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Trajectory of a point moving with acceleration perpendicular to velocity

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(Just some basic high-school physics stuff; to assure myself I can still do some elementary things. :P Essentially, showing that if a particle moves with acceleration perpendicular to velocity, or velocity perpendicular to position, then it traces out a circle. Stop reading here if this is obvious.)

Suppose a point moves in the plane such that its acceleration is always perpendicular to its velocity, and of the same magnitude. What is its path like?

To set up notation: let’s say the point’s position at time $t$ is $(p_x(t), p_y(t))$, its velocity is $(v_x(t), v_y(t)) = \left(\frac{d}{dt}p_x(t), \frac{d}{dt}p_y(t)\right)$, and its acceleration is $(a_x(t), a_y(t)) = \left(\frac{d}{dt}v_x(t), \frac{d}{dt}v_y(t)\right)$.

The result of rotating a point $(x,y)$ by 90° is $(-y, x)$. (E.g. see figure below)

So the fact that acceleration is at right angles to velocity means that $(a_x(t), a_y(t)) = (-v_y(t), v_x(t))$, or, to write everything in terms of the velocity,

\begin{aligned} \frac{d}{dt}v_x(t) &= -v_y(t) \\ \frac{d}{dt}v_y(t) &= v_x(t) \end{aligned}

where we can get rid of $v_x(t)$ by substituting the second equation (in the form $v_x(t) = \frac{d}{dt}v_y(t)$) into the first:

$v_y(t) = -\frac{d}{dt}v_x(t) = -\frac{d}{dt}\left(\frac{d}{dt}v_y(t)\right)$

or in other words

$v_y(t) = -\frac{d^2}{dt^2}v_y(t).$

By some theory about ordinary differential equations, which I don’t know (please help!) (but see the very related example you saw in high school, of simple harmonic motion), the solutions to this equation are $\sin(t)$ and $\cos(t)$ and any linear combination of those: the solution in general is

\begin{aligned} v_y(t) &= a \sin(t) + b \cos(t) \\ &= \sqrt{a^2 + b^2} \left(\frac{a}{\sqrt{a^2+b^2}}\sin(t) + \frac{b}{\sqrt{a^2+b^2}}\cos(t)\right) \\ &= R\sin (t + \alpha) \end{aligned}

where $R = \sqrt{a^2 + b^2}$ and $\alpha$ is the angle such that $\cos(\alpha) = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin(\alpha) = \frac{b}{\sqrt{a^2+b^2}}$. And the fact that $v_x(t) = \frac{d}{dt}v_y(t)$ gives $v_x(t) = R\cos(t + \alpha)$. So $(v_x(t), v_y(t)) = (R\cos(t + \alpha), R\sin(t + \alpha))$. Note that $(a_x(t), a_y(t)) = \left(\frac{d}{dt}v_x(t), \frac{d}{dt}v_y(t)\right) = (-R\sin(t+\alpha), R\cos(t+\alpha))$ is indeed perpendicular to $(v_x(t), v_y(t))$ as we wanted.

The actual trajectory $(p_x(t), p_y(t))$ can be got by integrating

$\left(\frac{d}{dt}p_x(t), \frac{d}{dt}p_y(t)\right) = (v_x(t), v_y(t)) = (R\cos(t + \alpha), R\sin(t + \alpha))$

to get $p_x(t) = R\sin(t + \alpha) + c_1$ and $p_y(t) = -R\cos(t + \alpha) + c_2$. This trajectory is a point moving on a circle centered at point $(c_1, c_2)$ and of radius $R$, with speed $R$ or unit angular speed. Note that velocity is also perpendicular to the point’s position wrt the centre of the circle, as velocity is tangential to the circle, as it should be.

With a suitable change of coordinates (translate the origin to $(c_1, c_2)$, then rotate the axes by $\frac{\pi}{2}+\alpha$, then scale everything so that $R = 1$), this is the familiar paremetrization $(\cos(t), \sin(t))$ of the circle.

Note: Just as we derived $(v_x(t), v_y(t)) = (R\cos(t + \alpha), R\sin(t + \alpha))$ from assuming that the acceleration is perpendicular to velocity, we can, by assuming that velocity is perpendicular to position, identically derive $(p_x(t), p_y(t)) = (R\cos(t + \alpha), R\sin(t + \alpha))$, i.e. that the point moves on a circle.

