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Torque with constant moment of inertia
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If it is desired to modify the rotational state of the body, the angular momentum must be modified.
The speed with which this occurs is called torque defined as the variation of the angular momentum in time and is vectorial since the variation of the angular momentum is. This was defined by Newton in his second principle in the case of rotation.
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Equations
$ a = r \alpha_0 $
a = r * alpha
$ \bar{\alpha} = \alpha_0 $
alpha_m = alpha_0
$ \bar{\alpha} \equiv \displaystyle\frac{ \Delta\omega }{ \Delta t }$
alpha_m = Domega / Dt
$ \Delta L = L - L_0 $
DL = L - L_0
$ \Delta\omega = \omega - \omega_0 $
Domega = omega - omega_0
$ \Delta p = p - p_0 $
Dp = p - p_0
$ \Delta t \equiv t - t_0 $
Dt = t - t_0
$ L = I \omega $
L = I * omega
$ L = r p $
L = r * p
$ \omega = \omega_0 + \alpha_0 ( t - t_0 )$
omega = omega_0 + alpha_0 * ( t - t_0 )
$ T = I \alpha_0 $
T = I * alpha
$ \theta = \theta_0 + \omega_0 ( t - t_0 )+\displaystyle\frac{1}{2} \alpha_0 ( t - t_0 )^2$
theta = theta_0 + omega_0 *( t - t_0 )+ alpha_0 *( t - t_0 )^2/2
$ \theta = \theta_0 +\displaystyle\frac{ \omega ^2- \omega_0 ^2}{2 \alpha_0 }$
theta = theta_0 +( omega ^2 - omega_0 ^2)/(2* alpha_0 )
$ T_m =\displaystyle\frac{ \Delta L }{ \Delta t }$
T_m = DL / Dt
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Angular Momentum
Equation
The moment ($p$) was defined as the product of the inertial Mass ($m_i$) and the speed ($v$), which is equal to:
| $ p = m_i v $ |
The analogue of the speed ($v$) in the case of rotation is the instantaneous Angular Speed ($\omega$), therefore, the equivalent of the moment ($p$) should be a the angular Momentum ($L$) of the form:
| $ L = I \omega $ |
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the inertial Mass ($m_i$) is associated with the inertia in the translation of a body, so the moment of Inertia ($I$) corresponds to the inertia in the rotation of a body.
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Angular momentum and moment relationship
Equation
Similar to the relationship that exists between linear velocity and angular velocity, represented by the equation:
| $ v = r \omega $ |
we can establish a relationship between angular momentum and translational momentum. However, in this instance, the multiplying factor is not the radius, but rather the moment. The relationship is expressed as:
| $ L = r p $ |
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Medium Torque
Equation
In the case of translation, the second principle defines how translational motion is generated with the definition of force
| $ F \equiv\displaystyle\frac{ \Delta p }{ \Delta t }$ |
In the case of rotation, within a time interval $\Delta t$, the angular momentum $\Delta L$ changes according to:
| $ T_m =\displaystyle\frac{ \Delta L }{ \Delta t }$ |
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Torque for constant moment of inertia
Equation
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_0 $ |
| $ T = I \alpha $ |
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 $ |
.
This relationship is the equivalent of Newton's second law for rotation instead of translation.
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Variation of angular momentum
Equation
Similar to the case of translation, where the third principle states that every action has an equal and opposite reaction:
| $ \Delta p = p - p_0 $ |
The analogous concept in rotation is
| $ \Delta L = L - L_0 $ |
.
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Angular velocity with constant angular acceleration
Equation
With the constant Angular Acceleration ($\alpha_0$), the angular Speed ($\omega$) forms a linear relationship with the time ($t$), incorporating the variables the initial Angular Speed ($\omega_0$) and the start Time ($t_0$) as follows:
| $ \omega = \omega_0 + \alpha_0 ( t - t_0 )$ |
If we assume that the mean Angular Acceleration ($\bar{\alpha}$) is constant, equivalent to the constant Angular Acceleration ($\alpha_0$), then the following equation applies:
| $ \bar{\alpha} = \alpha_0 $ |
Therefore, considering the difference in Angular Speeds ($\Delta\omega$) along with the angular Speed ($\omega$) and the initial Angular Speed ($\omega_0$):
| $ \Delta\omega = \omega - \omega_0 $ |
and the time elapsed ($\Delta t$) in relation to the time ($t$) and the start Time ($t_0$):
| $ \Delta t \equiv t - t_0 $ |
the equation for the mean Angular Acceleration ($\bar{\alpha}$):
| $ \bar{\alpha} \equiv \displaystyle\frac{ \Delta\omega }{ \Delta t }$ |
can be expressed as:
$\alpha_0 = \alpha = \displaystyle\frac{\Delta \omega}{\Delta t} = \displaystyle\frac{\omega - \omega_0}{t - t_0}$
Solving this, we obtain:
| $ \omega = \omega_0 + \alpha_0 ( t - t_0 )$ |
This equation represents a straight line in the angular velocity versus time plane.
