Could someone show me a simple and intuitive derivation of the Centripetal Acceleration Formula $a=v^2/r$, preferably one that does not involve calculus or advanced trigonometry?


8 Answers 8


Imagine an object steadily traversing a circle of radius $r$ centered on the origin. Its position can be represented by a vector of constant length that changes angle. The total distance covered in one cycle is $2\pi r$. This is also the accumulated amount by which the position has changed...

Now consider the velocity vector of this object: it can also be represented by a vector of constant length that steadily changes direction. This vector has magnitude $v$, so the accumulated change in velocity is $2 \pi v$.

The magnitude of acceleration is then $\frac{\text{change in velocity}}{\text{elapsed time}}$, which we can write as: $$a = \frac{2 \pi v}{\left(\frac{2\pi r}{v} \right)} = \frac{v^2}{r} \,.$$


Aside: that derivation is used in many algebra/trig-based textbooks.

  • $\begingroup$ Notice that after one full turn the change in position is also zero. What we are interested in here really the average value of the instantaneous acceleration, but to get it requires calculus (or at least the machinery of limits), which the OP doesn't want. So instead we consider not displacement but distance and also the equivalent of distance for velocity (which doesn't have a conventional name) rather than $\vec{v}_f - \vec{v}_i$. This gives the centripetal acceleration for all curves (knowing $r$ and $v$), but we must add the "transverse" acceleration by hand. $\endgroup$ Commented Jul 12, 2015 at 15:08
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    $\begingroup$ This is a great explanation but could someone explain why the accumulated change in velocity is 2*pi*v? $\endgroup$ Commented Jul 12, 2015 at 16:08
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    $\begingroup$ @Conceptuality So, by assumption the speed of the particle is constant, right? But as it goes around the circle its direction of motion changes. That means the velocity vector keeps the same length but turns through a whole circle. The tip of the velocity vector describes a circle of radius $v$, so the distance that the tip moved was $2 \pi v$. $\endgroup$ Commented Jul 12, 2015 at 16:15
  • $\begingroup$ @dmckee, in the similar triangle proof, the change of velocity is simply the distance between the tips of two velocity vectors draw tail-to-tail. But would "delta v" simply be zero since magnitude is constant?---------how do you really mean sure the change in direction anyway in degree? $\endgroup$ Commented Jul 22, 2015 at 16:17
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    $\begingroup$ @theenigma017 Angles are formally dimensionless (this is why angular velocity and angular frequency are the same measure in SI). Which is why $2 \pi r$ is a distance and not some other quantity, Which means that $2 \pi v$ is a speed. The first three comments may help, or you can just think of this as comparing distance and speed. $\endgroup$ Commented Oct 10, 2018 at 14:04

enter image description here

Pic 2

In right-angled-triangle ABC \begin{equation} \dfrac{|\dfrac{\Delta \mathbf{\vec v}}{2}|}{|\vec V|}=Sin\dfrac{\Delta \vec \theta}{2} \tag{01} \end{equation} If \begin{equation} \theta \tag{02} \end{equation} is small \begin{equation} v(t) \approx v(t + \delta t ) = \mathbf{\vec v} \tag{03} \end{equation} \begin{equation} \dfrac{|\dfrac{\delta \mathbf{\vec v}}{2}|}{|\vec V|}=Sin\dfrac{\delta \vec \theta}{2} \tag{03} \end{equation} For small angle \begin{equation} \delta \theta \approx Sin {\delta \theta} \tag{04} \end{equation} So on rearranging \begin{equation} \dfrac{\delta \mathbf{\vec v}}{2} =\dfrac{\delta \vec \theta}{2} \times \vec v \tag{05} \end{equation} \begin{equation} \delta \mathbf{\vec v} =\delta \vec \theta \times \vec v \tag{06} \end{equation} \begin{equation} \dfrac{\delta \mathbf{\vec v}}{\delta \mathbf{t}} =\dfrac{\delta \vec \theta}{\delta \mathbf{t}} \times \vec v \tag{07} \end{equation}

\begin{equation} \mathbf{\vec a}=\dfrac{\delta \mathbf{\vec v}}{\delta \mathbf{t}} \tag{08} \end{equation}

\begin{equation} \vec \omega =\dfrac{\delta \vec \theta}{\delta \mathbf{t}} \tag{09} \end{equation}

\begin{equation} \mathbf{\vec a}=\vec \omega \times \vec v \tag{10} \end{equation}

\begin{equation} \mathbf{a}=\omega \times v \tag{11} \end{equation} And since \begin{equation} \mathbf{v}=\omega \times r \tag{12} \end{equation} So \begin{equation} \mathbf{a}=\dfrac {v^{2}}{\mathbf{r}} \tag{13} \end{equation}

