In uniform circular motion, acceleration is expressed by the equation

$$a = \frac{v^2}{r}. $$

But this is a differential equation and solving it gets the result $$v = -\frac{r}{c+t}.$$

This doesn’t makes any sense. Should velocity be constant? Or at least something trigonometric?

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    $\begingroup$ In this case, the acceleration is not parallel, but perpendicular to the velocity, so $ a = dv/dt $ is not 1-dimensional. The differential equation above would imply that the acceleration would act in the same direction as the velocity $\endgroup$ Apr 5, 2022 at 20:31
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    $\begingroup$ @Comparative. Please don't answer in comments. Comments are not permanent and also note that an upvote on an answer gives more reputation points than an upvoted comment. $\endgroup$ Apr 5, 2022 at 21:33
  • $\begingroup$ @StephenG-HelpUkraine In which way are comments less permanent than answers? $\endgroup$ Apr 7, 2022 at 15:29
  • $\begingroup$ @Carl-FredrikNybergBrodda The system can wipe comments at any time. They can deleted or archived or moved to chat (and I don't think chat records are permanent either). Some of the interventions are automated (as I understand it). I've noticed many comments of my own disappear over the years. $\endgroup$ Apr 7, 2022 at 17:48

3 Answers 3


That equation is misleading.

You wrongly assume that $a$ is $dv/dt$ here, which it isn't.

Let us first rewrite the centripetal acceleration equation properly:

$$|\vec {\mathbf a}_N| = \dfrac {\left|\vec {\mathbf v}\right|^2} {R} \Longleftrightarrow a_N = \dfrac {v^2} R$$ where $a_N$ is the normal acceleration. The normal acceleration can be expressed as $\vec{\mathbf a} - \mathbf{\vec a}_T$ where $a_T$ is the tangential compenent of acceleration $a$. For uniform circular motion, $a_T=0$ so what we're left with is $$\left|\dfrac{d\mathbf{\vec v}}{dt}\right| = \dfrac{|\mathbf{\vec v}|^2}{R} \Longleftrightarrow a=\dfrac{v^2}R$$

Note that $a$ here is $a=\sqrt{a_x^2 +a_y^2}$ in 2-D. Since $\mathbf{\vec v} = \left( v_x, v_y\right)$, we can say $$a=\sqrt{\left(\dfrac{dv_x}{dt}\right)^2+\left(\dfrac{dv_y}{dt}\right)^2}.$$

Putting this into our centripital acceleration equation, we can get,

$$\sqrt{\left(\dfrac{dv_x}{dt}\right)^2+\left(\dfrac{dv_y}{dt}\right)^2}=\dfrac{v_x^2 + v_y^2 }R \tag{*}$$

This is the proper differential equation we are solving.

Note that since $a_T=0$, we have that $\dfrac d{dt} \left| \mathbf{\vec v}\right| = \dfrac d{dt} \sqrt{v_x^2 + v_y^2}=0$, so $v_x a_x + v_y a_y = \mathbf{\vec v}\cdot\mathbf{\vec a}=0$, which serves as our second equation to solve ($\star$).

Now, since you know that the solution to uniform circular motion is $\mathbf{\vec r} = \left( R\cos(\omega t), R\sin (\omega t\right))$, you can go ahead and verify that it indeed satisfies $\star$.

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    $\begingroup$ I suppose it might be worth mentioning that you also need $\vec{\mathbf{a}}\cdot\vec{\mathbf{v}}=0$ for uniform circular motion, which implies a second equation as well (which makes sense -- 2 equations for 2 unknown functions $v_x$ and $v_y$). $\endgroup$
    – Andrew
    Apr 6, 2022 at 1:25
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    $\begingroup$ Imho, the equation for the circular acceleration should be written as $\vec{a} = -\frac{\vec{v}^2}{\vec{R}^2}\vec{R}$. This gives both the correct magnitude (because $r = \frac{\vec{R}^2}{|\vec{R}|}$) and the correct direction of the acceleration. $a = \frac{v^2}{r}$ is simply what you get when you take the magnitude on both sides and simplify (and totally lose the direction information in the process). $\endgroup$ Apr 6, 2022 at 9:55
  • $\begingroup$ Cmaster my thoughts exactly, uniform circular motion is a solution to that differential equation. It certainly isn't the only solution. There are lots of non circular motion solutions to that equation, the proper one is with a -radial basis vector component $\endgroup$ Apr 6, 2022 at 14:47

