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Angular velocity is a fundamental concept in circular motion, representing how fast an object rotates or revolves around a central point. It is usually denoted by the symbol \( \omega \). Angular velocity is defined as the rate of change of the angular position of a rotating object, and it indicates how quickly an angle is swept out. Its units are typically radians per second. In our context, the expression \( \omega = \frac{v}{R} \) relates the linear speed \( v \) of the object moving in a circle of radius \( R \), highlighting how linear and angular parameters interrelate.

In vector notation for circular motion, the angular velocity vector \( \overrightarrow{\boldsymbol{\omega}} \) is aligned with the axis of rotation. For an object moving counterclockwise in the \( xy \)-plane, this vector points along the positive \( z \)-axis, represented by \( \omega \hat{\mathbf{k}} \). This direction adheres to the right-hand rule: if the fingers curl in the direction of rotation, the thumb will point along the direction of the angular velocity vector. Understanding this concept is pivotal as it helps us relate linear and angular motion, and solves problems involving rotational dynamics.

The velocity vector \( \overrightarrow{\mathbf{v}} \) describes the speed and direction of an object at a specific point in time. For circular motion, finding this vector involves differentiating the position vector of the object with respect to time. In our exercise, the position vector is given as \( \overrightarrow{\mathbf{r}} = \hat{\mathbf{i}} R \cos \omega t + \hat{\mathbf{j}} R \sin \omega t \).

Upon differentiating each component of this vector, we calculate:

• For the \( \hat{\mathbf{i}} \) component: \( \frac{d}{dt}(R \cos \omega t) = -R \omega \sin \omega t \)
• For the \( \hat{\mathbf{j}} \) component: \( \frac{d}{dt}(R \sin \omega t) = R \omega \cos \omega t \)

Thus, the velocity vector is \( \overrightarrow{\mathbf{v}} = R \omega(-\sin \omega t \hat{\mathbf{i}} + \cos \omega t \hat{\mathbf{j}}) \). The negative sign in the \( \hat{\mathbf{i}} \) component indicates that the velocity is tangent to the circle and points in the direction of motion. Visualizing the velocity vector tangential to the path of the object provides a clear understanding of its motion, which is constant in magnitude for circular motion due to the fixed radius and constant angular speed.

The cross product is a vector operation that results in a vector perpendicular to the plane formed by two input vectors. It is denoted by \( \times \) and follows specific rules to determine the direction and magnitude of the resulting vector. In the context of circular motion, the cross product is utilized to relate angular velocity \( \overrightarrow{\boldsymbol{\omega}} \) with linear velocity \( \overrightarrow{\mathbf{v}} \).

Mathematically, the expression \( \overrightarrow{\mathbf{v}} = \overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\mathbf{r}} \) confirms that the velocity is the result of this cross product, involving the angular velocity vector and position vector. Here's the setup in identifying the cross product for such vectors:

• \( \overrightarrow{\boldsymbol{\omega}} = \omega \hat{\mathbf{k}} \)
• \( \overrightarrow{\mathbf{r}} = \hat{\mathbf{i}} R \cos \omega t + \hat{\mathbf{j}} R \sin \omega t \)

To calculate \( \overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\mathbf{r}} \), use the determinant:

\[\begin{vmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \0 & 0 & \omega \R \cos \omega t & R \sin \omega t & 0\end{vmatrix}\]

Solving the determinant leads to \( \overrightarrow{\mathbf{v}} = - \omega R \sin \omega t \hat{\mathbf{i}} + \omega R \cos \omega t \hat{\mathbf{j}} \), which matches the velocity vector obtained earlier. This consistency not only proves the relation but also demonstrates the utility of cross product in relating rotational and linear quantities in physics, providing deeper insights into three-dimensional vector interactions.