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    $\begingroup$ With a small angle limit this is, of course, the correct derivation. But you might want to be explicit that $|\Delta \mathbf{v}| = v |\Delta \phi|$ is correct in that limit. I have mixed success in the classroom with this derivation (and with the one I used, too). $\endgroup$ Commented Jul 12, 2015 at 15:21
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    $\begingroup$ Why is $|\Delta\mathbf v|=v|\Delta\phi|$? $\endgroup$
    – Allawonder
    Commented Jan 18, 2018 at 19:42
  • $\begingroup$ V(vector) = an angle * velocity ? what is the intuition ? also used in above answer $\endgroup$ Commented Oct 10, 2018 at 11:25
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    $\begingroup$ @Allawonder did you work it out? I am stuck at the same question. $\endgroup$
    – Mihail
    Commented Sep 11, 2019 at 14:07

You can do this derivation by breaking the position of the orbiting particle down into components. It isn't short, but I think it's useful because it supplements the algebra with concrete physical analogies. I'll organize it into four parts: decomposition, oscillation, energy, and symmetry.


The position of a particle moving along a circular path can be described by two half-out-of-phase sine waves -- or equivalently, a sine wave and a cosine wave:

An animated illustration of the relationship between rotation and the sine function, as parameterized by time.


This is easy to derive: suppose the particle is moving at a constant angular velocity $\omega$ around a circle of radius $r$. Then $\theta = \omega t$, and basic trigonometry tells us that the position of the particle given $\theta$ is given by $x = r \cos \theta$ and $y = r \sin \theta$. We can substitute to get $x = r \cos (\omega t)$ and $y = r \sin (\omega t)$.


It turns out that there's another kind of movement that's described by sine waves: the oscillation of a weight on a spring. In such a system,

$$x(t) =A \cos \left(\sqrt{k \over m}t \right)$$

where $A$ is the amplitude (i.e. the difference between the spring's maximum length and its resting length), $k$ is the spring's force constant, from Hooke's law, and $m$ is, of course, the mass. Deriving this equation in the usual way requires not only calculus, but also differential equations -- so I'm going to ask you to take my word for it until a bit later.

For our purposes, this means that the mechanics of the orbiting particle can be simulated by two oscillating springs: one for the $x$ component, and one for the $y$ component that is identical to the first, but that is half-out-of-phase with it. To picture what this looks like, look back at the animation above, and pretend the blue dot and the red dot are attached to springs that are at $0$ when resting.

Now that we have this spring-based model, we can use it to determine the force being applied to the particle along the $x$ axis! By Hooke's law, the force a spring exerts on an attached weight is $F = -kx$. Let's try to use that formula to determine the acceleration of the particle when the $x$ spring is at its maximum length. We know its maximum length will be $r$ in this case -- this corresponds to the moment when the particle is at $x = r, y = 0$. And we know that $F = ma$. So by substitution, $ma = -kr$; divide by $m$, and

$$ a = -k \frac{r}{m}$$

Our work is half done. But now we have a new problem: we don't know what $k$ is -- after all, there's no actual spring, so we can't measure anything. We need to know what value of $k$ a spring will have if it moves the same way as the particle does along the $x$ axis. To solve this problem, we have to start thinking about conservation laws.


Let's think about what happens when the $x$ spring oscillates. When the spring is at $x = r$, it exerts the largest force on the particle, but the particle isn't moving at all. This is right at the peak of the wave. That means $v = 0$. From that point, the force that the spring applies to the particle will accelerate it from $0$ to its largest velocity $-v_\text{max}$.1 And by the time the particle reaches $x = 0$, the spring will be applying no force; that's the spring's resting state. Since the spring is applying no force, we've reached $-v_\text{max}$ -- the direction of future acceleration will be in the opposite direction, slowing the particle down until it reaches $x = -r$.

