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Centrifugal and centripetal acceleration
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Any object with velocity tends to move in a straight line. To follow a circular orbit, an object needs to "fall" radically from its straight path to the orbit's radius. This "fall" corresponds to centripetal acceleration (centri = center, petal = toward), as observed by an external observer.
On the other hand, if the object continues on its straight path instead of following the circular orbit, an observer within the rotating system would perceive the same acceleration, but moving away from the center. This is known as centrifugal acceleration (centri = center, fuga = moving away).
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Tangential speed
Image
If an object is subjected to a mode of maintaining a constant radius, it will rotate as indicated in the figure. Upon observing the figure, one would notice that the mass undergoes a translational motion with a tangential velocity that is equal to the radius times the angular velocity:
However, if the element connecting the object to the axis is cut, the object will continue to move tangentially in a straight line.
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Centrifugal and centripetal acceleration
Description
Any object with velocity tends to move in a straight line. To follow a circular orbit, an object needs to "fall" radically from its straight path to the orbit's radius. This "fall" corresponds to centripetal acceleration (centri = center, petal = toward), as observed by an external observer. On the other hand, if the object continues on its straight path instead of following the circular orbit, an observer within the rotating system would perceive the same acceleration, but moving away from the center. This is known as centrifugal acceleration (centri = center, fuga = moving away).
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Equations
In the case where the initial Angular Speed ($\omega_0$) is equal to the mean angular velocity ($\bar{\omega}$),
| $ \bar{\omega} = \omega_0 $ |
Therefore, with the difference of Angles ($\Delta\theta$), which is equal to the angle ($\theta$) divided by the initial Angle ($\theta_0$), we obtain:
| $ \Delta\theta = \theta_2 - \theta_1 $ |
And with the time elapsed ($\Delta t$), which is equal to the time ($t$) divided by the start Time ($t_0$), we obtain:
| $ \Delta t \equiv t - t_0 $ |
We can rewrite the equation for the mean angular velocity ($\bar{\omega}$) as:
| $ \bar{\omega} \equiv\displaystyle\frac{ \Delta\theta }{ \Delta t }$ |
This can be expressed as:
$\omega_0 = \omega = \displaystyle\frac{\Delta\theta}{\Delta t} = \displaystyle\frac{\theta - \theta_0}{t - t_0}$
Solving for it, we get:
| $ \theta = \theta_0 + \omega_0 ( t - t_0 )$ |
(ID 1023)
With the distance traveled in a time ($\Delta s$) it is with the position ($s$) and the starting position ($s_0$):
| $ \Delta s = s - s_0 $ |
and the time elapsed ($\Delta t$) is with the time ($t$) and the start Time ($t_0$):
| $ \Delta t \equiv t - t_0 $ |
The equation for average velocity:
| $ v_0 \equiv\displaystyle\frac{ \Delta s }{ \Delta t }$ |
can be written as:
$v_0 = \bar{v} = \displaystyle\frac{\Delta s}{\Delta t} = \displaystyle\frac{s - s_0}{t - t_0}$
thus, solving for it we get:
| $ s = s_0 + v_0 ( t - t_0 )$ |
(ID 3154)
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 $ |
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(ID 3324)
(ID 3324)
The definition of the mean angular velocity ($\bar{\omega}$) is considered as the angle variation ($\Delta\theta$),
| $ \Delta\theta = \theta_2 - \theta_1 $ |
and the time elapsed ($\Delta t$),
| $ \Delta t \equiv t - t_0 $ |
The relationship between both is defined as the mean angular velocity ($\bar{\omega}$):
| $ \bar{\omega} \equiv\displaystyle\frac{ \Delta\theta }{ \Delta t }$ |
(ID 3679)
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)
Since the centrifugal acceleration is equal to
| $ a_c =\displaystyle\frac{ v_0 ^2}{ r }$ |
with
| $ v_0 = r \omega_0 $ |
we can conclude that:
| $ a_c = r \omega ^2$ |
(ID 4384)
If the distance traveled is small ($v\Delta t\ll r$), the square root of the distance between the center and the body,
$\sqrt{r^2+(v\Delta t)^2}$
can be approximated as
$r+\displaystyle\frac{1}{2}\displaystyle\frac{v^2}{r}\Delta t^2$
which corresponds to a parabolic relationship with respect to time $\Delta t$. Therefore, the behavior can be described with an acceleration equal to:
| $ a_c =\displaystyle\frac{ v ^2}{ r }$ |
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Examples
(ID 15417)
If an object is subjected to a mode of maintaining a constant radius, it will rotate as indicated in the figure. Upon observing the figure, one would notice that the mass undergoes a translational motion with a tangential velocity that is equal to the radius times the angular velocity:
However, if the element connecting the object to the axis is cut, the object will continue to move tangentially in a straight line.
(ID 310)
If a body fixed to a rope of length $r$ rotates with a tangential velocity $v$, and the rope is cut, the body will continue to move in a straight line with a constant velocity $v$ due to inertia.
Orbit circulates of radio
In a time interval $\Delta t$, the body will have traveled a distance of $v\Delta t$ tangentially to its previous orbit. From the perspective of an observer on the axis of the rotating system, the distance is calculated using the Pythagorean theorem, by adding the square of the orbit radius to the square of the distance traveled:
$\sqrt{r^2+v^2\Delta t^2}$
(ID 1155)
If we study a catapult, we will notice that the projectile initially moves along the curve described by the spoon. This happens because the spoon is designed to hold the projectile. Once the arm stops, the projectile continues in a straight line tangential to the circle it was following.
If an object is not restrained and travels with a tangential velocity $v$, it will cover a distance of $v\Delta t$ in a time interval $\Delta t$, moving from point B to point C. However, if it continues to orbit, after the time interval $\Delta t$, it will reach point D. If the object reaches point C, from the perspective of an observer on Earth, there will be an acceleration that causes the object to move away from Earth (centrifugal acceleration), covering the distance $\Delta r$ in the time interval $\Delta t$.
For an observer in space, an object in orbit is constantly falling: instead of ending up at point C, it falls over the time interval $\Delta t$ covering the distance $\Delta r$ until it reaches point D. In both cases, we can represent the situation graphically, and using the Pythagorean theorem, we can see that the following equation must hold true:
$(r+\Delta r)^2=r^2+(v\Delta t)^2$
Expanding the equation, we get:
$2\Delta rr+\Delta r^2=v^2\Delta t^2$
Since the variation in radius $\Delta r$ is much smaller than the radius itself ($r\ll\Delta r$), we can conclude that:
$2\Delta rr=v^2\Delta t^2$
Solving for $\Delta r$, we find:
$\Delta r=\displaystyle\frac{1}{2}\displaystyle\frac{v^2}{r}\Delta t^2$
Comparing this equation with the equation $s=at^2/2$, we can conclude that the object is accelerating with an acceleration equal to $v^2/r$.
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