The equation that you've given is just a simplified version of the real thing that lacks all direction information. The real thing is this:

$$\vec{a} = -\frac{\vec{v}^2}{\vec{r}^2}\cdot\vec{r}$$

If you take the magnitude of the vector on both sides of this, the equation simplifies as

$$|\vec{a}| = \left| -\frac{\vec{v}^2}{\vec{r}^2}\cdot\vec{r} \right|$$ $$\Leftrightarrow |\vec{a}| = \frac{|\vec{v}|^2}{|\vec{r}|^2}\cdot|\vec{r}|$$ $$\Leftrightarrow |\vec{a}| = \frac{|\vec{v}|^2}{|\vec{r}|}$$ $$\Leftrightarrow a = \frac{v^2}{r}$$

which is precisely the magnitude equation you gave. However, the direction of $\vec{a}$ must be towards the center, i.e. in the opposing direction of $\vec{r}$ which is produced by the minus sign.

  • $\begingroup$ Presumably by ${\overrightarrow{x}}^{2}$ you mean $\overrightarrow{x} \cdot \overrightarrow{x}$, where $(\cdot)$ is the regular dot product? IMO worth clarifying. $\endgroup$
    – Jivan Pal
    Apr 7, 2022 at 11:45
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    $\begingroup$ @JivanPal Yes, obviously. The cross product would invariably yield the null vector, which does not make any sense at all. (Btw, I've used \vec{v}^2 ($\vec{v}^2$), not \overrightarrow{v}^2 ($\overrightarrow{v}^2$). It's always good to use the commands that represent a meaning instead of the commands that represent some graphics/layout.) $\endgroup$ Apr 7, 2022 at 12:57
  • $\begingroup$ Thanks for the LaTeX tip, I was not aware that \vec is supported here, and personally I usually remap \vec to bold and/or underline (British convention), so I'm just used to using \overrightarrow for the arrow. $\endgroup$
    – Jivan Pal
    Apr 7, 2022 at 16:01
  • $\begingroup$ your notation is overly complicated and confusing since $\vec v^2=v^2$ is a scalar and $\vec r^2=r^2$ is also a scalar. Inserting vector signs is unnecessary and obscures this elementary but important fact. $\endgroup$ Apr 7, 2022 at 21:59
  • $\begingroup$ @ZeroTheHero I didn't drop the vector and magnitude signs earlier to show that the $|\vec{r}|^2$ does indeed cancel with the $|\vec{r}|$. I've broken up the sequence into several lines now in the hope that it makes the transformation clearer. $\endgroup$ Apr 8, 2022 at 7:19

The acceleration points towards the center of rotation, the acceleration is missing a $-\hat{r}$

The equations of motion for circular motion of radius $R$ at a constant frequency are

$$\boldsymbol{R}(t)= R \cos(\omega t) \hat i + R \sin(\omega t) \hat j$$

Velocity is changing, otherwise acceleration would be zero. Speed however, is constant.

Differentiate this and you obtain acceleration,

$$\boldsymbol{a} = \frac{d\boldsymbol{R}}{dt}= -\omega R \sin(\omega t) \hat{i} + \omega R \cos(\omega t) \hat{j} $$

$$\| \boldsymbol{a} \| = \sqrt{\omega^2 R^2 \sin^2(\omega t) + \omega^2 R^2 \cos^2(\omega t)} = \omega R$$


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