Let's analyze and solve the problem step by step.

Problem Breakdown:

Given:

• The position vector of the particle $\vec{r}$ is: $\vec{r} = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}$ where $\omega$ is a constant.

We are asked to show:

1. (a) The velocity vector $\vec{v}$ of the particle is perpendicular to $\vec{r}$.
2. (b) $\vec{r} \times \vec{v}$ is a constant vector.

(a) Show that $\vec{v}$ is perpendicular to $\vec{r}$:

The velocity vector $\vec{v}$ is the time derivative of the position vector $\vec{r}$.

Given: $\vec{r} = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}$

Find the velocity vector by differentiating $\vec{r}$ with respect to $t$: $\vec{v} = \frac{d}{dt}\left[\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}\right]$

Applying the chain rule: $\vec{v} = -\omega \sin(\omega t) \hat{i} + \omega \cos(\omega t) \hat{j}$

Now, for two vectors to be perpendicular, their dot product must be zero: $\vec{r} \cdot \vec{v} = 0$

Compute the dot product: [ \vec{r} \cdot \vec{v} = \left[\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}\right] \cdot \left[-\omega \sin(\omega t) \hat{i} + \omega \cos(\omega t) \hat{j}\right] ]

This expands to: $\vec{r} \cdot \vec{v} = \cos(\omega t)(-\omega \sin(\omega t)) + \sin(\omega t)(\omega \cos(\omega t))$

Simplifying: $\vec{r} \cdot \vec{v} = -\omega \cos(\omega t) \sin(\omega t) + \omega \sin(\omega t) \cos(\omega t)$ $\vec{r} \cdot \vec{v} = 0$

Thus, the velocity vector $\vec{v}$ is perpendicular to the position vector $\vec{r}$.

(b) Show that $\vec{r} \times \vec{v}$ is a constant vector:

The cross product of $\vec{r}$ and $\vec{v}$ is: [ \vec{r} \times \vec{v} = \left[\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}\right] \times \left[-\omega \sin(\omega t) \hat{i} + \omega \cos(\omega t) \hat{j}\right] ]

Using the property of the cross product: $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = 0 \quad \text{and} \quad \hat{i} \times \hat{j} = \hat{k}$

Expanding the cross product: $\vec{r} \times \vec{v} = \cos(\omega t)\omega \cos(\omega t)(\hat{i} \times \hat{j}) + \sin(\omega t)(-\omega \sin(\omega t))(\hat{j} \times \hat{i})$

Simplifying this: $\vec{r} \times \vec{v} = \omega(\cos^2(\omega t) + \sin^2(\omega t))\hat{k}$

Using the trigonometric identity $\cos^2(\omega t) + \sin^2(\omega t) = 1$: $\vec{r} \times \vec{v} = \omega \hat{k}$

Thus, $\vec{r} \times \vec{v} = \omega \hat{k}$, which is a constant vector (since $\omega$ is constant and $\hat{k}$ is a unit vector in the z-direction).

Conclusion:

• (a) The velocity vector $\vec{v}$ is perpendicular to the position vector $\vec{r}$.
• (b) The cross product $\vec{r} \times \vec{v}$ is a constant vector.

Would you like more details or have any questions?

Here are 5 related questions to further explore this topic:

1. What is the physical interpretation of a particle's velocity being perpendicular to its position vector?
2. How does this result relate to circular motion?
3. Can the result of $\vec{r} \times \vec{v}$ being constant apply to non-uniform circular motion?
4. What happens if $\omega$ is not a constant?
5. How does the magnitude of the cross product relate to angular momentum?

Tip: The cross product of position and velocity vectors is often related to angular momentum in physics.