Centripetal acceleration
Centripetal (radial) acceleration is the acceleration that causes an object to move along a circular path, or turn. Whereas ordinary (tangential) acceleration points along (or opposite to) an object's direction of motion, centripetal acceleration points radially inward from the object's position, making a right angle with the object's velocity vector. In fact, because of its direction, centripetal acceleration is also referred to as "radial" acceleration.
Although an object moving in a circular orbit may be moving at a constant speed, since velocity is composed of a speed and direction, any process that changes the direction must contain an acceleration. Therefore, objects moving at a constant speed along a circular path have a nonzero centripetal acceleration.
Centripetal acceleration is always written in terms of the radius of the circular path, $r,$ and either the tangential velocity, $v,$ or angular velocity, $\omega:$
$a_c = \frac{v^2}{r} = \omega^2 r.$
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Orbital Motion
In Newton's theory of gravity, bodies follow trajectories that are conic sections (e.g. ellipses, parabolas). For example, planets in our solar system orbit the Sun on trajectories that are roughly ellipses. For orbits characterized by eccentricity close to zero, it is a pretty good approximation to model orbital motion as uniform circular motion.
Earth takes about 365 days to complete its orbit around the Sun.
Approximating the orbit as perfectly circular, find the centripetal acceleration of Earth.
Assumptions:
- Orbital radius (distance from the Sun to Earth) is $1.50\times 10^{11} \text{m}.$
- Orbital period (time Earth takes to orbit once) is 365 Earth days.
- 1 Earth day is 24 hours.
Either orbital speed $v$ or angular speed $\omega$ can be used to calculate the centripetal acceleration, but $v$ is easier with the variables given, since
$v = \frac{\Delta s}{\Delta t} = \frac{2\pi r}{T},$
where $\Delta s$ is the arc length traversed during one orbit and $T$ is the orbital period. We can compute the orbital speed $v$ from the assumptions:
$v = \frac{2\pi r}{T} = \frac{2 \pi \big(1.50\times 10^{11}\big)}{365\times24\times60\times 60}=2.99\times 10^{4} \text{ m/s}.$
The orbital speed $v$ is the tangential velocity in the expression for $a_c$, so
$a_c = \frac{v^2}{r} = \frac{ \big( 2.99\times 10^{4} \text{ m/s} \big)^2} {1.50\times 10^{11} \text{m}} = 5.96 \times 10^{-3} \text{m/s}^2.$
If there were a planet with the mass $6\times 10^{24}\text{kg}$ and angular velocity $7\times10^{-5}\text{ rad/s}$ of Earth, what would be the radius required in order for the free-fall acceleration on its surface to be $0 \text{ m/s}^2?$
Vector Depiction of Centripetal Acceleration
The above picture depicts a general situation of circular motion at an instant of time.
Let $r$ and $r'$ be the position vectors and $v$ and $v'$ be the velocitites of the object when it is at point $P$ and $P',$ respectively, as shown above. By definition, velocity is at a point along the tangent at that point in the direction of motion. Since the path is circular, $v$ and $v'$ are perpendicular to $r$ and $r',$ respectively. Thus, $\Delta v \perp \Delta r$ . Since the average acceleration is along $\Delta v$ $\left(\text{since }\, \overline{a}=\dfrac{\Delta v}{\Delta t}\right)$, $\overline{a} \perp \Delta r$ . $\Delta v$ is directed to the center, placed on the line bisecting angle between $r$ and $r'$ . As $\Delta t \rightarrow 0$ , the acceleration becomes instantaneous and is always directed towards the center. Thus, the magnitude of acceleration by definition is given by
$|a|=\lim_{\Delta t \rightarrow 0} \dfrac{|\Delta v|}{\Delta t}.$
The angle between position vectors $r$ and $r'$ is $\Delta \theta$ and that between $v$ and $v'$ is also $\Delta \theta$ . Thus $\Delta CPP' \sim \Delta GHI$ and since the corresponding sides of similar triangles are in proportion, we have
$\dfrac{\Delta v}{v}=\dfrac{\Delta r}{R} \Rightarrow \Delta v=v\dfrac{\Delta r}{R}.$
Thus the magnitude of acceleration becomes
$|a|=\lim_{\Delta t \rightarrow 0} \dfrac{|\Delta v|}{\Delta t}=\lim_{\Delta t \rightarrow 0} \dfrac{v|\Delta r|}{R\Delta t}=\dfrac{v}{R}\lim_{\Delta t \rightarrow 0} \dfrac{|\Delta r|}{\Delta t}.$
Since $\Delta t$ is small, $\Delta \theta$ is also small and we can take $PP'$ approxiamately equal to $|\Delta r|$ :
$|\Delta r| \approx v\Delta t \Rightarrow \dfrac{\Delta r}{\Delta t} \approx v \Rightarrow \lim_{\Delta t \rightarrow 0} \dfrac{|\Delta r|}{\Delta t}=v.$
Therefore, the magnitude of acceleration in circular motion is
$a_c=\left(\dfrac{v}{R}\right)v=\dfrac{v^2}{R}.$
Proof of Relationship between the Two Definitions
Acceleration is defined as the change in the velocity vector per unit time: $\vec{a} = \frac{\Delta \vec{v}}{\Delta t}.$
The angular velocity, $\omega$, is given by
$\omega = \frac{v}{R}.$
The centripetal acceleration is given by $a_c=\omega^2 r.$
To calculate the biker's acceleration, find the change in the velocity vector $\Delta \vec{v} = \vec{v}(\theta+\Delta \theta) - \vec{v}(\theta)$, as the bike moves through the angle $\Delta \theta.$
Zooming in on a small piece of the trajectory, inspect the vectors used to calculate:
To find the difference, align $\vec{v}(\theta+\Delta \theta)$ and $-\vec{v}(\theta)$ tip to tail:
First, observe that the difference $\Delta \vec{v}$ points directly toward the center of the circle.
Next, the angle between the two vectors is obviously $\Delta \theta$.
The length of the difference $\Delta \vec{v}$ is given (approximately) by the arc length swept by $\vec{v}$ as it rotates through $\Delta \theta$. Using $s = \theta r,$
$\Delta \vec{v} \approx \Delta \theta \lvert v \rvert.$
Dividing both sides by $\Delta t,$
$\frac{\Delta \vec{v}}{\Delta t} \approx \frac{\Delta \theta}{\Delta t} = \omega v.$
If the usual definition of $\omega$ is substituted,
$\vec{a} = \frac{\Delta \vec{v}}{\Delta t} \approx \frac{v^2}{R}.$
Consider $\Delta \theta$ to be extremely small so that the path of the bicycle is smooth. Thus the small error due to the arc length approximation goes away completely, and only the final result remains.
Therefore, the acceleration of the bike is given by $\boxed{a_c = \displaystyle \frac{v^2}{R}}$ and points directly toward the center of the circular path. $_\square$
While this demonstration is revealing, it was somewhat contrived. Many constraints are kept on the motion: the radius stayed constant, the rate of change of the radius remained constant, and there was no angular acceleration.
There is in fact a much simpler method for calculating all four of the accelerations that can occur in rotating reference frames that involves the use of De Moivre's theorem.
References
- Sadhu, D. The Helical Model – vortex solar system animation. Retrieved May 4, 2016, from http://www.djsadhu.com/the-helical-model-vortex-solar-system-animation/