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Rotational kinetic energy
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The kinetic energy of rotation is a function of the angular velocity achieved through the application of torque over a given time while traversing a given angle.
Thus, rotational kinetic energy is proportional to the moment of inertia of the object and the square of the angular velocity.
ID:(2080, 0)
Flow into a channel
Definition
In the case of flow towards a channel, the system can be modeled in a one-dimensional manner, where the height of the water column on the ground ($h$) is a function of the position of the water column on the ground ($x$) representing the flux density ($j_s$), and it satisfies the condition
| $ h j_s = h_0 j_{s0} $ |
with the flow at a reference point ($j_{s0}$) and the reference height of the water column ($h_0$) defining the water profile in the soil:
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The key to this equation is that the product of the height of the water column on the ground ($h$) and the flux density ($j_s$) must always remain constant. In this sense, if the height of the water column on the ground ($h$) increases, the flux density ($j_s$) decreases, and vice versa. Moreover, the sign remains the same; hence, flow towards the channel, i.e., negative flow, will occur only when the groundwater level is higher than that of the channel. As the liquid approaches the channel, the groundwater level decreases, leading to an increase in flow density.
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Flow height solution towards a channel
Image
The solution to the one-dimensional flow equation towards a channel, where the height of the water column on the ground ($h$) is calculated as a function of the reference height of the water column ($h_0$) and the position of the water column on the ground ($x$) at the channel's edge, along with the characteristic length of the flow in the ground ($s_0$), takes the following form:
| $ \displaystyle\frac{ h }{ h_0 } = \sqrt{1 + \displaystyle\frac{ 2 x }{ s_0 }} $ |
This solution is graphically represented in terms of the additional factors $h/h_0$ and $x/s_0$ as follows:
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The profile reveals that, away from the channel, the height of the water column is significantly high. However, due to water extraction by the channel, this height begins to decrease until it reaches the channel's edge. Dynamically, the flux density ($j_s$) determines the amount of water flowing into the channel, while the reference height of the water column ($h_0$) gradually adjusts until it reaches an equilibrium state. In other words, if the value of the reference height of the water column ($h_0$) is too low relative to the total amount of water arriving at the channel, it increases; and if it is too high, it decreases. In this way, the reference height of the water column ($h_0$) acquires the value that balances the amount of incoming water with the amount of water flowing through the channel.
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Flux density solution towards a channel
Note
The solution obtained for the height and the parameters the flow at a reference point ($j_{s0}$) and the reference height of the water column ($h_0$) reveals that the flux density ($j_s$) is given by:
| $ \displaystyle\frac{ j_s }{ j_{s0} } = \displaystyle\frac{1}{\sqrt{1 + \displaystyle\frac{ 2 x }{ s_0 }}} $ |
We can graphically represent the flux density ($j_s$) as a function of the additional factors $j_s/j_{s0}$ and $x/s_0$ as follows:
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It is noticeable that the flux density ($j_s$) continues to increase as we approach the channel, as the height of the water column on the ground ($h$) decreases. This increase is necessary to maintain the flow velocity in the flux density ($j_s$) or, alternatively, to increase it.
ID:(15110, 0)
Rotational kinetic energy
Description
In the case of the rotation of a body, it is possibleanalogously to translational motionto define a form of kinetic energy, but this time associated with rotation. Just as translational kinetic energy depends on mass and linear velocity, in rotational motion there is an equivalent role associated with inertia. Although this depends on mass, its value is also determined by the distance of that mass from the axis of rotation. This leads to the concept of the moment of inertia, which plays a role analogous to inertial mass. The moment of inertia is multiplied by the square of the angular velocity, and dividing that product by two yields the rotational kinetic energy.
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The definition of average angular acceleration is based on the angle covered
| $ \Delta\omega = \omega_2 - \omega_1 $ |
and the elapsed time
| $ \Delta t \equiv t - t_0 $ |
The relationship between the two is defined as the average angular acceleration
| $ \bar{\alpha} \equiv \displaystyle\frac{ \Delta\omega }{ \Delta t }$ |
within that time interval.
(ID 3234)
Since the moment is equal to
| $ L = I \omega $ |
it follows that in the case where the moment of inertia doesn't change with time,
$T=\displaystyle\frac{dL}{dt}=\displaystyle\frac{d}{dt}(I\omega) = I\displaystyle\frac{d\omega}{dt} = I\alpha$
which implies that
| $ T = I \alpha $ |
.
