Translational, Rotational and Vibrational Energy: Difference between revisions

From Physics Book
Jump to navigation Jump to search
 
(42 intermediate revisions by 2 users not shown)
Line 1: Line 1:
GABRIEL ALMEIDA SPRING 2022
==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 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.
Line 49: Line 51:


::<math>K_{rotational} =\frac{1}{2} I_{cm}{\omega}^2</math>
::<math>K_{rotational} =\frac{1}{2} I_{cm}{\omega}^2</math>
Another important rotational equation is:
::<math></math>


=====Moment of Inertia=====
=====Moment of Inertia=====
The moment of inertia of an object shows the difficulty of rotating an object, since the larger moment of inertia the more energy required to rotate the object at the same angular velocity as an object with smaller moment of inertia. It is 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:
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:


<math> I = m_1r_{\perp,1}^2 + m_2r_2{\perp,2}^2 + m_3r_{\perp,3}^2 +...</math>
::<math>I = m_1{r_{1}}^2 + m_2{r_{2}}^2 + m_3{r_{3}}^2 +...</math>
Units: <math> kg.m^2 </math>


Here <math> r_1, r_2, r_3 </math> represent the perpendicular distance from the point/axis of rotation.  
:::Here <math> r_1, r_2, r_3 </math> represent the perpendicular distance from the point/axis of rotation.


Moment of Inertia is actually calculated using Calculus. These values we obtain for standard objects are not once you will need to derive, but conceptually understanding how they come to be will help you understand Moment of Inertia and it's impact on a solid. Below are a few moment of inertia used most frequently:
:or


[[File:4c906c92cebe30d9486deb2a682acf561d23c9c1.png]]
::<math>I = \sum_{i} m_{i}{r_{i}}^2</math>


When you compare two bodies rotating with the same kinetic energy and one of them has a higher moment of Inertia, then the one with the higher moment of Inertia has a lower speed. If angular speed is lower and radius is larger that means mass must be small since Moment of Inertia also depends on mass and distance from axis of rotation. So we can deduce such relations which helps us understand how the moment of Inertia of an object actually affects the object in different ways.
:For a body with a uniform distribution of mass this can be turned into an integral:


=====Angular Speed=====
::<math>I = \int r^2 \ dm</math>
 
The units of rotational inertia are <math> kg \cdot m^2 </math>
 
[[File:4c906c92cebe30d9486deb2a682acf561d23c9c1.png|900px|center]]
 
=====Angular Speed and Acceleration=====
The angular speed is the rate at which the object is rotating. It is given in the following formula:
The angular speed is the rate at which the object is rotating. It is given in the following formula:
[[File:Angularvelocity.png|right|200px]]
[[File:Angularvelocity.png|right|200px]]
Line 76: Line 87:
::<math>v(r)= \omega r</math>
::<math>v(r)= \omega r</math>


===Computational Model===
The angular acceleration a rotating object goes through to change its angular speed is given by:
 
==Examples==


===Simple===
::<math>a(r) = \alpha r</math>
 
===Middling===
 
===Difficult===
 
==Connectedness==
 
==History==
 
==See also==
 
===Further reading===
 
===External links===
 
==References==
 
==Main Idea==
 
==Mathematical Method==
=== Point Particle System VS Extended System ===
 
When calculating the total energy of a system, it will sometimes contain one or more types of kinetic energies.
 
Let's think about a system modeled as a point particle system. In this model of a system, we think of the object as one point, located at its center of mass.  It becomes clear that there can only be translational kinetic energy since there cannot be any rotational kinetic energy. This is because there are no atoms rotating about the center of mass, since we are thinking of the center of mass as the entire object. All the forces act on the center of mass.
 
