A particle moves so that its position vector is given by $\vec{r} = \cos(\omega t)\hat{x} + \sin(\omega t)\hat

Question

A particle moves so that its position vector is given by $\vec{r} = \cos(\omega t)\hat{x} + \sin(\omega t)\hat{y}$, where $\omega$ is a constant. Which of the following is true?

Accepted Answer

1. **Velocity Calculation:** Velocity is the time derivative of the position vector $\vec{r}$.
   $$ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}[\cos(\omega t)\hat{x} + \sin(\omega t)\hat{y}] = -\omega \sin(\omega t)\hat{x} + \omega \cos(\omega t)\hat{y} $$
2. **Check Perpendicularity:** Calculate the dot product of $\vec{r}$ and $\vec{v}$.
   $$ \vec{r} \cdot \vec{v} = (\cos \omega t)(-\omega \sin \omega t) + (\sin \omega t)(\omega \cos \omega t) = -\omega \sin \omega t \cos \omega t + \omega \sin \omega t \cos \omega t = 0 $$
   Since the dot product is zero, velocity $\vec{v}$ is perpendicular to position vector $\vec{r}$.
3. **Acceleration Calculation:** Acceleration is the time derivative of velocity.
   $$ \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}[-\omega \sin(\omega t)\hat{x} + \omega \cos(\omega t)\hat{y}] $$
   $$ \vec{a} = -\omega^2 \cos(\omega t)\hat{x} - \omega^2 \sin(\omega t)\hat{y} = -\omega^2 [\cos(\omega t)\hat{x} + \sin(\omega t)\hat{y}] $$
   $$ \vec{a} = -\omega^2 \vec{r} $$
4. **Direction of Acceleration:** The negative sign indicates that acceleration $\vec{a}$ is directed opposite to the position vector $\vec{r}$. Since $\vec{r}$ points away from the origin, $\vec{a}$ is directed **towards the origin** .

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