1
$\begingroup$

Question:

The position $\mathbf r$ of a particle satisfies

$$ \ddot {\mathbf r} = f\dot {\mathbf r} \times \mathbf k$$

where $f$ is a known constant and $\mathbf k$ is the unit vector in the $z$ direction.

Given that $\dot {\mathbf r} \cdot \mathbf k \equiv 0$, show that the path of the particle is a circle.


Attempt:

Firstly, there are a few quantities that I know are constant:

$$\frac{d}{dt}\big(|\dot{\mathbf r}|^2 \big) = 2\ddot {\mathbf r} \cdot \dot {\mathbf r} = 2f(\dot{\mathbf r} \times \mathbf k) \cdot \dot{\mathbf r} = 0 \implies |\dot{\mathbf r}|^2 \equiv \text{constant}$$

$$\frac{d}{dt}\big(\mathbf r \cdot \mathbf k\big) = \dot{\mathbf r} \cdot \mathbf k =0 \implies {\mathbf r} \cdot \mathbf k \equiv \text{constant}$$

Moreover, I noticed that

$$\frac{d^2\dot{\mathbf r}}{dt^2} = f^2\dot{\mathbf r}$$

and solving this gives

$$\dot{\mathbf r}(t) = \mathbf c_1 \cos(ft)+\mathbf c_2 \sin (ft)$$

but I am unsure how to deduce from this that $\mathbf r$ traces out a circle.

$\endgroup$
3
$\begingroup$

Let $\mathbf u = \mathbf r + \frac1f (\dot{\mathbf r} \times \mathbf k)$, we have $$\dot{\mathbf u} = \dot{\mathbf r} + \frac1f(\ddot{\mathbf r} \times \mathbf k) = \dot{\mathbf r} + ((\dot{\mathbf r} \times \mathbf k) \times \mathbf k) = \dot{\mathbf r} + ((\dot{\mathbf r}\cdot \mathbf k) \mathbf k - \dot{\mathbf r}) = \mathbf 0$$ This means $\mathbf u$ is a constant. Notice $$\frac{d}{dt}|\mathbf r - \mathbf u|^2 = 2(\mathbf r-\mathbf u)\cdot \dot{\mathbf r} = -\frac{2}{f} (\dot{\mathbf r} \times \mathbf k)\cdot \dot{\mathbf r} = \mathbf 0$$ The expression $|\mathbf r-\mathbf u|^2 = R^2$ is a constant. The particle is moving on a sphere centered at $\mathbf u$ with radius $R$. Together with what you know $\mathbf r \cdot \mathbf k = $ constant. The particle is moving along the intersection of a sphere and a plane, i.e. a circle.

$\endgroup$
1
$\begingroup$

Insert the solution you found of the derived equations into the original equations. $$ -f\vec c_1\sin(ft)+f\vec c_2\cos(ft)=f(\vec c_1×\vec k)\cos(ft)+f(\vec c_2×\vec k)\sin(ft). $$ Comparing coefficients, $$ \vec c_2=~~\vec c_1×\vec k\\ \vec c_1=-\vec c_2×\vec k\\ $$ is only possible if both vectors are orthogonal to $\vec k$, of the same length and orthogonal to each other. With $\vec k=(0,0,1)$ you get $\vec c_1=(a,b,0)$ and $\vec c_2=(b,-a,0)$. This means that $\dot{\vec r}$ follows a circular curve, and that remains preserved under integration, the integration constant is the midpoint around which $\vec r$ circulates.

$\endgroup$
0
$\begingroup$

One way is to show that the curvature is constant. Recall that $$\kappa = \frac{||\mathbf{v}(t) \times \mathbf{a}(t)||}{||\mathbf{v}(t)||^3} = \frac{||\dot{\mathbf{r}}(t) \times \ddot{\mathbf{r}}(t)||}{||\dot{\mathbf{r}}(t)||^3} $$ Which should be straightforward since you have $\dot{\mathbf{r}}(t)$.

It might help to write: \begin{align} \dot{\mathbf{r}}(t) &= \mathbf{c_1}\cos(ft) + \mathbf{c_2}\sin(ft)\\ &= \left[\begin{array}{c}c_{11}\\c_{12}\\c_{13}\end{array}\right]\cos(ft) + \left[\begin{array}{c}c_{21}\\c_{22}\\c_{23}\end{array}\right]\sin(ft)\\ &=(c_1\cos(ft) + c_2\sin(ft))\hat{\mathbf{i}} + (c_3\cos(ft) + c_4\sin(ft))\hat{\mathbf{j}} + (c_5\cos(ft) + c_6\sin(ft))\hat{\mathbf{k}} \end{align} Then just do the grinding to get the final result.

EDIT: I just noticed that it might be easier to replace quantities like $\ddot{\mathbf{r}}(t) = f\dot{\mathbf{r}}(t) \times \mathbf{k}$ into the curvature expression as well.

EDIT2: It's been a while since I"ve dealt with this... Showing that torsion is $0$ is important too.

$\endgroup$
2
  • $\begingroup$ I am not quite familiar with the concept of "curvature", so could you please kindly explain why this quantity being constant implies that $\mathbf r$ traces out a circle? $\endgroup$ Nov 20 '18 at 2:14
  • $\begingroup$ @glowstonetrees It actually takes a little while to get there if you haven't seen curvature before! This answer can get you started on that path though if you want to head down that way: math.stackexchange.com/questions/1505638/… $\endgroup$
    – AlkaKadri
    Nov 20 '18 at 2:28

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.