Linear algebra based proof of circular motion formula. How do I get a by itself on the LHS?

Question

I haven't taken a course on linear algebra, but I know the very basics and started watching 3blue1brown's series on it as I will be taking it this spring. While watching the series, I realized I could use it to prove the circular motion formula $ {a} = \frac{v^2}{r} $ in a different and in my opinion easier way than the geometric proof I was shown in AP Physics C.

We know that ${\overrightarrow{v}}$ is perpendicular to $\overrightarrow{r}$, so $\overrightarrow{v} \cdot \overrightarrow{r} = 0$.

Differentiating both sides with respect to t, we get $\overrightarrow{a} \cdot \overrightarrow{r} + v^2 = 0 \Rightarrow \overrightarrow{a} \cdot \overrightarrow{r} = -v^2 $

Here's where I get confused. How do I cancel out $\overrightarrow{r}$ if you can't divide by vectors? I figured the best I can do is divide by $||\overrightarrow{r}||$ which gives

$\overrightarrow{a} \cdot \hat{r} = -\frac{v^2}{||\overrightarrow{r}||}$

This is true, but I'm trying to solve for $\overrightarrow{a}$ by itself.

Answer 1

You are actually pretty close, but you need additional assumptions as pointed out by the other answer. Recall that every vector can be written as the scalar product of its norm and a unit vector in its direction. If we write velocity this way, $\mathbf{v}=v\hat{\mathbf{u}}$. We can see that $\hat{\mathbf{u}}$ is tangent to the trajectory. If we differentiate with respect to $t$ to find the acceleration, we find $$\mathbf{a}=\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}=\frac{\mathrm{d}v}{\mathrm{d}t}\hat{\mathbf{u}}+v\frac{\mathrm{d}\hat{\mathbf{u}}}{\mathrm{d}t}=\frac{\mathrm{d}v}{\mathrm{d}t}\hat{\mathbf{u}}+\frac{v^2}{r}\hat{\mathbf{n}},$$

the last step follows from relating $\frac{\mathrm{d}\hat{\mathbf{u}}}{\mathrm{d}t}$ to the curvature of the trajectory. This relates to the answer by Gabriel Golfetti by providing you with the other clue you need: $\mathbf{a}_\perp=\frac{\mathrm{d}v}{\mathrm{d}t}\hat{\mathbf{u}}$, and this term vanishes due to the existence of a constant $\mathbf{\Omega}$ such that $\mathbf{v}=\mathbf{\Omega}\times \mathbf{r}$ and that $|\mathbf{r}|=\mathrm{constant}$ (why?), and we are left with $$\mathbf{a}=\frac{v^2}{r}\hat{\mathbf{n}},$$ after taking the magnitudes of both sides of the equation we are left with $$a=\frac{v^2}{r},$$ as we wanted to prove.

Answer 2

It seems to me that in the particular case of uniform circular motion: given the abundant symmetry of that motion it is possible to forgo vector notation altogether.

That said: it could be, I don't know, that someone who insists on exhaustive mathematical rigor will offer that it is necessary to apply more abstract considerations to establish that the position vector $r$ and the velocity vector $v$ are perpendicular.

However, I gather you take it as granted that the position vector and the velocity vector are perpendicular.

If that is granted: the same symmetry consideration gives that the acceleration vector is perpendicular to the velocity vector, anti-parallel to the position vector.

With that granted: the orientations of the vectors with respect to each other are all fixed. That means that including them in the notation does not actually add information. It will be sufficient to consider magnitude only.

In the case of uniform circular motion: the ratio of the magnitudes of the position vector and the velocity vector is a constant.

Let a point be circumnavigating an origin, at a radial distance of 1 unit of distance, one full revolution per unit of time. Then the corresponding velocity is $2\pi$ units of distance per unit of time. Hence the ratio of the magnitudes of velocity and position is $2\pi$:

$$ \frac{v}{r} = 2\pi \tag{1} $$

Next we capialize on symmetry:

We consider the initial point and the terminal point of the velocity vector. (Wikipedia: Euclidean vector) We can think of the rotation of the velocity vector as a rotation in a velocity space, with the initial point of the velocity vector as a fixed point. In that space the terminal point of the velocity vector moves along a circle. Hence: the distance over which the terminal point of the velocity vector moves is equal to the circumference of a circle whose radius equals the magnitude of the velocity vector. The magnitude of the acceleration vector corresponds to that circumference.

We have this symmetry:

$$ v = \frac{ds}{dt} \qquad , \qquad a = \frac{dv}{dt} \tag{2} $$

Acceleration is to velocity what velocity is to position.

$$ \frac{v}{r} = 2\pi \qquad , \qquad \frac{a}{v} = 2\pi \tag{3} $$

In uniform circular motion the ratio of acceleration to velocity is the same as the ratio of velocity to position.
(The ratios of the respective magnitudes of the vectors are the same.)

$$ \frac{a}{v} = \frac{v}{r} \tag{4} $$

Hence: $$ a = \frac{v^2}{r} \tag{5} $$

Answer 3

You don't need $\mathbf a$ by itself. You've gone as far as you can, actually, because decomposing $\mathbf a=\mathbf a_\parallel+\mathbf a_\perp$, you have just shown that $\mathbf a_\parallel=-\frac{v^2}{r^2}\mathbf r$. To be more precise, the condition you arrived at guarantees that the motion (in three dimensions) lives on the surface of the sphere $r=\text{const}$. The perpendicular component is completely arbitrary, and you need other assumptions to show uniform circular motion. In three dimesions, that assumtion is always equivalent to the existence of a constant $\mathbf\Omega$ such that $\mathbf v=\mathbf\Omega\times\mathbf r$ (convince yourself that such a vector always exists in the general case of arbitrary motion on a sphere, but is allowed to vary with time, and therefore its constantness is important)

Answer 4

circular motion ,$~r~$ radius $~\omega=\dot\phi~$ angular velocity \begin{align*} &\vec{r}=r\,\hat{e}_r\\ &\vec{v}=\omega\,r\hat{e}_t\quad, v=\omega\,r\\ &\vec{a}=\omega\,r\hat{e}_t-\omega^2\,r\hat{e}_r\quad\Rightarrow \vec{a}\cdot\hat e_r=-\omega^2\,r \end{align*} where $~\hat{e}_r\perp\hat{e}_t~$

your equation

\begin{align*} &\vec{a}\cdot\vec{r}=-\vec{v}\cdot\vec{v}\quad\text{or }\quad \frac{1}{r}\vec{a}\cdot\vec{r}=-\frac{1}{r}\vec{v}\cdot\vec{v}\quad,r\ne 0\quad\Rightarrow\\ &\vec{a}\cdot\hat{e}_r=-\frac{v^2}{r}=-\frac{\omega^2\,r^2}{r}=-\omega^2\,r\quad\surd \end{align*}

thus, just divide your equation with $~r~$ you obtain the correct result