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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:

Image 1 Image 2

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?


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

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

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  • $\begingroup$ Note that you can write $\vec{v} = \omega \vec{h}$ for some vector $h$ (which is of course just $\vec{v}/\omega$ in the formula given above). Then by the product rule $\frac{d\vec{v}}{dt} = \frac{d\omega}{dt} \vec{h} + \omega\frac{d\vec{h}}{dt}$. The last term is the 'old' result (the one you get when $\omega$ is constant). The new part is the first term. $\endgroup$ – Winther Jun 23 '14 at 23:12
  • $\begingroup$ Your derivatives are not technically correct because you did not differentiate r, i.e. you treated it as if it were a constant. Why should it be? $\endgroup$ – Bryson S. Jun 24 '14 at 2:35
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    $\begingroup$ You're very close. Write down what form the unit tangent and radial vectors are in terms of $\theta$ and $r$, and the rest should pop out. $\endgroup$ – Jerry Schirmer Jun 24 '14 at 3:07
  • $\begingroup$ @BrysonS. Which variable did I treat as a constant? Are you talking about $\theta(t)$? I represented its derivative as $\omega(t)$ $\endgroup$ – 1110101001 Jun 24 '14 at 4:08
  • $\begingroup$ You did not differentiate $r$ to get $\dot{r}$. The correct velocity equation should be $\boxed{\overset{\rightharpoonup }{v}(t)=[\dot{r}cos(\theta (t))-r \sin (\theta (t))\omega(t)]\cdot\hat{i}+[\dot{r}sin(\theta (t))+r \cos (\theta (t))\omega(t)]\cdot\hat{j}}$ $\endgroup$ – Bryson S. Jun 24 '14 at 4:37
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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.

Coord

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.

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  • $\begingroup$ There is also another way to derive this result by taking the derivatives of the polar unit vectors themselves as applied to the velocity equation. If you want me to add that solution as well just let me know. $\endgroup$ – Bryson S. Jun 24 '14 at 16:56
  • $\begingroup$ Just so you know for the future: we prefer it if you not edit your post too many times. When you have a minor edit, e.g. just a few characters, see if you can sit on it for a while (a few hours, a day) until you have several such edits to make at once. It's not like there's a strict rule about this, but if you find yourself editing your posts more than 5 or 6 times, it's probably worth trying to condense the edits. $\endgroup$ – David Z Jun 26 '14 at 2:32
  • $\begingroup$ @David Z Noted. $\endgroup$ – Bryson S. Jun 26 '14 at 13:00
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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.

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  • $\begingroup$ I tried doing so and I got stuck with a messy soup of variables. I updated my opening post with my progress. How should I proceed from there? $\endgroup$ – 1110101001 Jun 23 '14 at 23:00
  • $\begingroup$ Sorry I didn't see the request not to add more detail. $\endgroup$ – Bryson S. Jun 24 '14 at 2:15
  • $\begingroup$ @BrysonS.: no worries. $\endgroup$ – Jerry Schirmer Jun 24 '14 at 3:05
  • $\begingroup$ This should be a comment nothing more $\endgroup$ – Your Majesty Jun 22 '15 at 0:51
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    $\begingroup$ @LoveLearning: complete answers to homework questions violate the spirit of the site. $\endgroup$ – Jerry Schirmer Jun 22 '15 at 3:40
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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.

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protected by Qmechanic Feb 23 '15 at 21:49

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