Prove that a particle moves along a circle

Question

Assume that the particle has the velocity of $v(t)$, and $v = (wk) \times r, w>0$, where $v(t) = r'(t)$. Apparently, $k$ is the unit direction vector of the vertical axis in the 3d-space.

I need to prove that the particle moves along a circle, with a constant angular speed $w$. I tried proving that the path has constant curvature, because if it is a circle, than the curvature should be constant everywhere, but:

$$\kappa(t) = \frac{\lVert r''(t) \times r'(t) \rVert}{\lVert r'(t) \rVert^3} = \frac{\lVert wk \rVert}{\lVert (wk) \times r \rVert^3} = \frac{w}{w^3\big(\lVert r(t)\rVert^2 - (k \cdot r(t))^2\big)^{3/2}}$$

I cannot simplify it further to arrive at a constant. How can I prove that the movement is along a circle?

Answer 1

You can dot multiply your condition by $\mathbf{r}(t)$. The triple product on the left will be zero and you obtain $$ \mathbf{r}'(t)\cdot \mathbf{r}(t)=\frac{1}{2}\frac{d}{dt}\|\mathbf{r}(t)\|^2=0 $$ and therefore $r=\|\mathbf{r}(t)\|$ is a constant. So the point is within a constant distance from the origin.

Next, you dot multiply by $w\mathbf{k}$. Again the left hand side is zero and you get $$ \mathbf{r}'(t)\cdot w\mathbf{k}=0 $$ that is $\mathbf{v}(t)$ is parallel to the $XY$ plane. Therefore your motion is circular on a plane parallel to the $XY$-plane. Finally, taking modules and taking into account that $\mathbf{r}(t)$ forms a constant angle $\alpha$ with $\mathbf{k}$ you have $$ v=\|\mathbf{v}(t)\|=wr\sin\alpha=wd $$ where $d$ is the distance from the particle to the $z$-axis which means exactly that your angular velocity is $w$.

Answer 2

Let's write $\vec r=x\hat i+y\hat j+z\hat k$, where $x$, $y$ and $z$ are functions of time. Then $$\vec v=x'\hat i+y'\hat j+z'\hat k=(\omega\hat k)\times(x\hat i+y\hat j+z\hat k)=\omega x\hat j-\omega y\hat i+0\hat k$$ From here you get $$\begin{align}x'&=-\omega y\\y'&=\omega x\\z'&=0\end{align}$$ The last equation tells you that the $z$ coordinate does not change, so the motion is in a plane perpendicular to $\hat k$.

Now take the derivative of the $x'$ and $y'$ with respect to time: $$x''=-\omega y'=-\omega^2x$$ and $$y''=\omega x'=-\omega^2 y$$ The solution of the first equation is $$x=A_x\sin(\omega t+\phi_x)$$ Similarly $$y=A_y\sin(\omega t+\phi_y)$$ Taking the derivatives you get $$x'=\omega A_x\cos(\omega t+\phi_x)=-\omega A_y\sin(\omega t+\phi_y)$$ and $$y'=\omega A_y\cos(\omega t+\phi_y)=\omega A_x\sin(\omega t+\phi_x)$$ From here you get $A_x=A_y$ and $\phi_x-\phi_y=\frac \pi 2$. Then the equations of motion are $$\begin{align}x&=A\cos(\omega t+\phi)\\y&=A\sin(\omega t+\phi)\\z&=z_0\end{align}$$ This is a circle of radius $A$ in the plane $z=z_0$, traversed with angular velocity $\omega$.

Answer 3

With

$r = (x, y, z), \tag 1$

we have

$wk \times r = (0, 0, w) \times (x, y, z) = -wy i + w x j = (-wy, wx, 0); \tag 2$

thus

$r' = v = wk \times r \tag 3$

becomes

$(x', y', z') = (-wy, wx, 0), \tag 4$

that is,

$x' = -wy, \tag 5$

$y' = wx, \tag 6$

$z' = 0; \tag 7$

from (7) we infer that

$z = z_0, \; \text{ a constant}; \tag 8$

thus the particle remains in the plane $(x, y, z_0)$; furthermore (5) and (6) imply

$(x^2 + y^2)' = 2xx' + 2yy' = -2wxy +2wxy = 0, \tag 9$

that is,

$x^2 + y^2 = \text{ a constant}, \tag 9$

which together with (8) shows that $r(t)$ lies in the circle of radius $\sqrt{x^2 + y^2}$ centered at $(0, 0, z_0)$ in the plane $z = z_0$; finally, we observe that (5)-(6) imply that

$(x')^2 + (y')^2 = w^2(x^2 + y^2) = w^2r^2 \ne 0 \tag{10}$

provided the circle is non-trivial; thus the particle traverses the entire circumference, and so indeed travels in a circular path.

These things may also be seen by writing, from (5) and (6),

$x'' = -wy' = -w^2 x, \tag{11}$

that is,

$x'' + w^2 x = 0; \tag{12}$

we also have

$y'' = wx' = -wy, \tag{13}$

$y'' + wy = 0; \tag{14}$

if we now pick some initial point such as $(x_0, y_0)$ then we obtain the initial derivatives from (5), (6):

$x'_0 = -wy_0, \tag{15}$

$y'_0 = w x_0; \tag{16}$

the solution for $x$ is well-known to be

$x(t) = x_0 \cos (wt) + \dfrac{x'_0}{w} \sin (wt) = x_0 \cos (wt) - y_0 \sin (wt), \tag{17}$

whilst that for $y$ is

$y(t) = y_0 \cos (wt) + \dfrac{y'_0}{w} \sin (w t) = y_0 \cos (wt) + x_0 \sin (w t). \tag{18}$

The reader may easily verify that

$x^2(t) + y^2(t) = x_0^2 + y_0^2, \tag{18}$

showing the motion lies in a circle. In fact we may write (17)-(18) in matrix-vector form:

$\begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = \begin{bmatrix} \cos (wt) & -\sin (wt) \\ \sin (wt) & \cos (wt) \end{bmatrix} \begin{pmatrix} x_0 \\ y_0 \end{pmatrix}; \tag{19}$

the reader may indeed recognize the $2 \times 2$ matrix on the right as belonging to $SO(2)$ for each $t$; as $t$ increases (or decreases), the entire circle is swept out by $(x(t), y(t))^T$.