<math> \triangle E_{system} = W </math>
<math> W =  F_{net} . \triangle r_{CM} </math>
<math> \triangle E_{system} = \triangle K_{translational} </math>
<math> \triangle K_{translational} = F_{net} . \triangle r_{CM} </math>
 
Viewing a system as a point particle system allows us to easily calculate the translational kinetic energy. This translational kinetic energy will be the same in the extended system as it is in the point particle system. We can use this value to calculate other terms in the general equation in the extended system. In the extended system we add up all the forces that might have different points of application. In the extended system, all the atoms, their rotation about the axis, and the general movement of the center of mass is considered in the extended/real system.There may be spring potential energy and rest energy included for both point particle and real system depending on the example. We start with the general energy equation:
 
<math> \triangle E_{system} = W </math>
<math>W = F_1r_1 + F_2r_2 + F_3r_3 +... +F_nr_n  </math>
<math> \triangle K_{translational} + \triangle K_{rotational} + \triangle K_{vibrational} = W </math>
 
[[File:Pointparticlevsrealsystem.png]]
 
Here are some videos that go into more depth about point particle vs real system:
[https://youtu.be/vL5yTCyRMGk]
 
==A Computational Model==
 
=== Vpython ===
There are not many applications of vpython for this chapter because it necessitates such a conceptual way of thinking. However, it is possible to make a program that calculates the rotational energy given displacements for each force and the magnitude of the force. The mass of the object and the initial conditions would have to be given as well.
 
===Visualizing translational and rotational kinetic energy ===
It is important to really think about translational kinetic energy as the movement of the center of mass and only that. In the point particle system, only translational kinetic energy is present because we are visualizing the object only as its center of mass, and that is therefore the only kind of movement we have. Let's think about a baton. The baton can be twirled and tossed. If you think of the baton only in terms of its center of mass, you realize that the center of mass only has one kind of movement: being tossed. It is useful to think of the baton as a crushed particle.


===Computational Model===
Here is a rotating rod computational model example:


In the extended system, however, we consider the movement of the center of mass as well as the movement of all the atoms around it. Since the atoms of the baton that are not its center rotate when it is twirled, there is rotational energy. Its center of mass is still being transferred, there is therefore also translational energy. Here, we don't think the baton as a dot since the object is being both twirled and tossed.
https://trinket.io/glowscript/31d0f9ad9e


==Examples==
==Examples==


===Simple===
===Simple===
''Problem statement'':


(a) A player lobs a mid-court pass with a 624-g basketball, which covers 15 m in 2 s. What is the basketball’s horizontal translational kinetic energy while in flight? (b) An average molecule of air, in the basketball in part (a), has a mass of 29 u, and an average speed of 500 m/s, relative to the basketball. There are about  <math> 3×10^{23}</math> molecules inside it, moving in random directions, when the ball is properly inflated. What is the average translational kinetic energy of the random motion of all the molecules inside, relative to the basketball? (c) How fast would the basketball have to travel relative to the court, as in part (a), so as to have a kinetic energy equal to the amount in part (b)?
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>.  


"Strategy":
:'''a) What is the basketball’s average translational kinetic energy while in flight?


In part (a), first find the horizontal speed of the basketball and then use the definition of kinetic energy in terms of mass and speed, <math> K=1/2mv^2 </math>. Then in part (b), convert unified units to kilograms and then use <math>K=1/2mv^2 </math> to get the average translational kinetic energy of one molecule, relative to the basketball. Then multiply by the number of molecules to get the total result. Finally, in part (c), we can substitute the amount of kinetic energy in part (b), and the mass of the basketball in part (a), into the definition <math> K=1/2mv^2 </math>, and solve for v.
::Since the ball is moving relative to the gym, we can describe its average velocity, and thus translational kinetic energy as:


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


''Solution'':
:::<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>


a.The horizontal speed is (15 m)/(2 s), so the horizontal kinetic energy of the basketball is
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.


