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

Question

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

Accepted Answer

1. **Velocity Vector ($\mathbf{v}$):** Velocity is the time derivative of the position vector $\mathbf{r}$.
   $$ \mathbf{v} = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}[\cos(\omega t)\hat{x} + \sin(\omega t)\hat{y}] = -\omega \sin(\omega t)\hat{x} + \omega \cos(\omega t)\hat{y} $$
2. **Check Perpendicularity ($\mathbf{v} \cdot \mathbf{r}$):** Calculate the dot product of velocity and position vectors.
   $$ \mathbf{v} \cdot \mathbf{r} = [-\omega \sin(\omega t)] [\cos(\omega t)] + [\omega \cos(\omega t)] [\sin(\omega t)] $$
   $$ \mathbf{v} \cdot \mathbf{r} = -\omega \sin(\omega t)\cos(\omega t) + \omega \sin(\omega t)\cos(\omega t) = 0 $$
   Since the dot product is zero, the velocity is **perpendicular** to the position vector $\mathbf{r}$.
3. **Acceleration Vector ($\mathbf{a}$):** Acceleration is the time derivative of velocity.
   $$ \mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}[-\omega \sin(\omega t)\hat{x} + \omega \cos(\omega t)\hat{y}] $$
   $$ \mathbf{a} = -\omega^2 \cos(\omega t)\hat{x} - \omega^2 \sin(\omega t)\hat{y} = -\omega^2 [\cos(\omega t)\hat{x} + \sin(\omega t)\hat{y}] $$
   $$ \mathbf{a} = -\omega^2 \mathbf{r} $$
4. **Direction of Acceleration:** The relation $\mathbf{a} = -\omega^2 \mathbf{r}$ shows that acceleration is antiparallel to the position vector $\mathbf{r}$. Since $\mathbf{r}$ is directed outward from the origin, acceleration is directed **towards the origin** (Centripetal Acceleration) .

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