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Torque
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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.
ID:(599, 0)
Torque with constant moment of inertia
Description
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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(ID 1072)
(ID 1072)
(ID 1072)
As the mean Speed ($\bar{v}$) is with the distance traveled in a time ($\Delta s$) and the time elapsed ($\Delta t$), equal to
| $ \bar{v} \equiv\displaystyle\frac{ \Delta s }{ \Delta t }$ |
and with the distance traveled in a time ($\Delta s$) expressed as an arc of a circle, and the radius ($r$) and the angle variation ($\Delta\theta$) are
| $ \Delta s=r \Delta\theta $ |
and the definition of the mean angular velocity ($\bar{\omega}$) is
| $ \bar{\omega} \equiv\displaystyle\frac{ \Delta\theta }{ \Delta t }$ |
then,
$v=\displaystyle\frac{\Delta s}{\Delta t}=r\displaystyle\frac{\Delta\theta}{\Delta t}=r\omega$
Since the relationship is general, it can be applied for instantaneous values, resulting in
| $ v = r \omega $ |
.
(ID 3233)
As the mean Speed ($\bar{v}$) is with the distance traveled in a time ($\Delta s$) and the time elapsed ($\Delta t$), equal to
| $ \bar{v} \equiv\displaystyle\frac{ \Delta s }{ \Delta t }$ |
and with the distance traveled in a time ($\Delta s$) expressed as an arc of a circle, and the radius ($r$) and the angle variation ($\Delta\theta$) are
| $ \Delta s=r \Delta\theta $ |
and the definition of the mean angular velocity ($\bar{\omega}$) is
| $ \bar{\omega} \equiv\displaystyle\frac{ \Delta\theta }{ \Delta t }$ |
then,
$v=\displaystyle\frac{\Delta s}{\Delta t}=r\displaystyle\frac{\Delta\theta}{\Delta t}=r\omega$
Since the relationship is general, it can be applied for instantaneous values, resulting in
| $ v = r \omega $ |
.
(ID 3233)
As the mean Speed ($\bar{v}$) is with the distance traveled in a time ($\Delta s$) and the time elapsed ($\Delta t$), equal to
| $ \bar{v} \equiv\displaystyle\frac{ \Delta s }{ \Delta t }$ |
and with the distance traveled in a time ($\Delta s$) expressed as an arc of a circle, and the radius ($r$) and the angle variation ($\Delta\theta$) are
| $ \Delta s=r \Delta\theta $ |
and the definition of the mean angular velocity ($\bar{\omega}$) is
| $ \bar{\omega} \equiv\displaystyle\frac{ \Delta\theta }{ \Delta t }$ |
then,
$v=\displaystyle\frac{\Delta s}{\Delta t}=r\displaystyle\frac{\Delta\theta}{\Delta t}=r\omega$
Since the relationship is general, it can be applied for instantaneous values, resulting in
| $ v = r \omega $ |
.
(ID 3233)
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)
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
| $ \alpha_0 \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 $ |
(ID 3236)
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_2 - \omega_1 $ |
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 ($\bar{\alpha}$):
| $ \alpha_0 \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 )$ |
(ID 3237)
(ID 3251)
(ID 3251)
(ID 3251)
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)
(ID 3324)
(ID 3324)
(ID 3324)
The relationship between the angular Momentum ($L$) and the moment ($p$) is expressed as:
| $ L = r p $ |
Using the radius ($r$), this expression can be equated with the moment of Inertia ($I$) and the angular Speed ($\omega$) as follows:
| $ L = I \omega $ |
Then, substituting with the inertial Mass ($m_i$) and the speed ($v$):
| $ p = m_i v $ |
and
| $ v = r \omega $ |
it can be concluded that the moment of inertia of a particle rotating in an orbit is:
| $ I = m_i r ^2$ |
(ID 3602)
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$ |
(ID 3682)
(ID 3683)
If we start from the starting position ($s_0$) and want to calculate the distance traveled in a time ($\Delta s$), we need to define a value for the position ($s$).
In a one-dimensional system, the distance traveled in a time ($\Delta s$) is simply obtained by subtracting the starting position ($s_0$) from the position ($s$), resulting in:
| $ \Delta s = s - s_0 $ |
(ID 4352)
(ID 4355)
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 }$ |
(ID 4386)
Si se deriva en el tiempo la relaci n para el momento angular
| $ L = r p $ |
para el caso de que el radio sea constante
$T=\displaystyle\frac{dL}{dt}=r\displaystyle\frac{dp}{dt}=rF$
por lo que
| $ T = r F $ |
(ID 4431)
(ID 9875)
Since the moment ($p$) is defined with the inertial Mass ($m_i$) and the speed ($v$),
| $ p = m_i v $ |
If the inertial Mass ($m_i$) is equal to the initial mass ($m_0$), then we can derive the momentum with respect to time and obtain the force with constant mass ($F$):
$F=\displaystyle\frac{d}{dt}p=m_i\displaystyle\frac{d}{dt}v=m_ia$
Therefore, we conclude that
| $ F = m_i a $ |
(ID 10975)
Examples
(ID 15527)
(ID 15530)
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 $ |
(ID 4353)
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)
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 $ |
.
