0
$\begingroup$

Let acceleration be $$\overrightarrow a= a_T \overrightarrow T + a_N \overrightarrow N$$ where $\overrightarrow T$ is the unit tangent component and $\overrightarrow N$ is the unit normal component.

If we define $$\ v=\lVert \overrightarrow v(t) \rVert$$

Then the tangential and normal components of acceleration are given by,

$$\ a_T = v'= \frac{\overrightarrow r\ '(t)\ \cdot \overrightarrow r\ ''(t)}{\lVert\overrightarrow r\ '(t)\rVert} \ \ \ \ and\ \ \ \ \ \ a_N=kv^2 = \frac{\lVert \overrightarrow r\ '(t)\ \times \overrightarrow r\ ''(t)\rVert}{\lVert \overrightarrow r\ '(t)\rVert }$$

Where k is the curvature and $\overrightarrow T(t)$ is unit tangent vector given by $$\ k= \frac{\lVert \overrightarrow T\ '(t)\rVert}{\lVert \overrightarrow r\ '(t)\rVert}\ \ \ \ \ \ \overrightarrow T(t)=\frac{\overrightarrow r\ '(t)}{\lVert\overrightarrow r\ '(t)\rVert}$$

Where do these equations come from? Why are they defined in terms of the speed $\ v=\lVert\overrightarrow v(t)\rVert$? is there a geometric interpretation of this definition?

edit: I'm still a little confused, but this is what I have gathered thus far.

The acceleration vector $\overrightarrow a$ is a non-zero vector so it can be represented as a linear combination of $\overrightarrow T$ and $\overrightarrow N$ such that $\overrightarrow a$ lies in the plane formed by $\overrightarrow T$ and $\overrightarrow N$. Therefore, $\overrightarrow a$ can also be represented as the sum of the projection of $\overrightarrow a$ onto $\overrightarrow T$ and the projection of $\overrightarrow a $ onto $\overrightarrow N$. $$\overrightarrow a(t) = Proj_{\overrightarrow T}\overrightarrow a + Proj_{\overrightarrow N}\overrightarrow a$$ $$\overrightarrow a(t)= \frac{\overrightarrow a\cdot\overrightarrow T}{(\lVert \overrightarrow T\rVert)^2}\overrightarrow T + \frac{\overrightarrow a\cdot\overrightarrow N}{(\lVert \overrightarrow N\rVert)^2}\overrightarrow N$$ $$\overrightarrow a(t)=(\overrightarrow a\cdot\overrightarrow T)\overrightarrow T + (\overrightarrow a\cdot\overrightarrow N)\overrightarrow N$$ $$\overrightarrow a(t)= a_{T}\overrightarrow T + a_{N}\overrightarrow N$$ (From this point on, I'm dropping the overhead arrows to represent vectors.) I understand how the tangential component $\ a_{T}$ can be derived from the dot product of $\ a\cdot\ T$ but I am still unclear about the Normal component.

How do you prove:$$\ a\cdot N = \frac{\lVert v\times a\rVert}{\lVert v\rVert}=\sqrt{\lVert a \rVert^2-a_{T}^2}$$

$\endgroup$
1
$\begingroup$

This really isn't a definition, but rather a computation, decomposing the acceleration vector into its tangential and normal components. The unit tangent vector, curvature, and normal vector should not change when we reparametrize the curve; indeed, they are usually defined assuming the particle moves at constant speed $1$. The curvature tells us the rate at which the unit tangent vector changes (turns) when we move at speed $1$, and the unit normal vector $\vec N$ gives the direction of that change. That is, using $s$ to give arclength along the curve, $$\frac{d\vec T}{ds} = \kappa\vec N.$$ Note, also, that $v = ds/dt$ (why?).

Now, the unit tangent vector is given by the equation $$\vec v(t) = \|\vec v(t)\| \vec T(t),$$ so, differentiating, and using the chain rule, \begin{align*} \vec a(t) = \vec v'(t) &= \frac{d\vec v}{dt} = v'(t) \vec T(t) + v(t) \frac{d\vec T}{dt} = v'(t)\vec T(t) + v(t) \frac{d\vec T}{ds}\,\frac{ds}{dt}\\ &= v'\vec T + kv^2 \vec N. \end{align*}

You can now see that $a_T = \vec a\cdot \vec T = \vec r''\cdot\left(\dfrac{\vec r'}{\|\vec r'\|}\right)$ and $|a_N| = \|\vec a\times \vec T\| = \left\|\vec r''\times\left(\dfrac{\vec r'}{\|\vec r'\|}\right)\right\|$, so $a_N = kv^2 = \dfrac{\|\vec r''\times \vec r'\|}{\|\vec r'\|}$. (Note that $kv^2\ge 0$, so we know that in fact $a_N\ge 0$ and the absolute value is unnecessary.)

$\endgroup$
  • $\begingroup$ by $\overrightarrow N$ do you mean the normal vector or the unit normal vector where, $$\overrightarrow N = \frac{\overrightarrow T\ '(t)}{\lVert \overrightarrow T\ '(t)\rVert}$$ $\endgroup$ – Nicholas Cousar Mar 8 '18 at 21:14
  • $\begingroup$ Right, I mean the unit vector in the direction of $d\vec T/ds$, which is, after all, just a scalar multiple of $d\vec T/dt$. $\endgroup$ – Ted Shifrin Mar 8 '18 at 21:17
  • $\begingroup$ A couple of criticisms: I would write $\hat T$ and $\hat N$ rather than $\vec T$ and $\vec N$. Also $\vec a\times\hat T=-a_N\hat B$ where $\hat B$ is the unit binormal vector. But you answer is OK otherwise. $\endgroup$ – user5713492 Mar 10 '18 at 6:16
  • $\begingroup$ @user5713492: Physicists use the hat for unit vectors. Mathematicians do not. But the other error was a significant typo. Thanks. I've edited. (I'm not sure the OP knows about the binormal.) $\endgroup$ – Ted Shifrin Mar 10 '18 at 7:12
  • $\begingroup$ I was unclear on how you derived $\ a_{N}$ but I think I get it now. $\endgroup$ – Nicholas Cousar Mar 10 '18 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.