Proof of centripetal acceleration formula ($a_c = v^2/r$) for non-uniform circular motion

Question

The formula for centripetal (radial) acceleration is well known, and there exist many proofs for it: $$||a_c|| = \frac{||v||^2}{r}$$

However, all the proofs I've seen rely on the fact that it is uniform circular motion and the magnitude of the tangential velocity vector does not change. For instance, take the classic proof using similar triangles — the similarity can only be established if the final tangential velocity vector and the initial one are of the same length.

In addition, take this calculus based proof from Khan Academy, outlined as follows:

For this proof to work, $(d\theta / dt)$ must be considered a constant, $\omega$, that does not depend on time. In the case of non-uniform circular motion, however, this is not always true as since there exists a tangential acceleration along with a radial one, $\omega$ must depend on time and is not necessarily a constant value.

Intuitively I understand that the centripetal/radial acceleration depends only on the difference in orientation between two tangential velocity vectors, and that their magnitudes do not matter -- hence the formula intuitively holds true in the non-uniform case. However, how would you go about modifying either of the proofs presented so that they are still valid in this case? Or alternatively, is there another proof that holds valid even when there exists tangential acceleration?

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As per the suggestion of one of the answers, I let $\omega$ vary with time and took its derivative as $\alpha(t)$. This is my work so far. Unfortunately, I am stuck after the last step.

\begin{align} \overset{\rightharpoonup }{p}(t) &= r \cos (\theta (t))\cdot\hat{i}+r \sin (\theta (t))\cdot\hat{j} \\ \overset{\rightharpoonup }{v}(t) &= -r \sin (\theta (t))\cdot\omega(t)\cdot\hat{i}+r \cos (\theta (t))\cdot\omega(t)\cdot\hat{j} \\ \overset{\rightharpoonup }{a}(t) &= (-r \cos(\theta(t))\cdot\omega(t)^2 - r \sin(\theta(t))\cdot\alpha(t))\hat{i} \\ &+ (-r \sin(\theta(t))\cdot\omega(t)^2 + r \cos(\theta(t))\cdot\alpha(t))\hat{j} \end{align}

From here on, $\theta(t)$ is represented as just $\theta$ for brevity and clarity

$$\overset{\rightharpoonup }{a}(t) = -\omega(t)^2(r \cos\theta\cdot\hat{i} + r \sin\theta\cdot\hat{j}) -\alpha(t)(r \sin\theta\cdot\hat{i} - r \cos\theta\cdot\hat{j})$$

Answer 1

The proper derivation of the centripetal acceleration—without assuming any kinematic variables are constant—requires a solid understanding of both the stationary Cartesian unit vectors $\hat{i}$ and $\hat{j}$ as well as the rotating polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$. The Cartesian unit vectors $\hat{i}$ and $\hat{j}$ are stationary and always aligned with the X and Y axes respectively, while the polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$ rotate with an angular velocity of $\omega=\|\dot{\theta}\|$ and point in the directions of increasing radius and angle (respectively). The included graphic below shows the two basis vector pairs overlaid on top of one another.

The position vector of the object is obviously defined as:

$\vec{p}(t)=x\hat{i}+y\hat{j}=rcos(\theta)\hat{i}+rsin(\theta)\hat{j}$,

with

$\|\vec{p}(t)\|=\sqrt{(rcos{\theta})^2+(rsin{\theta})^2}=\sqrt{r^2(sin^2(\theta)+cos^2(\theta))}=r\sqrt{(1)}=r$

Less obviously, it can be shown that the polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$ can be expressed solely in terms of the Cartesian unit vectors $\hat{i}$ and $\hat{j}$ and the angular position $\theta$ as,

$\boxed{\hat{e}_r=cos(\theta)\hat{i}+sin(\theta)\hat{j}}$ and $\boxed{\hat{e}_\theta=-sin(\theta)\hat{i}+cos(\theta)\hat{j}}$.

These two equations are extremely important, as they will be the key to expressing the Cartesian acceleration in polar coordinates, of which one of the terms will be our desired $v^2/r=\omega^2r$ centripetal acceleration. Moving forward, the vector acceleration of the object in Cartesian coordinates is simply

$\vec{a}(t)=\frac{d^2}{dt^2}\left[\vec{p}(t)\right]=\ddot{x}\hat{i}+\ddot{y}\hat{j}$.

Starting with $x=rcos(\theta)$ and $y=rsin(\theta)$ and differentiating once, we have

$\boxed{\dot{x}=\dot{r}cos(\theta)-r\dot{\theta}sin(\theta)}$ and $\boxed{\dot{y}=\dot{r}sin(\theta)+r\dot{\theta}cos(\theta)}$.

