Theory

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Energy Units

Definition

Las unidades de la energía se han nombrado en honor a James Joule que descubrió la equivalencia entre energía térmica y mecánica. La unidad es igual a\\n\\n

$J=\displaystyle\frac{kg,m^2}{s^2}$

ID:(637, 0)



Kinetic Energy

Image

ID:(1137, 0)



Potential Energy

Note

ID:(1139, 0)



Total Energy

Quote

ID:(1140, 0)



Energy necessary for Walking

Exercise

ID:(58, 0)



Theory

Storyboard

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$\omega$
omega
Angular Speed
rad/s
$\vec{F}$
&F
Force
N
$m_i$
m_i
Inertial Mass
kg
$K_i$
K_i
Kinetic energy 1
J
$K_r$
K_r
Kinetic energy of rotation
J
$I$
I
Moment of Inertia
kg m^2
$\Delta\vec{s}$
&Ds
Path traveled (vector)
m
$\vec{s}$
&s
Posición (vector)
m
$V$
V
Potential Energy
J
$U_i$
U_i
Potential energy 1
J
$v$
v
Speed
m/s
$K$
K
Total Kinetic Energy
J
$K_t$
K_t
Translational Kinetic Energy
J
$W$
W
Work
J
$\Delta W$
DW
Work fraction
J

Calculations


First, select the equation:   to ,  then, select the variable:   to 

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

 Variable   Given   Calculate   Target :   Equation   To be used



Equations

The energy required for an object to change its angular velocity from $\omega_1$ to $\omega_2$ can be calculated using the definition

equation=12550

Applying Newton's second law, this expression can be rewritten as

$\Delta W=I \alpha \Delta\theta=I\displaystyle\frac{\Delta\omega}{\Delta t}\Delta\theta$



Using the definition of angular velocity

equation=3679

we get

$\Delta W=I\displaystyle\frac{\Delta\omega}{\Delta t}\Delta\theta=I \omega \Delta\omega$



The difference in angular velocities is

$\Delta\omega=\omega_2-\omega_1$



On the other hand, angular velocity itself can be approximated with the average angular velocity

$\omega=\displaystyle\frac{\omega_1+\omega_2}{2}$



Using both expressions, we obtain the equation

$\Delta W=I \omega \Delta \omega=I(\omega_2-\omega_1)\displaystyle\frac{(\omega_1+\omega_2)}{2}=\displaystyle\frac{I}{2}(\omega_2^2-\omega_1^2)$



Thus, the change in energy is given by

$\Delta W=\displaystyle\frac{I}{2}\omega_2^2-\displaystyle\frac{I}{2}\omega_1^2$



This allows us to define kinetic energy as

equation

The work variance ($\Delta W$) required for an object to change from the initial Angular Speed ($\omega_0$) to the angular Speed ($\omega$) is obtained by applying a the torque ($T$) that produces an angular displacement the difference of Angles ($\Delta\theta$), according to:

equation=12550

Applying Newton's second law for rotation, in terms of the moment of inertia for axis that does not pass through the CM ($I$) and the mean Angular Acceleration ($\bar{\alpha}$):

equation=3253

this expression can be rewritten as:

$\Delta W = I \alpha \Delta\theta$



or, using the difference in Angular Speeds ($\Delta\omega$) and the time elapsed ($\Delta t$):

equation=3234

we get:

$\Delta W = I\displaystyle\frac{\Delta\omega}{\Delta t} \Delta\theta$



Using the definition of the mean angular velocity ($\bar{\omega}$) and the time elapsed ($\Delta t$):

equation=3679

results in:

$\Delta W = I\displaystyle\frac{\Delta\omega}{\Delta t} \Delta\theta = I\omega \Delta\omega$



where the difference in Angular Speeds ($\Delta\omega$) is expressed as:

equation=3681

On the other hand, the angular velocity can be approximated by the average angular velocity:

$\bar{\omega}=\displaystyle\frac{\omega_1 + \oemga_2}{2}$



By combining both expressions, we obtain the equation:

$\Delta W = I \omega \Delta\omega = I(\omega_2 - \omega_1) \displaystyle\frac{(\omega_1 + \omega_2)}{2} = \displaystyle\frac{I}{2}(\omega_2^2 - \omega_1^2)$



Therefore, the change in energy is expressed as:

$\Delta W = \displaystyle\frac{I}{2}\omega_2^2 - \displaystyle\frac{I}{2}\omega_1^2$



This allows us to define the rotational kinetic energy as:

equation

Expanding the concept to a longer path involves summing up the energy required for each path element:

$\bar{W}=\displaystyle\sum_i \vec{F}_i\cdot\Delta\vec{s}_i$



However, the value of this equation represents just an average energy required or generated. The precise energy is obtained when the steps become very small, allowing the force to be considered constant within them:

$W=\displaystyle\sum_i \mbox{lim}_{\Delta\vec{s}_i\rightarrow\vec{0}}\vec{F}_i\cdot\Delta\vec{s}_i$



In this limit, the energy corresponds to the integral along the traveled path, giving us:

equation


Examples

The concept of energy was initially introduced in thermodynamics with the purpose of quantifying the amount of heat that could be converted into mechanical work. In a representative experiment, friction was generated by sliding a surface against a cable subjected to a force. This cable traveled a the distance traveled in a time ($\Delta s$) which, when multiplied by the applied force the force with constant mass ($F$), resulted in the generated work the work variance ($\Delta W$):

equation=3202

Since both the force with constant mass ($F$) and the distance traveled in a time ($\Delta s$) are actually vectors, this expression can be generalized using the scalar product between the force ($\vec{F}$) and the path traveled (vector) ($\Delta\vec{s}$), yielding the work fraction ($\Delta W$):

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In other words, only the component of the force that acts in the direction of the displacement effectively contributes to the energy transfer.

Las unidades de la energ a se han nombrado en honor a James Joule que descubri la equivalencia entre energ a t rmica y mec nica. La unidad es igual a\\n\\n

$J=\displaystyle\frac{kg,m^2}{s^2}$

Carnot was the first to describe energy in terms of the path and the necessary force to traverse it. Progressing along a path with a force requires or generates energy. This corresponds to the equation:

equation=1136

In the continuous limit, the sum can be represented as an integral:

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The translational Kinetic Energy ($K_t$) is determined based on the speed ($v$) and the inertial Mass ($m_i$), according to:

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5288 is associated with 6290 and not with 8762, even though they are numerically equal. The energy that an object possesses is a direct consequence of the inertia that had to be overcome to set it in motion.

The kinetic energy of rotation ($K_r$) is a function of the angular Speed ($\omega$) and of a measure of inertia represented by the moment of inertia for axis that does not pass through the CM ($I$):

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The total Kinetic Energy ($K$) can have translational and/or rotational components. Therefore, it is expressed as the sum of the translational Kinetic Energy ($K_t$) and the kinetic energy of rotation ($K_r$):

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In a more complex system, the total kinetic energy is equal to the sum of the kinetic energies of the individual parts

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In a more complex system, the total potential energy is equal to the sum of the potential energies of the individual parts

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