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Inertia and Centrifuge Acceleration
Definition 
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 $v^2/r$.
ID:(313, 0)
Aceleración Centrifuga
Description
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Equations
Since the centrifugal acceleration is equal to
| $ a_c =\displaystyle\frac{ v ^2}{ r }$ |
with
| $ v = r \omega $ |
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 }$ |
(ID 4735)
Examples
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 $v^2/r$.
(ID 313)
If we express the tangential velocity in terms of angular velocity, the centrifugal acceleration is given by:
| $ a_c = r \omega ^2$ |
(ID 4384)
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:
| $ a_c =\displaystyle\frac{ v ^2}{ r }$ |
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.
(ID 4735)
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:
| $ a_p =\displaystyle\frac{ v ^2}{ r }$ |
Unlike centrifugal acceleration, the centripetal acceleration is measurable for the object that is literally 'falling' towards the center.
(ID 4383)
ID:(657, 0)