(ID 1072)
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 $ |
.
(ID 1072)
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)
We can calculate the distance traveled in a time ($\Delta s$) from the starting position ($s_0$) and the position ($s$) using the following equation:
| $ \Delta s = s - s_0 $ |
(ID 4352)
Acceleration corresponds to the change in velocity per unit of time.
Therefore, it is necessary to define the speed Diference ($\Delta v$) in terms of the speed ($v$) and the initial Speed ($v_0$) as follows:
| $ dv \equiv v - v_0 $ |
(ID 4355)
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 )$ |
This equation represents a straight line in the angular velocity versus time plane.
(ID 3237)
Similar to the case of translation, where the third principle states that every action has an equal and opposite reaction:
| $ dp = p - p_0 $ |
The analogous concept in rotation is
| $ \Delta L = L - L_0 $ |
.
(ID 9875)
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:
| $ dp = p - p_0 $ |
(ID 3683)
Since the perimeter of a circle is $2\pi r$, ERROR:6294 along the circle will correspond to the arc spanned by ERROR:5059, so:
| $ s = r \theta $ |
(ID 3324)
Since the perimeter of a circle is $2\pi r$, ERROR:6294 along the circle will correspond to the arc spanned by ERROR:5059, so:
| $ s = r \theta $ |
(ID 3324)
Since the perimeter of a circle is $2\pi r$, ERROR:6294 along the circle will correspond to the arc spanned by ERROR:5059, so:
| $ s = r \theta $ |
(ID 3324)
If we divide the relationship between the distance traveled in a time ($\Delta s$) and the radius ($r$) by the angle variation ($\Delta\theta$),
| $ \Delta s=r \Delta\theta $ |
and then divide it by the time elapsed ($\Delta t$), we obtain the relationship that allows us to calculate the speed ($v$) along the orbit, known as the tangential velocity, which is associated with the angular Speed ($\omega$):
| $ v = r \omega $ |
(ID 3233)
If we divide the relationship between the distance traveled in a time ($\Delta s$) and the radius ($r$) by the angle variation ($\Delta\theta$),
| $ \Delta s=r \Delta\theta $ |
and then divide it by the time elapsed ($\Delta t$), we obtain the relationship that allows us to calculate the speed ($v$) along the orbit, known as the tangential velocity, which is associated with the angular Speed ($\omega$):
| $ v = r \omega $ |
(ID 3233)
If we divide the relationship between the distance traveled in a time ($\Delta s$) and the radius ($r$) by the angle variation ($\Delta\theta$),
| $ \Delta s=r \Delta\theta $ |
and then divide it by the time elapsed ($\Delta t$), we obtain the relationship that allows us to calculate the speed ($v$) along the orbit, known as the tangential velocity, which is associated with the angular Speed ($\omega$):
| $ v = r \omega $ |
(ID 3233)
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 $ |
(ID 3236)
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 $ |
.
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.
(ID 3251)
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 $ |
.
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.
(ID 3251)
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 $ |
.
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.
(ID 3251)
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 $ |
.
(ID 1072)
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$ |
This expression corresponds to the general form of a parabola.
(ID 3682)
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 }$ |
(ID 4386)
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)
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 }$ |
.
(ID 9876)
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)
The force ($F$) is defined as the momentum variation ($\Delta p$) by the time elapsed ($\Delta t$), which is defined by the relationship:
| $ F \equiv\displaystyle\frac{ \Delta p }{ \Delta t }$ |
(ID 3684)
Since the relationship between angular momentum and torque is
| $ L = r p $ |
its temporal derivative leads us to the torque relationship
| $ T = r F $ |
The body's rotation occurs around an axis in the direction of the torque, which passes through the center of mass.
(ID 4431)
In the case where the inertial Mass ($m_i$) equals the initial mass ($m_0$),
| $ m_g = m_i $ |
the derivative of momentum will be equal to the mass multiplied by the derivative of the speed ($v$). Since the derivative of velocity is the instant acceleration ($a$), we have that the force with constant mass ($F$) is
| $ F = m_i a $ |
(ID 10975)
For a particle of mass the point Mass ($m$) orbiting around an axis at a distance the radius ($r$), the relationship can be established by comparing the angular Momentum ($L$), expressed in terms of the moment of Inertia ($I$) and the moment ($p$), which results in:
| $ I = m_i r ^2$ |
.
(ID 3602)
ID:(599, 0)