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Angular velocity is a key concept in understanding uniform circular motion. It represents how quickly a particle rotates around a circular path. In mathematical terms, angular velocity, denoted by \( \omega \), is defined as the rate of change of the angular position of an object with respect to time. This can be calculated using the formula:

• \( \omega = \frac{2\pi}{T} \)

where \( T \) is the period of the motion (the time taken for one complete cycle around the circle).
In the exercise, the period \( T \) is given as 7.00 seconds. By substituting this into our formula, we can calculate the angular velocity:

• \( \omega = \frac{2\pi}{7.00} \approx 0.897 \) rad/s

This tells us how many radians the particle travels through per second as it moves along its circular path.

In uniform circular motion, the velocity vector is constantly changing direction while maintaining a constant speed. The velocity vector represents the instantaneous speed and direction of a particle at any point along its path. It is always tangent to the path of motion because velocity must be perpendicular to the radius at any point on the circle.
To find the magnitude of the velocity vector, you use the radius \( r \) and the angular velocity \( \omega \):

• \( v = \omega r \)

In our example, the magnitude of the position vector or radius is found to be \( \sqrt{13} \) m, hence:
\( v = \left(\frac{2\pi}{7.00}\right) \sqrt{13} \approx 4.98 \) m/s.
The direction is given by rotating the position vector by 90 degrees clockwise, resulting in components.

• \( \vec{v} = \frac{2\pi}{7.00}(-3 \hat{i} + 2 \hat{j}) \approx (-2.69 \hat{i} + 1.79 \hat{j}) \) m/s

Unit-vector notation is a way of expressing vectors in terms of their components along the coordinate axes. This is useful for visualizing the direction and magnitude of a vector in a multi-dimensional space.
In the context of circular motion, we can represent vectors such as the position \( \vec{r} \) and velocity \( \vec{v} \) using unit vectors \( \hat{i} \) and \( \hat{j} \) along the x and y axes, respectively.
For the given problem, the position vector is expressed as:

• \( \vec{r} = (2.00 \, \mathrm{m}) \hat{i} - (3.00 \, \mathrm{m}) \hat{j} \)

This tells us that the particle is 2 m in the positive x-direction and 3 m in the negative y-direction from the origin. Additionally, the calculated velocity vector:

• \( \vec{v} = (-2.69 \hat{i} + 1.79 \hat{j}) \) m/s

informs us that the velocity is directed predominantly downwards in the third quadrant of the coordinate plane.

The period of motion refers to the time taken for a particle to complete one full rotation around a circular path. This is a fundamental aspect of uniform circular motion, as it helps determine other quantities like angular velocity.
Given in the exercise, the period \( T \) is 7.00 seconds. This means that the particle takes exactly 7 seconds to make one complete circle. Knowing the period enables us to determine the speed and frequency of the circular motion using formulas:

• Frequency: \( f = \frac{1}{T} = \frac{1}{7.00} \approx 0.143 \text{ Hz} \)
• Angular velocity: \( \omega = \frac{2\pi}{T} \)

Understanding the period is crucial because it is directly related to how fast the particle is moving along its path and how frequently the motion repeats itself.