Exercise 8b

\[\vec{r}(t) = \left( \begin{matrix} \cos (\omega t - \phi) \\ \sin(\omega t - \phi) \end{matrix} \right)\] \[\vec{v}(t) = \left( \begin{matrix} - \omega \sin (\omega t - \phi) \\ \omega \cos(\omega t - \phi) \end{matrix} \right)\] \[ \left| \vec{v} \right| = \sqrt{ (- \omega \sin (\omega t - \phi))^2 + (\omega \cos(\omega t - \phi))^2}\] \[= \omega \sqrt{ (\sin (\omega t - \phi))^2 + (\cos(\omega t - \phi) )^2} = \omega \] \[\vec{a}(t) = \left( \begin{matrix} - \omega^2 \cos (\omega t - \phi) \\ - \omega^2 \sin(\omega t - \phi) \end{matrix} \right) = - \omega^2\, \vec{r}(t)\]

Animation

position vector: red
velocitiy vector: blue
accelaration vector: green

The position vector moves on a circle with angular velocity \(\omega\). The velocity vector is always perpendicular to the position vector. The acceleration vector is oriented in the opposite direction of the position vector (centripetal force).

\(\phi\) was set to zero (does only a phase shift with respect to time).