rotational motion and energy changes.

---

Connectedness

Personal Connection: Dance and sports like tennis involve rotation and motion, similar to energy concepts discussed here.

Academic Connection: Important in physics, engineering, and chemistry for analyzing motion and energy systems.

Industrial Applications:

  • Flywheels for energy storage
  • Rotating machinery
  • Engines and turbines

---

History

The concept of kinetic energy developed over time through contributions from scientists such as Aristotle, Leibniz, Bernoulli, and Gaspard-Gustave Coriolis. The term “kinetic energy” was later coined by Lord Kelvin.

Why This Matters for Exams

Most physics problems:

  • Combine translation and rotation
  • Require identifying ALL forms of energy

Missing one energy component often leads to incorrect answers.

---

Summary

Kinetic energy in real systems consists of multiple components. By separating it into translational, rotational, and vibrational parts, we can more accurately understand and analyze motion.

External Links

References

All problem examples, youtube videos, and images are from the websites referenced below:

Retrieved from "http://www.physicsbook.gatech.edu/index.php?title=Translational,_Rotational_and_Vibrational_Energy&oldid=48302"