(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

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

and the definition of the mean angular velocity ($\bar{\omega}$) is

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

.

(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

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

and the definition of the mean angular velocity ($\bar{\omega}$) is

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

.

(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

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

and the definition of the mean angular velocity ($\bar{\omega}$) is

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

.

(ID 3233)

The definition of average angular acceleration is based on the angle covered

and the elapsed time

The relationship between the two is defined as the average angular acceleration

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

and the mean Angular Acceleration ($\bar{\alpha}$) equals the difference in Angular Speeds ($\Delta\omega$) and the time elapsed ($\Delta t$) as per

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 ($\alpha_0$)

and assuming that the mean Acceleration ($\bar{a}$) equals the constant Acceleration ($a_0$)

then the following equation is obtained:

(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:

Therefore, considering the difference in Angular Speeds ($\Delta\omega$) along with the angular Speed ($\omega$) and the initial Angular Speed ($\omega_0$):

and the time elapsed ($\Delta t$) in relation to the time ($t$) and the start Time ($t_0$):

the equation for the mean Angular Acceleration ($\bar{\alpha}$):

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:

(ID 3237)

(ID 3251)

(ID 3251)

(ID 3251)

Since the moment is equal to

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

.

(ID 3253)

(ID 3324)

(ID 3324)

(ID 3324)

The relationship between the angular Momentum ($L$) and the moment ($p$) is expressed as:

Using the radius ($r$), this expression can be equated with the moment of Inertia ($I$) and the angular Speed ($\omega$) as follows:

Then, substituting with the inertial Mass ($m_i$) and the speed ($v$):

and

it can be concluded that the moment of inertia of a particle rotating in an orbit is:

(ID 3602)

(ID 3680)

(ID 3681)

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:

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 ($\theta_0$):

(ID 3682)

(ID 3683)

(ID 3684)

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:

(ID 4352)

(ID 4353)

(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 ($\alpha_0$):

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:

which results in the following equation:

(ID 4386)

Si se deriva en el tiempo la relaci n para el momento angular

para el caso de que el radio sea constante

$T=\displaystyle\frac{dL}{dt}=r\displaystyle\frac{dp}{dt}=rF$

por lo que

(ID 4431)

(ID 9875)

(ID 9876)

Since the moment ($p$) is defined with the inertial Mass ($m_i$) and the speed ($v$),

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

(ID 10975)