# Uniform circular motion in a plane: differentiation of exponentials of imaginary vectors

The position $r$ of a point in circular motion about the origin is given by

$(1)\qquad \mathbf r=r\mathbf e_1e^{i\mathbf e_3(\omega t+\phi)}=\mathbf e_1r\Psi,$

where $r$ is the radius of rotation, $\omega$ is the angular frequency, and $\phi$ is the rotational phase angle. The velocity $\mathbf v$ and acceleration $\mathbf a$ of the particle is obtained by differentiating its position $\mathbf r$ with respect to time:

$(2a)\qquad\mathbf v=\frac{d\mathbf r}{dt}=r\mathbf e_1\frac{d}{dt}\Psi=\omega r\mathbf e_1i\mathbf e_3\Psi=\omega r\mathbf e_2\Psi,$
$(2b)\qquad\mathbf a=\frac{d\mathbf v}{dt}=\omega r\mathbf e_2\frac{d}{dt}\Psi=\omega^2 r\mathbf e_2i\mathbf e_3\Psi=-\omega^2 r\mathbf e_1\Psi.$

These may be rewritten as

$(3a)\qquad\mathbf v=\mathbf ri\mathbf e_3\omega,$
$(3b)\qquad\mathbf a=\mathbf vi\mathbf e_3\omega=\mathbf r(i\mathbf e_3\omega)^2=-\omega^2\mathbf r.$

Notice that the position vector $r$ of the particle is perpendicular to its velocity $\mathbf v$ and opposite to its acceleration $\mathbf a$. In terms of rectangular coordinates, these quantities are

$(4a)\qquad\mathbf r=r\cos\theta\mathbf e_1+r\sin\theta\mathbf e_2,$
$(4b)\qquad\mathbf v=-\omega r\sin\theta\mathbf e_1+\omega r\cos\theta\mathbf e_2,$
$(4c)\qquad\mathbf a=-\omega^2 r\cos\theta\mathbf e_1-\omega^2 r\sin\theta\mathbf e_2.$

About Quirino M. Sugon Jr
Theoretical Physicist in Manila Observatory