Differentiating again, we will have

$\ddot{x}=\ddot{r}cos(\theta)-\dot{r}\dot{\theta}sin(\theta)-\dot{r}\dot{\theta}sin(\theta)-r\frac{d}{dt}\left[\dot{\theta}sin(\theta)\right]$

$=\ddot{r}cos(\theta)-2\dot{r}\dot{\theta}sin(\theta)-r\left[\ddot{\theta}sin(\theta)+{\dot{\theta}}^2cos(\theta)\right]$, such that

$\boxed{\ddot{x}=(\ddot{r}-r\dot{\theta}^2)cos(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})(-sin(\theta))}$.

Similarly, the y acceleration $\ddot{y}$ becomes

$\ddot{y}=\ddot{r}sin(\theta)+\dot{r}\dot{\theta}cos(\theta)+\dot{r}\dot{\theta}cos(\theta)+r\frac{d}{dt}\left[\dot{\theta}cos(\theta)\right]$

$=\ddot{r}sin(\theta)+2\dot{r}\dot{\theta}cos(\theta)+r\left[\ddot{\theta}cos(\theta)-{\dot{\theta}}^2sin(\theta)\right]$, such that

$\boxed{\ddot{y}=(\ddot{r}-r\dot{\theta}^2)sin(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})cos(\theta)}$.

Now, we must plug these scalar derivatives into our formulation for the vector acceleration. In Cartesian coordinates, this is

$\vec{a}(t)=\ddot{x}\hat{i}+\ddot{y}\hat{j}=\{(\ddot{r}-r\dot{\theta}^2)cos(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})(-sin(\theta))\}\hat{i}+\{(\ddot{r}-r\dot{\theta}^2)sin(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})(cos(\theta))\}\hat{j}$

which can be rearranged into the following form:

$\vec{a}(t)=(\ddot{r}-r\dot{\theta}^2)\{cos(\theta)\hat{i}+sin(\theta)\hat{j}\}+(r\ddot{\theta}+2\dot{r}\dot{\theta})\{-sin(\theta)\hat{i}+cos(\theta)\hat{j}\}$

But as we have already seen, this is simply equal to

$\boxed{\boxed{\vec{a}(t)=(\ddot{r}-r\dot{\theta}^2)\hat{e}_r+(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{e}_\theta}}$

As we can now appreciate from carrying out the full derivation, there are actually two components each to both the radial and tangential accelerations. The $\ddot{r}$ term is straightforwardly equal to the second derivative of the position vector magnitude. The second term, $r\dot{\theta}^2$, is our long sought-after centripetal acceleration $r\dot{\theta}^2=\omega^2r=v^2/r$, and (as expected) it points in the negative radial direction. The tangential terms are perhaps a bit less intuitive. The $r\ddot{\theta}$ term is the acceleration that occurs whenever the radius and angular acceleration $\ddot{\theta}$ are both non-zero (imagine the tangential acceleration of a turbine blade of a jet engine as the engine spools up). The final term $2\dot{r}\dot{\theta}$ is what's commonly known as the Coriolis acceleration, and it occurs whenever the radius and angle change simultaneously. It arises because, for a given angular velocity, the arc length travelled every second increases with radius (tangential velocity increases with radius). Thus, an object with a given angular velocity will have different tangential velocities at different local radii of rotation. If the radius changes with time ($\dot{r}\not=0$) and the angular velocity $\dot{\theta}$ is not equal to zero, then the tangential velocity will change with time, which is by definition a tangential acceleration.

Answer 2

I'm outlining this and stating the final result so that the OP gets the fun of figuring this out themselves. Future responders, please don't work this out

All you have to do is allow $\omega(t)$ to be a function of time. You'll get extra ${\dot \omega} = \alpha$ terms in your equation, and you'll get a final result that says that

$${\vec a} = {\vec a}_{T} + {\vec a}_{C}$$

Where $\vec a_{T}$ is proportional to $\alpha r$ and points tangentially to the circle and $a_{C}$ is proportional to $\frac{v^{2}}{r}$ and points radially inward.

Answer 3

The answer by Bryson S. is solid, thorough and vey good, as is Jerry Schirmer's hint. This is merely another way of looking at the problem.

We can consider, as Jerry Schirmer points out, two components of acceleration; a tangential and a normal component. Before we begin, note that velocity always points tangential to the path a particle travels. This is easy to see intuitively (imagine travelling on a road) and can be proven from the definition of velocity.