<math>1/2(0.624kg)(7.5m/s^2)=17.6J </math>
:'''b) What is the average translational kinetic energy of the random motion of all the molecules inside, relative to the basketball?


b.The average translational kinetic energy of a molecule is
::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:


<math>1/2(29u)(1.66×10^{−27}kg/u)(500m/s)^2=6.02×10^{−21}J </math>
:::<math>K_{avg_{g}} = \frac{1}{2}m_{g}{v_{avg_{g}}}^2</math>


and the total kinetic energy of all the molecules is
::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>:


<math>(3×10^{23})(6.02×10^{−21}J)=1.80kJ</math>
:::<math>m_{g} = m_{A}A = 29 \times 1.66 \times 10^{-27} = 4.814 \times 10^{-26} \ kg</math>


c.v=76.0m/s.
::Thus, the translational kinetic energy of an average molecule relative to the ball is:


===Middling===
:::<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>
''Problem statement'':
In the figure which is in the representations section you observe that 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 mass 5 kg and radius 10 cm and an outer ring with mass 2 kg and radius 25 cm; 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. 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?


"Facts":
::Multiplying this kinetic energy by the number of molecules will give us our answer:


Inner ring of wheel has mass of 5 kg and radius 10 cm
:::<math>K_{avg_{total}} = N K_{avg_{g}} = 3 \times 10^{23} \times 6.0175 \times 10^{-21} = 1,805.25 \ J</math>
Outer ring has a mass of 2 kg and radius of 25 cm
String with negligible mass is wrapped around the outer rung and pulled out


"Assumption and Approximations":
::The kinetic energy possessed by the gas relative to the ball is <math>1,805.25</math> Joules.


The rod has to be sufficiently thin so your not worried about contributions to the moment of inertia from parts of the rod that are further away from the center of mass.
:'''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)?
Stationary axle is nearly frictionless so that the wheel turns freely.
The spokes of the wheel have negligible mass.
String wrapped around the outer ring has negligible mass.


"Representations"
::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:


System: Wheel and string
:::<math>K_{ball} = \frac{1}{2}m_{ball}v^2</math>
Surroundings: Your hand, axle, Earth


[[File:Rotational_kietnic_energy_problem.jpg]]
::Solving for <math>v</math> gives:


''Solution'':
:::<math>v = \sqrt{\frac{2K_{ball}}{m_{ball}}} = \sqrt{\frac{2 \times 1,805.25}{0.624}} = 76.07 \ \frac{m}{s}</math>


From the Energy Principle: <math>E_f=E_i+W</math>
===Middling===


Substitute in equation for rotational energy for energy initial and energy final.
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.


<math>1/2Iω_f^2=1/2Iω_i^2+W </math>
:'''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?'''


<math>W=1/2I(ω_f^2−ω_i^2) </math>
::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.


What is the moment of inertia?
::We are given the initial and final frequencies of rotation for the wheel:
Group this sum into a part that includes just the atoms of the inner ring and another part that includes just the atoms of the outer ring:


<math>I=(mr{⊥1}^2 + m_2r_{⊥2}^2+ ⋅⋅⋅)inner+ (m_1r_{⊥1}^2+ m_2r_{⊥2}^2)outer + ⋅⋅⋅</math>
:::<math>f_{0} = 4 \ s^{-1} \quad \And \quad f_{f} = 7 \ s^{-1}</math>


The total inertia is then the inertia inner + the inertia outer (I=Iinner+Iouter) This is a general result: The moment of inertia of a composite object is the sum of the moments of inertia of the individual pieces, because we have to add up all the contributions of all the atoms. We already determined that the moment of inertia of a ring is <math>MR^2</math> (all the atoms are at the same perpendicular distance R from the center), so the moment of inertia of this wheel is:
::The frequency of a rotation is related to its period by:


<math>I=M_{inner}R_{inner}^2+M_{outer}R_{outer}^2</math>
:::<math>T = \frac{1}{f}</math>


Substitute the corresponding values for the variables:
::Therefore, we have the initial and final period of rotation:


<math>I=(5kg)(.1m)^2+ (2kg)(.25m)^2= (0.050+0.125)kg⋅m^2=0.175kg⋅m^2 </math>
:::<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>