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Written by S

Sun, 2013-04-07 at 23:38:01 +05:30

Posted in mathematics

8 Responses

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1. I tried to solve this problem independent of a coordinate system. I used a tensor, but it turns out you don’t really need one:

Let $\underbar{Q}$ be an orthogonal tensor that rotates its argument(operand?) by 90 degrees, so that $det(\underbar{Q}) = 1$ and $\underbar{a} \cdot \underbar{Q} \underbar{a} = \underbar{0} \forall \underbar{a}$. Then $\dot{\underbar{v}} = \underbar{Q} \underbar{v}$ by your definition.

Also, $\underbar{v} \cdot \underbar{Q} \underbar{v} = 0$, so $\underbar{v} \cdot \dot{\underbar{v}} = 0$. This can be written as $\frac{1}{2} \dot{v^2} = 0$, so $\underbar{v}$ has constant magnitude.

Finally, $\dot{\underbar{v}} = \underbar{Q} \underbar{v}$ can be integrated to give $\underbar{v} = \underbar{Q} (\underbar{r} - \underbar{r0})$ because $\underbar{Q}$ is not time dependent. Choosing our origin carefully so that $\underbar{r0} = 0$, We get $\underbar{v} = \underbar{Q} \underbar{r}$, the same as the relation between $\dot{\underbar{v}}$ and $\underbar{v}$. The same step as above gives $\frac{1}{2}\dot{r^2} = 0$, so $\underbar{r}$ has constant magnitude as well.

In two dimensions, $r=const$ tells us that it moves in a circle about the origin, $v=const$ tells us that the circular motion is uniform.

Incidentally, I think setting $\underbar{a} \cdot \underbar{Q} \underbar{a} = k a^2$ where unlike above, $k$ is not zero, describes a (linear) spiral instead. (Not sure, just spitballing.) An interesting question is what the spirals look like when $k(t)$ is not a constant.

About the solution of the ODE, the best I could do was note that it’s linear/homogeneous, so $exp(\alpha t)$ must be a solution for some $\alpha$. Solving the resulting polynomial gives $\alpha = \pm i$, which are sines/cosines. This simply shifts the problem one link up the chain of reasoning, but there’s my engineering education for you.

(global-set-key (kbd “C-`”) ‘underbar)

Karthik

Mon, 2013-04-08 at 14:22:44 +05:30

• Ah, jeez. I don’t know why I thought the comments would handle TeX input.

Karthik

Mon, 2013-04-08 at 14:23:39 +05:30

• WordPress.com uses $latex ...$ syntax (without the space between $and latex) instead of$ ... \$ syntax… I’ve edited your comment; now to read it. :-)

(The two expressions with \underbar{r_0} gave “Formula does not parse” errors, so I’ve changed them to \underbar{r0} (ugly-looking, but still better than raw))

S

Mon, 2013-04-08 at 15:06:10 +05:30

• Thanks, that’s useful to know. Does WP have Latex support (in this form) built in or have you installed a plugin of some sort?

I see now some Emacs gunk also made its way into the comment. The perils of composing blog comments in the scratch buffer, it seems. :-)

Karthik

Mon, 2013-04-08 at 15:40:48 +05:30

• Read it — thanks, that’s clean. Yes, this makes sense, and roughly this sort of reasoning went into the previous post on sin and cos, which is ultimately unsurprising (both are about circles).

About the ODE: are you saying that because it’s a linear/homogeneous differential equation, the solution must be of the form $\exp(\alpha t)$? I don’t know (or if I did, I don’t remember — I don’t even remember if I did know!) why this should be so. This is the k=1 case in this, but the argument in that article that “In order for the sum of multiple derivatives of a function to sum up to zero, the derivatives must cancel each other out and the only way for them to do so is for the derivatives to have the same form as the initial function” seems unsatisfying to me.

S

Mon, 2013-04-08 at 15:25:03 +05:30

• The ODE: If it’s linear, homogeneous and has constant coefficients, then the solution can be expressed using $\exp(\alpha t)$. I don’t remember how I know this either! Wikipedia’s explanation is about the best I’ve found in the past.

I do believe there’s a theorem that says an $n^{th}$ order linear homogeneous ODE (variable coefficients) has an $n$ dimensional solution space, so the $n$ independent exponentials do form a basis.

Karthik

Mon, 2013-04-08 at 15:54:35 +05:30

2. […] writing a series of posts leading up to an understanding of the exponential function (here, here, here), but it seems to have got abandoned. Consider this one a contribution to that […]