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.

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Variation of angular momentum

Equation

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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 $

$L$
Angular Momentum
$kg m^2/s$
$L_0$
Initial Angular Momentum
$kg m^2/s$
$dL$
Variation of Angular Momentum
$kg m^2/s$

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Variation of angular momentum (vector)

Equation

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Just as in the one-dimensional case, the change in angular momentum is

$ \Delta L = L - L_0 $



this can be generalized to more dimensions as

$ \Delta\vec{L} = \vec{L} - \vec{L}_0 $

$\vec{L}$
Angular Momentum (Vector)
$kg m^2/s$
$\vec{L}_0$
Initial angular momentum (vector)
$kg m^2/s$
$dL$
Variation of Angular Momentum
$kg m^2/s$

.

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Medium Torque

Equation

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In the case of translation, the second principle defines how translational motion is generated with the definition of force

$ F_m \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 }$

$\Delta t$
Time elapsed
$s$
$T$
Torque
$N m$
$dL$
Variation of Angular Momentum
$kg m^2/s$

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Medium torque (vector)

Equation

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Since the average torque in one dimension is

$ T_m =\displaystyle\frac{ \Delta L }{ \Delta t }$



this expression can be generalized for multiple dimensions using

$ \vec{T}_m =\displaystyle\frac{ \Delta\vec{L} }{ \Delta t } $

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Instantaneous Torque

Equation

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El torque medio calculado con la variación del momento angular \Delta L y el tiempo transcurrido \Delta t se define con:

$ T_m =\displaystyle\frac{ \Delta L }{ \Delta t }$



por ello el torque instantaneo se puede definir en el limite de tiempo infinitesimal:

$ T =\displaystyle\frac{d L }{d t }$

$L$
Angular Momentum
$kg m^2/s$
$t$
Time
$s$
$T$
Torque
$N m$

Con la definición de torque medio:

$ T_m =\displaystyle\frac{ \Delta L }{ \Delta t }$

\\n\\nse puede pasar al limite instantáneo en la medida que se consideren tiempos infinitesimales \Delta t\rightarrow 0 por lo que\\n\\n

$T\equiv \lim_{t\rightarrow 0}\displaystyle\frac{\Delta L}{\Delta t}\equiv \displaystyle\frac{dL}{dt}$



con lo que el torque instantaneo se define con la derivada del momento angular:

$ T =\displaystyle\frac{d L }{d t }$

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Instantaneous torque (vector)

Equation

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Given that the average torque in more than one dimension is

$ \vec{T}_m =\displaystyle\frac{ \Delta\vec{L} }{ \Delta t } $



the instantaneous torque is the torque in the limit as $\Delta t$ approaches infinitesimal, which is equivalent to the derivative of angular momentum

$ \vec{T} =\displaystyle\frac{ d\vec{L} }{ dt } $

$\vec{L}$
Angular Momentum (Vector)
$kg m^2/s$
$t$
Time
$s$
$\vec{T}$
Torque (Vector)
$Pa$

.

ID:(10988, 0)



Torque for constant moment of inertia

Equation

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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 $

$\alpha$
Instantaneous Angular Acceleration
$rad/s^2$
$I$
Moment of Inertia
$kg m^2$
$T$
Torque
$N m$

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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Torque for constant moment of inertia (vector)

Equation

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In the case where the moment of inertia is constant, the derivative of angular momentum is equal to

$ \vec{L} = I \vec{\omega} $



which implies that the torque is equal to

$ \vec{T} = I \vec{\alpha} $

$\vec{\alpha}$
Instantaneous Angular Acceleration (vector)
$rad/s^2$
$I$
Moment of Inertia
$kg m^2$
$\vec{T}$
Torque (Vector)
$Pa$

Since the moment is equal to

$ \vec{L} = I \vec{\omega} $



it follows that in the case where the moment of inertia does not vary with time,

$\vec{T}=\displaystyle\frac{d\vec{L}}{dt}=\displaystyle\frac{d}{dt}(I\vec{\omega}) = I\displaystyle\frac{d\vec{\omega}}{dt} = I\vec{\alpha}$



which implies that

$ \vec{T} = I \vec{\alpha} $

.

This relationship is the equivalent of Newton's second law for translation, but applied to rotation in the vectorial context.

ID:(10989, 0)



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