We need to convert revolutions per second into radians per second:
::The period of a rotation is related to the body's angular speed by:


<math>ω_i=(4revs)(2πradians/rev)=25.1 radians/s</math>
:::<math>T = \frac{2\pi}{\omega} \quad \therefore \quad \omega = \frac{2\pi}{T}</math>


<math> ω_f=(7revs)(2πradians/rev)=44.0 radians/s </math>
::Therefore, we can calculate the initial and final angular speeds of the wheel:


You, the Earth, and the axle will exert forces on the system. How much work does the Earth do? Zero, because the center of mass of the wheel doesn't move. How much work does the axle do? If there is negligible friction between the axle and the wheel, the axle does no work, because there is no-displacement of the axle's force. Therefore only you do work, and the work that you do is:
:::<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>


<math>W=1/2(0.175kg⋅m^2)(44.02+25.12)radians^2s^2=114J</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:


===Difficult===
:::<math>I = mr^2</math>
''Problem statement''
You're playing with a yo-yo of mass m on a low-mass string. You pull up on the string with a force of magnitude F, and your hand moves up a distance d. During this time the mass falls a distance h (and some of the string reels off the yo-yo's axle). (a) What is the change in translational kinetic energy of the yo-yo? (b) What is the change in the rotational kinetic energy of the yo-yo, which spins faster?


"Assumptions and Approximations"
::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:
able to maintain constant force when pulling up on yo-yo
no slippage of string around axis
sprinkle turns the same amount as string that has unraveled
no wobble included
string has no mass


:::<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>


a.
::Using the Energy Principle, and assuming the only work done on the wheel is due to the string being pulled, we can say:
''Facts and Representations''
Initial State: point particle with initial translational kinetic energy
Final State: Point Particle with final translational Kinetic energy
Point Particle System
System: Point Particle of mass m
Surroundings: Hand and earth


[[File:Yoyopointparticle.jpg]]
:::<math>\Delta E_{wheel} = W_{on wheel}</math>


"Solution"
:::<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>


From the Energy Principle ( when dealing with a point particle it only has Ktrans):  
:::<math>\Delta K_{rotational_{wheel}} = W_{wheel}</math>


ΔKtrans= ∫F⃗ net⋅dr⃗ cm
::The rotational kinetic energy of the wheel is given by:


Substituting in for the forces acting on the yo-yo for Fnet and the change in position in the y direction for the centre of mass for dr⃗ cm we get:
:::<math>K_{rotational} = \frac{1}{2} I_{wheel} \omega_{wheel}</math>


ΔKtrans=(F−mg)ΔyCM
::Thus, the change in rotational kinetic energy of the wheel will be:


As indicated in diagram in the b section of the representation:  
:::<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>


ΔyCM=−h (Substitute in −h for yCM)
::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:


ΔKtrans=(F−mg)(−h)
:::<math>W_{you} = \Delta K_{r} = 113.89 \ J</math>


Multiply across by a minus and you get an equation for ΔKtrans that looks like:


ΔKtrans=(mg−F)h
===Difficult===
Image: https://drive.google.com/file/d/185xWvF0iMpX-dUNc91ZOhCxQIq7-hHM3/preview


b.
"Facts and Representation"
Initial State: Initial rotational and translational kinetic energy
Final State: Final rotational and translational kinetic energy
Real system
System: Mass and String
Surroundings:  Earth and hand


[[File:Extenedsystemyoyo.jpg]]


"Solution"
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.   
From the energy principle we know:


ΔEsys = Wsurr


In this case we know that the change in energy in the system is due to the work done by the hand and the work done by the Earth.