So this means that at $x = r$, we have $F = -F_\text{max}$, $a = -a_\text{max}$, and $v = 0$. And at $x = 0$, we have $F = 0$, $a = 0$, and $v = -v_\text{max}$.

Here's what's happening: the energy in the system is moving back and forth between maximum kinetic energy (at $x = 0$ when the spring is not stretched or compressed at all) and maximum potential energy (at $x = r$, when the spring is fully stretched). And because of conservation of energy, these two maxima must be equal; in other words, $E_\text{max}$ = $P_\text{max}$.

The formula for kinetic energy is $E = \frac{1}{2} mv^2$ -- that's basic Newtonian mechanics. What's the formula for $P$ in this case? This is the hardest part of the derivation. The potential energy stored in the spring is equal to the amount of negative work done to stretch it to $x = r$. So we have to remember the formula for work: $W = Fd$, where $d$ stands for distance traveled -- i.e. $x$, assuming that we start at $x = 0$. But then we have a problem. $F = -kx$ is not constant -- it's a function of $x$.

In general, this would mean we have to do calculus. But luckily enough, $F = -kx$ is a linear function, and so the value we want is equal to the area of the triangle formed by the $x$ axis and the line $F = -kx$:

The integral of -kxdx from x = 0 to x = 1, with k = 1.


In the above chart, $k = 1$, the distance traveled $d = x_\text{max} = 1$, and the area of the given triangle represents the value you get when you multiply $F(x)$ by distance traveled, adjusting for changes in the value of $F$ as the distance increases. But since the height of the triangle is $-kx_\text{max}$ and the base of the triangle is $x_\text{max}$, we can just use plain old geometry. The area of a triangle is $\rm \frac{1}{2} base \times height$ -- or, here, because $x_\text{max} = r$

$$W = -\frac{1}{2} kr^2$$

And since potential energy is equal to negative work:

$$P = \frac{1}{2} kr^2$$

Isn't it interesting how similar that is to $E = \frac{1}{2} mv^2$? If you think about that for a while, and consider that $E + P$ has to be a constant value, you might start to realize why an oscillating spring follows a sinusoidal path! (Hint: write the formula for a circle, but use $\sqrt{E + P}$ instead of $r$, and $v$ instead of $y$.)

But you can come back to that later. We're very close! Setting $v = v_\text{max}$:

$$E + P = E_\text{max} = P_\text{max} = \frac{1}{2} kr^2 = \frac{1}{2} mv^2$$

What happens when we solve for $k$?

\begin{align}\frac{1}{2} kr^2 &= \frac{1}{2} mv^2\\ \implies kr^2 &= mv^2\\ \implies k &= \frac{mv^2}{r^2}\end{align}

Now we can substitute that into our formula for acceleration above:

\begin{align}a &= -k \frac{r}{m}\\ &= -\frac{mv^2}{r^2} \frac{r}{m}\\ &= -\frac{v^2}{r}\end{align}

You might be wondering why the negative sign appears in this version. But recall that the acceleration is technically in the opposite direction of the displacement. So when $x = r, y = 0$, the acceleration is in the direction of $-r$. If it were otherwise, the particle would be accelerating outwards!2


The last step of this derivation requires a trick. We began by breaking movement in two dimensions into movement along two one-dimensional components. We then used imaginary springs to describe the particle's movement along those two components. And now, we face a final question: how do we pick our $x$ and $y$ components?

They have to be at right angles to one another, but that's only half the battle -- we have to find the "right place" to start, the "real" $x$ coordinate. The trouble is that we can't. The circular path along which the particle moves is rotationally symmetrical. There's nothing about the circle to tell us where it "starts" or "ends."

This means that the above line of reasoning holds no matter where we start. We can pick any point on the circle as the $x=1, y=0$ point, and the above will be valid. So wherever the particle is, we simply set that point as our $x = r, y = 0$ point, and everything else falls into place.

If we wanted to make more work for ourselves, we could work out the details trigonometrically, using the above formulas, adjusting them for the $y$ axis, and then recombining the $x$ and $y$ values using vector algebra. But we don't need to -- the symmetry argument is more powerful in this case.