(ID 3253)
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:
| $ \Delta W = T \Delta\theta $ |
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 ($\bar{\alpha}$):
| $ T = I \bar{\alpha} $ |
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 ($\Delta t$):
| $ \bar{\alpha} \equiv \displaystyle\frac{ \Delta\omega }{ \Delta t }$ |
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 ($\Delta t$):
| $ \bar{\omega} \equiv\displaystyle\frac{ \Delta\theta }{ \Delta t }$ |
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:
| $ \Delta\omega = \omega_2 - \omega_1 $ |
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:
| $ K_r =\displaystyle\frac{1}{2} I \omega ^2$ |
(ID 3255)
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:
| $ \Delta W = T \Delta\theta $ |
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 ($\bar{\alpha}$):
| $ T = I \bar{\alpha} $ |
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 ($\Delta t$):
| $ \bar{\alpha} \equiv \displaystyle\frac{ \Delta\omega }{ \Delta t }$ |
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 ($\Delta t$):
| $ \bar{\omega} \equiv\displaystyle\frac{ \Delta\theta }{ \Delta t }$ |
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:
| $ \Delta\omega = \omega_2 - \omega_1 $ |
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:
| $ K_r =\displaystyle\frac{1}{2} I \omega ^2$ |
(ID 3255)
(ID 4440)
(ID 10301)
Using the classical definition of energy, with the work variance ($\Delta W$) calculated from the force with constant mass ($F$) and the distance traveled in a time ($\Delta s$) according to:
| $ \Delta W = F \Delta s $ |
in the case of rotation, the force is perpendicular to the radius ($r$), generating the torque ($T$), which is expressed as:
| $ T = r F $ |
Since the arc length is equal to the radius ($r$) multiplied by the difference of Angles ($\Delta\theta$):
| $ \Delta s=r \Delta\theta $ |
we obtain:
$\Delta W = \vec{F}\cdot\Delta\vec{s} = F\Delta s = F r\Delta\theta = T\Delta\theta$
That is:
| $ \Delta W = T \Delta\theta $ |
(ID 12550)
Examples
(ID 15604)
(ID 15606)
The work variance ($\Delta W$), in the case of rotation, can be calculated from the rotation generator, which is the torque ($T$) multiplied by the angle variation ($\Delta\theta$):
| $ \Delta W = T \Delta\theta $ |
(ID 12550)
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 ($I$):
| $ K_r =\displaystyle\frac{1}{2} I \omega ^2$ |
(ID 3255)
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 ($I$):
| $ K_r =\displaystyle\frac{1}{2} I \omega ^2$ |
(ID 3255)
La variaci n del trabajo en el tiempo se denomina la potencia. Por lo general es una limitante ya que indica la velocidad que es un sistema capaz de crear/absorber energ a.
| $ \Delta W = W - W_0 $ |
(ID 4440)
The rate at which angular velocity changes over time is defined as the mean Angular Acceleration ($\bar{\alpha}$). To measure it, we need to observe the difference in Angular Speeds ($\Delta\omega$) and the time elapsed ($\Delta t$).
The equation describing the mean Angular Acceleration ($\bar{\alpha}$) is as follows:
| $ \bar{\alpha} \equiv \displaystyle\frac{ \Delta\omega }{ \Delta t }$ |
(ID 3234)
Acceleration is defined as the change in angular velocity per unit of time.
Therefore, the angular acceleration the difference in Angular Speeds ($\Delta\omega$) can be expressed in terms of the angular velocity the angular Speed ($\omega$) and time the initial Angular Speed ($\omega_0$) as follows:
| $ \Delta\omega = \omega_2 - \omega_1 $ |
(ID 3681)
To describe the rotation of an object, we need to determine the angle variation ($\Delta\theta$). This is achieved by subtracting the initial Angle ($\theta_0$) from the angle ($\theta$), which is reached by the object during its rotation:
| $ \Delta\theta = \theta_2 - \theta_1 $ |
(ID 3680)
To describe the motion of an object, it is necessary to calculate the elapsed time. This magnitude is obtained by measuring the start Time ($t_0$) and the time ($t$) of the object's motion. The duration is determined by subtracting the final time from the initial time.
When the time ($t$) is very similar to the start Time ($t_0$), the elapsed time is considered infinitesimal and is denoted as the infinitesimal Variation of Time ($dt$):
| $dt \equiv t - t_0 $ |
(ID 10301)
In the case where the moment of inertia is constant, the derivative of angular momentum is equal to
| $ L = I \omega $ |
which implies that the torque is equal to
| $ T = I \alpha $ |
This relationship is the equivalent of Newton's second law for rotation instead of translation.
(ID 3253)
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