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.

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About Quirino M. Sugon Jr
Theoretical Physicist in Manila Observatory

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