1. Here "largest" really means "most negative," because the particle is moving in the negative $x$ direction. These values will really be $-F_\text{max}$ and $-v_\text{max}$. Later, when the particle is moving in the opposite direction, these values will be positive.

2. Making sense of the signs takes a lot of subtle detail work; in particular, you have to understand why negative work becomes positive potential energy. It also helps to think about $r$ as a vector (which has a direction) instead of a magnitude (which doesn't). Fortunately, physical intuition provides a reliable guide in this case; if something turns out obviously wrong, double-check your thinking.


In order to move through a concave path, an agent has to impart force to otherwise a linearly-moving object. The object , by virtue of its motion, under the absence of any external force, always travels or tends to travel in the direction of the velocity vector at the concerned instant.

So, when the object has to transverse a curve trajectory, the main requisite is the introduction of a force that manipulates the direction of velocity such that the resultant locus is the required curvilinear path, otherwise the object would travel straight.

The direction of the force is evidently the direction of the acceleration or the limit of change of velocity with respect to time. So, to find the direction, let's think of an infinitesimal situation.

Let for a short amount of time $\Delta t$, the distance traveled is $v(t)\Delta t$ along a circular arc of radius $r$. The angle transversed is then $$\Delta\theta = \dfrac{v(t) \Delta t}{r}$$.

Imagine the bisector of the angle. Now, consider the changes in velocity parallel & perpendicular to this bisector. Initially the velocity has a component $v\sin(\frac{\Delta\theta}{2})$ away from the center & $v\cos(\frac{\Delta\theta}{2})$ transversely.Afterwards, it has a component $v\sin(\frac{\Delta\theta}{2})$ toward the center & $v\cos(\frac{\Delta\theta}{2})$ transversely as before. Thus the change of velocity is of magnitude $2v\sin(\frac{\Delta\theta}{2})$ toward the center of the arc.

As $\Delta\theta$ is vanishingly small, $\sin(\frac{\Delta\theta}{2})$ becomes indistinguishable as $\dfrac{\Delta\theta}{2}$. Thus, we can put $$|\Delta v(t)| = v^2 \dfrac{\Delta\theta}{r}$$. And the direction is towards the center. Thus the force is rotating the position vector along the curved trajectory and the change is radially inward regardless of whether it is traced clockwise or counter-clockwise.

The picture becomes more vivid if we calculate using polar co-ordinate.

First, we write the position vector as $\mathbf{r} = r\cdot \mathbf e_r$ . Now consider the change of $\mathbf{r}$ with time. Its change during $\Delta t$ is $r\Delta\theta \cdot \mathbf e_{\theta}$. $e_r \, \textrm{and} \, e_{\theta}$ are mutually perpendicular, the first being outward radially from the center . Therefore velocity is $$\mathbf v = \dfrac{\mathrm d\mathbf{r}}{\mathrm dt} = r\frac{\mathrm d\theta}{\mathrm dt}\cdot \mathbf e_{\theta} = \omega r \cdot \mathbf e_{\theta}$$.

By putting $r = 1$, we get $$\frac{\mathrm d}{\mathrm dt} (\mathbf e_r) = \omega \mathbf e_{\theta}\;.$$

Similarly, a change of $\theta$ implies a change of $\mathbf e_{\theta}$.It can be seen that $$\frac{\mathrm d}{\mathrm dt} (\mathbf e_{\theta}) = -\omega \cdot \mathbf e_r\;.$$ Now we differentiate the velocity, $$\mathbf{a} = \omega r \dfrac{\mathrm d}{\mathrm dt} (\mathbf e_{\theta}) = -{\omega}^2 r\cdot \mathbf e_r \;.$$ This result falls down automatically with the correct direction which is opposite of $\mathbf e_r$ i.e. toward the center radially.


enter image description here

$$\vec{a}_{avg}=\frac{∆\vec{v}}{∆t}$$ So, $$ a_{avg}=\frac{|∆\vec{v}|}{∆t}$$ From (a) and (b) in the figure, it is clear that the velocities triangle is similar to the triangle $OP_{1}P_{2}$ So $$\frac{|∆\vec{v}|}{v_{1}}=\frac{|∆\vec{s}|}{R}$$ $$\frac{a_{avg}∆t}{v_{1}}=|∆\vec{s}|/R$$ then $$ a_{avg}=\frac{|∆\vec{s}|v_{1}}{R∆t}$$ However, as t goes to zero, we have $|∆\vec{s}|=∆s$ and $\frac{∆s}{∆t}=v_{1}$. Therefore, $$ a=\frac{v_{1}^{2}}{R}$$ However, since speed is constant we can simply drop the subscript 1 to get

$$ a=\frac{v^{2}}{R}$$


You said to not include calculus, but I will allow myself to take simple derivatives since velocity and acceleration are calculus quantities anyway.