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Variation of angular speeds
Equation
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 - \omega_0 $ |
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Elapsed time
Equation
To describe the motion of an object, we need to calculate the time elapsed ($\Delta t$). This magnitude is obtained by measuring the start Time ($t_0$) and the the time ($t$) of said motion. The duration is determined by subtracting the initial time from the final time:
| $ \Delta t \equiv t - t_0 $ |
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Momentum Difference
Equation
According to Galileo, objects tend to maintain their state of motion, meaning that the momentum
$\vec{p} = m\vec{v}$
should remain constant. If there is any action on the system that affects its motion, it will be associated with a change in momentum. The difference between the initial momentum $\vec{p}_0$ and the final momentum $\vec{p}$ can be expressed as:
| $ \Delta p = p - p_0 $ |
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Angle at Constant Angular Acceleration
Equation
Given that the total displacement corresponds to the area under the angular velocity versus time curve, in the case of a constant Angular Acceleration ($\alpha_0$), it is determined that the displacement the angle ($\theta$) with the variables the initial Angle ($\theta_0$), the time ($t$), the start Time ($t_0$), and the initial Angular Speed ($\omega_0$) is as follows:
| $ \theta = \theta_0 + \omega_0 ( t - t_0 )+\displaystyle\frac{1}{2} \alpha_0 ( t - t_0 )^2$ |
In the case of the constant Angular Acceleration ($\alpha_0$), the angular Speed ($\omega$) as a function of the time ($t$) follows a linear relationship with the start Time ($t_0$) and the initial Angular Speed ($\omega_0$) in the form of:
| $ \omega = \omega_0 + \alpha_0 ( t - t_0 )$ |
Given that the angular displacement is equal to the area under the angular velocity-time curve, in this case, one can add the contributions of the rectangle:
$\omega_0(t-t_0)$
and the triangle:
$\displaystyle\frac{1}{2}\alpha_0(t-t_0)^2$
This leads us to the expression for the angle ($\theta$) and the initial Angle ($\theta_0$):
| $ \theta = \theta_0 + \omega_0 ( t - t_0 )+\displaystyle\frac{1}{2} \alpha_0 ( t - t_0 )^2$ |
This expression corresponds to the general form of a parabola.
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Braking angle as a function of angular velocity
Equation
In the case of the constant Angular Acceleration ($\alpha_0$), the function of the angular Speed ($\omega$) with respect to the time ($t$), along with additional variables the initial Angular Speed ($\omega_0$) and the start Time ($t_0$), is expressed by the equation:
| $ \omega = \omega_0 + \alpha_0 ( t - t_0 )$ |
From this equation, it is possible to calculate the relationship between the angle ($\theta$) and the initial Angle ($\theta_0$), as well as the change in angular velocity:
| $ \theta = \theta_0 +\displaystyle\frac{ \omega ^2- \omega_0 ^2}{2 \alpha_0 }$ |
If we solve for time in the equation of the angular Speed ($\omega$) that includes the variables the initial Angular Speed ($\omega_0$), the time ($t$), the start Time ($t_0$), and the constant Angular Acceleration ($\alpha_0$):
| $ \omega = \omega_0 + \alpha_0 ( t - t_0 )$ |
we obtain the following expression for time:
$t - t_0 = \displaystyle\frac{\omega - \omega_0}{\alpha_0}$
This solution can be substituted into the equation to calculate the angle ($\theta$) using the initial Angle ($\theta_0$) as follows:
| $ \theta = \theta_0 + \omega_0 ( t - t_0 )+\displaystyle\frac{1}{2} \alpha_0 ( t - t_0 )^2$ |
which results in the following equation:
| $ \theta = \theta_0 +\displaystyle\frac{ \omega ^2- \omega_0 ^2}{2 \alpha_0 }$ |
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Acceleration and Angular Acceleration
Equation
If we divide the relationship between the mean Speed ($\bar{v}$), the radio ($r$), and the mean angular velocity ($\bar{\omega}$), expressed in the following equation:
| $ v = r \omega $ |
by the value of the time elapsed ($\Delta t$), we can obtain the factor that allows us to calculate the angular acceleration along the orbit:
| $ a = r \alpha_0 $ |
| $ a = r \alpha $ |
Given that the mean Acceleration ($\bar{a}$) equals the speed Diference ($\Delta v$) and the time elapsed ($\Delta t$) according to
| $ \bar{a} \equiv\displaystyle\frac{ \Delta v }{ \Delta t }$ |
and the mean Angular Acceleration ($\bar{\alpha}$) equals the difference in Angular Speeds ($\Delta\omega$) and the time elapsed ($\Delta t$) as per
| $ \bar{\alpha} \equiv \displaystyle\frac{ \Delta\omega }{ \Delta t }$ |
it follows that
$\bar{a}=\displaystyle\frac{\Delta v}{\Delta t}=r\displaystyle\frac{\Delta\omega}{\Delta t}=\bar{\alpha}$
Assuming that the mean Angular Acceleration ($\bar{\alpha}$) is equal to the constant Angular Acceleration ($\alpha_0$)
| $ \bar{\alpha} = \alpha_0 $ |
and assuming that the mean Acceleration ($\bar{a}$) equals the constant Acceleration ($a_0$)
| $ a_0 = \bar{a} $ |
then the following equation is obtained:
| $ a = r \alpha $ |
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Mean Angular Acceleration
Equation
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 }$ |
The definition of average angular acceleration is based on the angle covered
| $ \Delta\omega = \omega - \omega_0 $ |
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.
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Constant angular acceleration
Equation
If the acceleration does not vary, the mean Angular Acceleration ($\bar{\alpha}$) will be equal to the constant Angular Acceleration ($\alpha_0$), which is expressed as:
| $ \bar{\alpha} = \alpha_0 $ |
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