# Why does solving the differential equation for circular motion lead to an illogical result?

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?

• 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 Apr 5, 2022 at 20:31
• @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. Apr 5, 2022 at 21:33
• @StephenG-HelpUkraine In which way are comments less permanent than answers? Apr 7, 2022 at 15:29
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## 3 Answers

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 $$\star$$.

• 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 $v_y$). Apr 6, 2022 at 1:25
• 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). Apr 6, 2022 at 9:55
• 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 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.

• Presumably by ${\overrightarrow{x}}^{2}$ you mean $\overrightarrow{x} \cdot \overrightarrow{x}$, where $(\cdot)$ is the regular dot product? IMO worth clarifying. Apr 7, 2022 at 11:45
• @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.) Apr 7, 2022 at 12:57
• 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. Apr 7, 2022 at 16:01
• 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. Apr 7, 2022 at 21:59
• @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. 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,

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

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