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Centrifugal and centripetal acceleration
Storyboard 
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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Centrifuge acceleration
Note
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:
$\sqrt{r^2+v^2\Delta t^2}$
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Inertia and Centrifuge Acceleration
Quote
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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Centrifugal and centripetal acceleration
Storyboard
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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In the case where the initial Angular Speed ($\omega_0$) is equal to the mean angular velocity ($\bar{\omega}$),
Therefore, with the difference of Angles ($\Delta\theta$), which is equal to the angle ($\theta$) divided by the initial Angle ($\theta_0$), we obtain:
And with the time elapsed ($\Delta t$), which is equal to the time ($t$) divided by the start Time ($t_0$), we obtain:
We can rewrite the equation for the mean angular velocity ($\bar{\omega}$) as:
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:
With the distance traveled in a time ($\Delta s$) it is with the position ($s$) and the starting position ($s_0$):
and the time elapsed ($\Delta t$) is with the time ($t$) and the start Time ($t_0$):
The equation for average velocity:
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:
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
The definition of the mean angular velocity ($\bar{\omega}$) is considered as the angle variation ($\Delta\theta$),
and the time elapsed ($\Delta t$),
The relationship between both is defined as the mean angular velocity ($\bar{\omega}$):
Since the centrifugal acceleration is equal to
with
we can conclude that:
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:
Examples
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.
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.
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:
$\sqrt{r^2+v^2\Delta t^2}$
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 $v^2/r$.
Bodies tend, by inertia, to move in a straight line at a constant velocity. Therefore, if a body orbits around another, it deviates from its straight path and 'falls' into an orbit. Similarly, if there is nothing to hold a body, it will start moving away from the orbit, experiencing, for an object at the center of the rotating system, an apparent acceleration that moves it away from the center, known as centrifugal acceleration. The acceleration is defined as:
Centrifugal acceleration is an acceleration observed by a system on the axis of rotation when an object moves away (flees) at a constant velocity. For the object moving away, such acceleration does not exist.
If we express the tangential velocity in terms of angular velocity, the centrifugal acceleration is given by:
When an object orbits at a radius $r$ with a tangential velocity $v$, it maintains a constant distance from the center equal to the radius.
For an observer external to the system, the body, which would otherwise travel in a straight line due to inertia, deviates from this trajectory while maintaining the distance from the center. From the perspective of this observer, the body is accelerating towards the center (centripetal acceleration) of the orbit. Unlike centrifugal acceleration, the object is experiencing a real acceleration. The magnitude of this acceleration is equal to the centrifugal acceleration but with the opposite sign. Therefore, the magnitude of the centripetal acceleration is:
Unlike centrifugal acceleration, the centripetal acceleration is measurable for the object that is literally 'falling' towards the center.
If we divide the relationship between the distance traveled in a time ($\Delta s$) and the radius ($r$) by the angle variation ($\Delta\theta$),
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 ($\omega$):
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:
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:
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:
The mean Speed ($\bar{v}$) can be calculated from the distance traveled in a time ($\Delta s$) and the time elapsed ($\Delta t$) using:
If the speed is constant, the velocity will be equal to the initial Speed ($v_0$). In this case, the distance traveled as a function of time can be calculated using the difference between the position ($s$) and the starting position ($s_0$), divided by the difference between the time ($t$) and the start Time ($t_0$):
The corresponding equation defines a straight line in space-time.
To estimate the displacement of an object, it's necessary to know its the angular Speed ($\omega$) as a function of the time ($t$). Therefore, the the mean angular velocity ($\bar{\omega}$) is introduced, defined as the ratio between the angle variation ($\Delta\theta$) and the time elapsed ($\Delta t$).
To measure this, a system like the one shown in the image can be used:
To determine the average angular velocity, a reflective element is placed on the axis or on a disk with several reflective elements, and the passage is recorded to estimate the length of the arc $\Delta s$ and the angle associated with the radius $r$. Then the time difference when the mark passes in front of the sensor is recorded as $\Delta t$. The average angular velocity is determined by dividing the angle traveled by the time elapsed.
The equation that describes the average angular velocity is:
It should be noted that the average velocity is an estimation of the actual angular velocity. The main problem is that:
If the angular velocity varies during the elapsed time, the value of the average angular velocity can be very different from the average angular velocity.
Therefore, the key is:
Determine the velocity in a sufficiently short elapsed time to minimize its variation.
In the case where the angular velocity is constant, the mean angular velocity ($\bar{\omega}$) coincides with the value of the initial Angular Speed ($\omega_0$), so
In this scenario, we can calculate the angle traveled as a function of time by recalling that it is associated with the difference between the current and initial angles, as well as the current and initial time. Therefore, the angle ($\theta$) is equal to the initial Angle ($\theta_0$), the initial Angular Speed ($\omega_0$), the time ($t$), and the start Time ($t_0$) as shown below:
The equation represents a straight line in angle-time space.
When the angular velocity is constant, it is trivial that the average angular velocity is equal to that constant angular velocity. In other words, the initial Angular Speed ($\omega_0$) is equal to the mean angular velocity ($\bar{\omega}$):
When the velocity is constant, then trivially the average velocity is equal to that constant velocity. In other words, the constant velocity ($v_0$) is equal to the mean Speed ($\bar{v}$):
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:
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:
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:
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