ΔEsys=Whand+WEarth
<math> I = (0.5)*(240)*(0.5)^2 = 30 Kg*m^2 </math>


Because we are dealing with the real system in this scenario the change in energy is equal to the change in translational kinetic energy + the change in rotational kinetic energy.
Energy consumption in a day:


ΔKtrans+ΔKrot=Whand+WEarth
<math> E{k} = 1000 * 24* 3600 = 8.64 * 10^7 J </math>


Substitute in the work represented by force by distance for both the hand and the Earth.
<math> E = (1 / 2)Iw^2 </math>


ΔKtrans+ΔKrot=Fd+(−mg)(−h)
Energy = angular times inertia:


From part (a) of the problem we can substitute in (mg−F)h for ΔKtrans as the translational kinetic energy will be the same.
<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>


ΔKtrans=(mg−F)h
==Connectedness==
 
'''1. How is this topic connected to something that you are interested in?'''
Substituting this into our equation leaves us with:


(mg−F)h+ΔKrot=Fd+mgh
*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.


Solve for change in rotational 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.


ΔKrot=F(d+h)
'''2. How is it connected to your major?'''


==Connectedness==
*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.  
'''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 that different parts of kinetic energy. You could break up different parts of dance and compare it to kinetic energy.  


'''2.How is it connected to your major?'''
'''3. Is there an interesting industrial application?'''
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.


'''3.Is there an interesting industrial application?'''
*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.
There are many machines that use kinetic energy for power and we will probably see in a few years from now the use of rational, translational, and vibrational energy to power anything from phones to computers.


==History==
==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.'


Kinetic energy was first set apart from potential energy by Aristotle. However, it wasn't until 1929 that Gaspard-Gustave Coriolis showed the first signs of understanding of kinetic energy the way that we do today by focusing on the transfer on energy in rotating water wheels.
==See also==
 
== See also ==
 
For more specific information on point particle systems and extended systems:
http://www.physicsbook.gatech.edu/Point_Particle_Systems ;
http://www.physicsbook.gatech.edu/Real_Systems
 
For more specific information on the conservation of energy:
http://www.physicsbook.gatech.edu/Conservation_of_Energy
 
For more information on potential energy and when it is present:
http://www.physicsbook.gatech.edu/Potential_Energy
 
For information on how to take this concept further to calculate thermal energy in some instances:
http://www.physicsbook.gatech.edu/Thermal_Energy
 
===External links===


https://www.youtube.com/watch?v=5qwW8WI1gkw&feature=youtu.be
===Further Reading===
*[[Point Particle Systems]]<br>
*[[Real Systems]]<br>
*[[Conservation of Energy]]<br>
*[[Potential Energy]]<br>
*[[Thermal Energy]]<br>
*[[Internal Energy]]<br>
*[[Center of Mass]]<br>


https://youtu.be/Cobhu3lgeMg
===External Links===
 
*https://www.youtube.com/watch?v=5qwW8WI1gkw&feature=youtu.be<br>
https://www.youtube.com/watch?v=craljBk-E5g&feature=youtu.be
*https://youtu.be/Cobhu3lgeMg<br>
 
*https://www.youtube.com/watch?v=craljBk-E5g&feature=youtu.be<br>
https://youtu.be/XlFlZHfAZeE
*https://youtu.be/XlFlZHfAZeE<br>
 
*https://youtu.be/vL5yTCyRMGk<br>
https://youtu.be/vL5yTCyRMGk
*https://en.wikipedia.org/wiki/Kinetic_energy<br>
*https://en.wikipedia.org/wiki/Moment_of_inertia<br>
*https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/a/rotational-inertia<br>
*https://youtu.be/vL5yTCyRMGk


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


All images used on this page do not belong to me.
*http://www.robjorstad.com/Phys161/161Lab/161RotationalKinematicsSim.pdf
All problem examples, youtube videos, and images are from the websites referenced below.
 
http://p3server.pa.msu.edu/coursewiki/doku.php?id=183_notes:energy_sep


https://cnx.org/contents/1Q9uMg_a@6.4:V7Fr-AEP@3/103-Relating-Angular-and-Trans
*http://p3server.pa.msu.edu/coursewiki/doku.php?id=183_notes:energy_sep


https://en.wikipedia.org/wiki/Gaspard-Gustave_de_Coriolis
*https://cnx.org/contents/1Q9uMg_a@6.4:V7Fr-AEP@3/103-Relating-Angular-and-Trans


*https://en.wikipedia.org/wiki/Gaspard-Gustave_de_Coriolis


[[Category:Energy]]
*https://newton.ph.msstate.edu/fox/public_html/ph2213/examples10-core.pdf

Latest revision as of 12:21, 18 April 2022

GABRIEL ALMEIDA SPRING 2022

Main Idea

In many cases, analyzing the kinetic energy of an object is in fact more difficult than just applying the formula [math]\displaystyle{ 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.