For this proof I will assume two things

  1. The radius is constant, i.e. $\vec r\cdot \vec r=r^2=c$
  2. The acceleration is in opposite direction to the position vector, i.e. $\vec a\cdot \vec r=-ar$. It is centripetal acceleration after all. I think this should be provable from 1. with some more work.

Now I will use the identity $\frac{d}{dt}(\vec a\cdot \vec b)=\frac{d\vec a}{dt}\cdot\vec b+\vec a\cdot\frac{d\vec b}{dt}$ on 1. to get \begin{align} \frac{d}{dt}(\vec r\cdot \vec r)&=0\\ \vec v\cdot \vec r+\vec r\cdot \vec v&=0\\ \vec v\cdot\vec r&=0. \end{align} This is a well-known proof to show the velocity is orthogonal to the position vector. Applying the derivative once more we get \begin{align} \frac{d}{dt}(\vec v\cdot\vec r)&=0\\ \vec a\cdot\vec r+v^2&=0 \end{align} Now we can use 2.: \begin{align} -ar+v^2&=0\\ a=\frac{v^2}{r} \end{align}

Note that I didn't assume that $a$ or $v$ are constant, so this equality should hold whenever 1. and 2. are true.

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    $\begingroup$ Perhaps the shortest derivation uses the cross product: $$\vec v=\vec {\omega} \times \vec r$$ $$\text {so}\ \ \ \ \ \ \vec a=\vec {\omega} \times \vec v +\dot{\vec {\omega}} \times \vec r.$$But $\dot{\vec {\omega}}=0$, so $$\vec a=\vec {\omega} \times \vec v =\omega v(-\hat r)=-\frac{v^2}r\ \hat r.$$ $\endgroup$ Commented Apr 27, 2023 at 14:21
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    $\begingroup$ Though if we let the rotation occur in the complex plane: $$z=|z|e^{i\omega t}$$ So $$\frac{dz}{dt}=i \omega |z|e^{i\omega t}\ \ \ \ \ \text{and} \ \ \ \ \ \frac {d_2 z}{dt^2}=-\omega^2 |z| e^{i\omega t}=-\omega^2 z.$$ $\endgroup$ Commented Apr 27, 2023 at 15:03

Here is a rather quirky derivation that doesn't use calculus notation. I believe that the general idea dates back to Huygens (c 1659).

A small arc of a circle is almost indistinguishable from part of a parabola. You can show using Pythagoras's theorem that if $x$ is the semi-length of the chord that runs between the ends of an arc of a circle of radius $r$, and $h$ is the sagitta of the arc (the distance between the centre of the arc and the centre of the chord) then $$x^2=2rh\ \ \ \ \ \ \ \ \text {provided $h$ is small enough to neglect $h^2$.}$$ In order to follow the path, a body at the midpoint of the arc at $t=0$ and moving parallel to the chord at speed $u$ must have an acceleration $a=\frac {u^2}r$ at right angles to the chord. We know this from ordinary projectile theory... At time $t$ from passing the centre of the arc, $$x=ut\ \ \ \ \ \text{and}\ \ \ \ \ h=\tfrac 12 at^2\ \ \ \ \ \text{so}\ \ \ \ \ \ x^2=2\left(\frac{u^2}a\right) h$$ Comparing with our first equation, $r=\frac{u^2}a$, that is $a=\frac {u^2}r$.


A most elementary way to look at what is changing in the radial direction only.

The time it takes from $v$ at rim to stationary state $v=0$ at center traversing radius length $r$ along radius is $$ T=\frac{r}{v-0} =\frac{r}{v}$$

Note that this is not a uniform motion along a straight line in the tangential direction.

Acceleration along the radius (centripetal) then is

$$ T=\frac{v-0}{T} =\frac{v}{r/v} = \frac{v^2}{r}. $$


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