$\mathbf a = \frac{d}{dt} \mathbf v = \frac{d}{dt} v \mathbf T = \mathbf T \frac{dv}{dt} + v\frac{d\mathbf T}{dt}$

Now, curvature, a geometric property of curves is defined as following: $\kappa = |\frac{d\mathbf T}{ds}|$, where $s$ is the arc length of any curve. Curvature is useful in this case because we can simplify $\frac{d\mathbf T}{dt}$ using the chain rule into $\frac{d\mathbf T}{dt} = \frac{d\mathbf T}{ds} \frac{ds}{dt} = \kappa \frac{d\mathbf T}{ds}$. Now, using the fact that $\mathbf T$ is a unit vector and thus has a constant magnitude, you can take the dot product of $\mathbf T$ and $\frac{d\mathbf T}{ds}$ and show that the result is zero, i.e. that $\mathbf T$ and $\frac{d\mathbf T}{ds}$ are perpendicular. From this, using the unit normal vector (perpendicular to the tangent and pointing toward the concave side), we get $\frac{d\mathbf T}{dt} = v\frac{d\mathbf T}{ds} = v^2 \kappa \mathbf N$. We could have also gotten this result by applying Frenet's equations directly.

$\mathbf a = (\frac{dv}{dt})\mathbf T + (v^2 \kappa) \mathbf N$

Now, the radius of curvature is in fact the reciprocal of curvature; while this is the definition for non-circular curves, it can be proven that this is the case for circles by taking a reference frame with the center of the circle as the origin, then splitting up the tangent unit vector into its components then writing the angle in terms of arclenght then differentiating then finding the magnitude of the resulting vector (if proof is wanted, then ask for it in the comments).

From here, we get

$\mathbf a = (\frac{dv}{dt})\mathbf T + (\frac{v^2}{r}) \mathbf N$

And from here the result you seek follows.

Answer 4

From Wikipedia,

As a generalization of the uniform circular motion case, suppose the angular rate of rotation is not constant. The acceleration now has a tangential component, as shown in the image below. This case is used to demonstrate a derivation strategy based on a polar coordinate system.

image source: Wikipedia

Let $\textbf{r}(t)$ be a vector that describes the position of a point mass as a function of time. Since we are assuming circular motion, let $\textbf{r}(t) = R·\textbf{u}_r$, where $R$ is a constant (the radius of the circle) and $\textbf{u}_r$ is the unit vector pointing from the origin to the point mass. The direction of $\textbf{u}_r$ is described by $θ$, the angle between the x-axis and the unit vector, measured counterclockwise from the x-axis. The other unit vector for polar coordinates, $\textbf{u}_θ$ is perpendicular to $\textbf{u}_r$ and points in the direction of increasing $θ$. These polar unit vectors can be expressed in terms of Cartesian unit vectors in the x and y directions, denoted $i$ and $j$ respectively.

$$ \mathbf{u}_{r}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j} $$

and $$ \mathbf{u}_{\theta}=-\sin \theta \mathbf{i}+\cos \theta \mathbf{j} $$ One can differentiate to find velocity: $$ \begin{aligned} \mathbf{v} &=r \frac{\mathrm{d} \mathbf{u}_{\mathrm{r}}}{\mathrm{d} t}=r \frac{\mathrm{d}}{\mathrm{d} t}(\cos \theta \mathbf{i}+\sin \theta \mathbf{j}) \\ &=r \frac{d \theta}{d t}(-\sin \theta \mathbf{i}+\cos \theta \mathbf{j}) \\ &=r \frac{\mathrm{d} \theta}{\mathrm{d} t} \mathbf{u}_{\theta} \\ &=\omega r \mathbf{u}_{\theta} \end{aligned} $$ where $\omega$ is the angular velocity $d \theta / d t$.

This result for the velocity matches expectations that the velocity should be directed tangentially to the circle, and that the magnitude of the velocity should be $rω$. Differentiating again, and noting that

$$ \frac{\mathrm{d} \mathbf{u}_{\theta}}{\mathrm{d} t}=-\frac{\mathrm{d} \theta}{\mathrm{d} t} \mathbf{u}_{\mathrm{r}}=-\omega \mathbf{u}_{\mathrm{r}} $$

we find that the acceleration, $\mathbf{a}$ is: $$ \mathbf{a}=r\left(\frac{\mathrm{d} \omega}{\mathrm{d} t} \mathbf{u}_{\theta}-\omega^{2} \mathbf{u}_{\mathrm{r}}\right) $$ Thus, the radial and tangential components of the acceleration are: $$ \mathbf{a}_{\mathrm{r}}=-\omega^{2} r \mathbf{u}_{\mathrm{r}}=-\frac{|\mathbf{v}|^{2}}{r} \mathbf{u}_{\mathrm{r}} \quad \text { and } \quad \mathbf{a}_{\theta}=r \frac{\mathrm{d} \omega}{\mathrm{d} t} \mathbf{u}_{\theta}=\frac{\mathrm{d}|\mathbf{v}|}{\mathrm{d} t} \mathbf{u}_{\theta} $$ where $|\mathbf{v}|=r \omega$ is the magnitude of the velocity (the speed).

These equations express mathematically that, in the case of an object that moves along a circular path with a changing speed, the acceleration of the body may be decomposed into a perpendicular component that changes the direction of motion (the centripetal acceleration), and a parallel, or tangential component, that changes the speed.

References:

• Centripetal force, Wikipedia.