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.

Here is a link to a video which explains kinetic energy in detail: [3]

Mathematical Model

Total Kinetic Energy

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.

[math]\displaystyle{ K_{total} = K_{translational} + K_{relative} }[/math]

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

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

Translational Kinetic Energy

"Translation" means:

To move from one location to another location

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]\displaystyle{ r_{CM} = \cfrac{m_1r_1 + m_2r_2+m_3r_3 + ...}{m_1 + m_2 +m_3} }[/math]
[math]\displaystyle{ v_{CM} = \cfrac{m_1v_1 + m_2v_2+m_3v_3 + ...}{m_1+ m_2 +m_3} }[/math]
Here is a link to a video if you want to refresh your knowledge on center of mass: [4]

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]\displaystyle{ K_{translational} = \cfrac{1}{2}M_{total}v_{CM}^2 }[/math]

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.

[math]\displaystyle{ E_{vibrational} = K_{vibrational} + U_{s} }[/math]

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:

[math]\displaystyle{ E_{total} = K_{trans} + K_{vibrational} + U_{s} +E_{rest} }[/math]
[math]\displaystyle{ K_{vibrational} = E_{total} - (K_{trans} + U_{s} +E_{rest}) }[/math]

Rotational Kinetic Energy

Here are links to two videos that cover rotational kinetic energy and moment of interia: [1][2]

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]\displaystyle{ K_{rotational} =\frac{1}{2} I_{cm}{\omega}^2 }[/math]

Another important rotational equation is:

[math]\displaystyle{ }[/math]
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:

[math]\displaystyle{ I = m_1{r_{1}}^2 + m_2{r_{2}}^2 + m_3{r_{3}}^2 +... }[/math]
Here [math]\displaystyle{ r_1, r_2, r_3 }[/math] represent the perpendicular distance from the point/axis of rotation.
or
[math]\displaystyle{ I = \sum_{i} m_{i}{r_{i}}^2 }[/math]
For a body with a uniform distribution of mass this can be turned into an integral:
[math]\displaystyle{ I = \int r^2 \ dm }[/math]

The units of rotational inertia are [math]\displaystyle{ kg \cdot m^2 }[/math]

Angular Speed and Acceleration

The angular speed is the rate at which the object is rotating. It is given in the following formula:

[math]\displaystyle{ \omega = \cfrac{2\pi}{T} }[/math], where
[math]\displaystyle{ T = }[/math] the period of the rotation

The angle in which a disk turns is [math]\displaystyle{ 2 \pi }[/math] in a time [math]\displaystyle{ 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:

[math]\displaystyle{ v(r)= \omega r }[/math]

The angular acceleration a rotating object goes through to change its angular speed is given by:

[math]\displaystyle{ a(r) = \alpha r }[/math]

Computational Model

Here is a rotating rod computational model example:

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

Examples

Simple

A player throws a mid-court pass horizontally with a [math]\displaystyle{ 624g }[/math] basketball. This pass covers [math]\displaystyle{ 15 \ m }[/math] in [math]\displaystyle{ 2 \ s }[/math].

a) What is the basketball’s average translational kinetic energy while in flight?
Since the ball is moving relative to the gym, we can describe its average velocity, and thus translational kinetic energy as:
[math]\displaystyle{ v_{avg} = \frac{d}{t} = \frac{15}{2} = 7.5 \ \frac{m}{s} }[/math]
[math]\displaystyle{ K_{avg_{b}} = \frac{1}{2}m{v_{avg}}^2 = \frac{1}{2} \times 0.624 \times (7.5)^2 = 17.55 \ J }[/math]

