Translational, Rotational and Vibrational Energy

Claimed by Ferguson Beardsley Fall 2017

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.

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: [1]

Mathematical Method

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 of the center of mass. [2]

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

Calculating 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 to 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][4]

Here is a link to video if you want to refresh knowledge one center of mass: [5]

The motion of canter of mass is described by the velocity of center of mass. Using the total mass and the velocity of the center of mass, we :

[math]\displaystyle{ K_{translational} = \cfrac{1}{2}M_{total}v_{CM}^2 }[/math][6]

Calculating Vibrational Kinetic Energy

The total energy due to vibrations is the sum of the potential energy associated with interaction 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]

Calculating Rotational Kinetic Energy

Rotational kinetic energy is the energy due to the rotation about center of mass and can be calculated by finding the angular momentum and inertia, which be discussed in greater detail in the next two sections. The equation used to find kinetic rotational energy is below:

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

Here are links to two videos that cover rotational kinetic energy and moment of interim: [7][8]

Calculating Moment of Inertia

The moment of inertia of an object shows the difficulty of rotating an object, since the larger moment of inter 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:

[math]\displaystyle{ I = m_1r_{\perp,1}^2 + m_2r_2{\perp,2}^2 + m_3r_{\perp,3}^2 +... }[/math] Units: [math]\displaystyle{ kg.m^2 }[/math]

Here [math]\displaystyle{ 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

File:Https://cnx.org/resources/4c906c92cebe30d9486deb2a682acf561d23c9c1

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.

Angular Speed

The angular speed is that rate at which the object is rotating. It is given in the following formula: [math]\displaystyle{ \omega = \cfrac{2\pi}{T} }[/math]

The angle which the disk turns is 2 pi in a time T. It is measured in units in radians per second. The tangential velocity of an object is related to some distance r and 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)=wr }[/math]

File:Users/fergie/Desktop/angularvelocity.pngample.jpg

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]\displaystyle{ \triangle E_{system} = W }[/math] [math]\displaystyle{ W = F_{net} . \triangle r_{CM} }[/math] [math]\displaystyle{ \triangle E_{system} = \triangle K_{translational} }[/math] [math]\displaystyle{ \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]\displaystyle{ \triangle E_{system} = W }[/math] [math]\displaystyle{ W = F_1r_1 + F_2r_2 + F_3r_3 +... +F_nr_n }[/math] [math]\displaystyle{ \triangle K_{translational} + \triangle K_{rotational} + \triangle K_{vibrational} = W }[/math]

Here are some videos that go into more depth about point particle vs real system: [9]

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.

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.

Examples

Simple

Problem statement:

Calculate the rotational kinetic energy of a wheel of radius 100cm, mass 10kg, with and angular velocity of 22 radians/s.

Solution:

We know how to calculate the moment of inertia for a disk, and the moment of inertia for a wheel will be the same since all the atoms are at the same distance from the center. Therefore, [math]\displaystyle{ I = MR^2 = (10)(1)^2 = 10 kgm^2 }[/math]

From there, we can easily calculate the rotational kinetic energy: [math]\displaystyle{ K_{rotational} = \cfrac{1}{2}I\omega^2 = \cfrac{1}{2}(10)(22)^2 = 2420 }[/math] Joules

Middling

Problem statement:

A string is wrapped around a disk of radius 0.15m and mass 3kg. The disk is initially at rest, but you pull the string with a force of 10N along a smooth surface. The disk moves a distance d = 0.1m and your hand pulls through a distance h = 0.2m. What is the speed of the center of mass of the disk after having pulled the string?

Solution:

The problem states that the disk is moving on a smooth surface, so there is no friction here. Since the problem asks about the speed of the center of mass, we will consider the point particle system first. We start with:

[math]\displaystyle{ \triangle K_{translational} = F_{net} . \triangle r_{CM} }[/math]

We know that the translational kinetic energy is [math]\displaystyle{ K = \cfrac{1}{2}mv_{CM}^2 }[/math]

So we end up with:

[math]\displaystyle{ \cfrac{1}{2}mv_{CM,f}^2 - \cfrac{1}{2}mv_{CM,i}^2 = F_{net} . \triangle r_{CM} }[/math]

Because initially the disk is still, we can reduce this equation to:

[math]\displaystyle{ \cfrac{1}{2}mv_{CM,f}^2 = F_{net} . \triangle r_{CM} }[/math]

[math]\displaystyle{ v_{CM,f}^2 = \cfrac{2. F_{net} . \triangle r_{CM}}{m} }[/math]

[math]\displaystyle{ v_{CM,f} = \sqrt{\cfrac{2. F_{net} . \triangle r_{CM}}{m}} }[/math]

[math]\displaystyle{ v_{CM,f} = \sqrt{\cfrac{ (2)(10)(0.1)}{3}} }[/math]

[math]\displaystyle{ v_{CM,f} = 0.816 }[/math] m/s

Difficult

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" 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

a. 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

"Solution" From the Energy Principle ( when dealing with a point particle it only has Ktrans):

ΔKtrans= ∫F⃗ net⋅dr⃗ cm

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:

ΔKtrans=(F−mg)ΔyCM

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

ΔyCM=−h (Substitute in −h for yCM)

ΔKtrans=(F−mg)(−h)

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

ΔKtrans=(mg−F)h

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

"Solution" 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

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.

ΔKtrans+ΔKrot=Whand+WEarth

Substitute in the work represented by force by distance for both the hand and the Earth.

ΔKtrans+ΔKrot=Fd+(−mg)(−h)

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

ΔKtrans=(mg−F)h

Substituting this into our equation leaves us with:

(mg−F)h+ΔKrot=Fd+mgh

Solve for change in rotational kinetic energy:

ΔKrot=F(d+h)

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 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? 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 rational, translational, and vibrational energy to power anything from phones to computers.

History

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. The term was later coined by William Thomson.

References

All images used on this page do not belong to me. 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