An average molecule of air in the basketball, has a mass of [math]\displaystyle{ 29 \ u }[/math], and an average speed of [math]\displaystyle{ 500 \ \frac{m}{s} }[/math], relative to the basketball. There are about [math]\displaystyle{ 3 \times 10^{23} }[/math] molecules inside it, moving in random directions, when the ball is properly inflated.

b) What is the average translational kinetic energy of the random motion of all the molecules inside, relative to the basketball?
Since the average speed of a molecule is [math]\displaystyle{ 500 \ \frac{m}{s} }[/math], we can calculate the average translational kinetic energy of a given molecule:
[math]\displaystyle{ K_{avg_{g}} = \frac{1}{2}m_{g}{v_{avg_{g}}}^2 }[/math]
The mass of a molecule in kilograms [math]\displaystyle{ (m_{g}) }[/math] will be the atomic mass [math]\displaystyle{ (m_{A}) }[/math] times a conversion factor [math]\displaystyle{ (A) }[/math]:
[math]\displaystyle{ m_{g} = m_{A}A = 29 \times 1.66 \times 10^{-27} = 4.814 \times 10^{-26} \ kg }[/math]
Thus, the translational kinetic energy of an average molecule relative to the ball is:
[math]\displaystyle{ 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]
Multiplying this kinetic energy by the number of molecules will give us our answer:
[math]\displaystyle{ 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]\displaystyle{ 1,805.25 }[/math] Joules.
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)?
The basketball would have to have a translational kinetic energy equal to [math]\displaystyle{ 1,805.25 }[/math] Joules. To do this the ball's speed would have to satisfy:
[math]\displaystyle{ K_{ball} = \frac{1}{2}m_{ball}v^2 }[/math]
Solving for [math]\displaystyle{ v }[/math] gives:
[math]\displaystyle{ v = \sqrt{\frac{2K_{ball}}{m_{ball}}} = \sqrt{\frac{2 \times 1,805.25}{0.624}} = 76.07 \ \frac{m}{s} }[/math]

Middling

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]\displaystyle{ 5 \ kg }[/math] and a radius of [math]\displaystyle{ 10 \ cm }[/math], and an outer ring with a mass of [math]\displaystyle{ 2 \ kg }[/math] and a radius of [math]\displaystyle{ 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?
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]\displaystyle{ f_{0} = 4 \ s^{-1} \quad \And \quad f_{f} = 7 \ s^{-1} }[/math]
The frequency of a rotation is related to its period by:
[math]\displaystyle{ T = \frac{1}{f} }[/math]
Therefore, we have the initial and final period of rotation:
[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ \Delta E_{wheel} = W_{on wheel} }[/math]
[math]\displaystyle{ \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]\displaystyle{ \Delta K_{rotational_{wheel}} = W_{wheel} }[/math]
The rotational kinetic energy of the wheel is given by:
[math]\displaystyle{ K_{rotational} = \frac{1}{2} I_{wheel} \omega_{wheel} }[/math]
Thus, the change in rotational kinetic energy of the wheel will be:
[math]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ I = (0.5)*(240)*(0.5)^2 = 30 Kg*m^2 }[/math]

Energy consumption in a day:

[math]\displaystyle{ E{k} = 1000 * 24* 3600 = 8.64 * 10^7 J }[/math]

[math]\displaystyle{ E = (1 / 2)Iw^2 }[/math]

Energy = angular times inertia:

[math]\displaystyle{ w = √(2E/I) }[/math] [math]\displaystyle{ w = √(2*8.64*10^7/30) = 2400 rad/sec }[/math] [math]\displaystyle{ 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.

2. How is it connected to your major?

  • 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.

3. Is there an interesting industrial application?

  • 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.

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]\displaystyle{ 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

Further Reading

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=39523"