91Ó°ÊÓ

Angular acceleration is a measure of how quickly the angular velocity of an object changes with time. In the given problem, the particle has a constant angular acceleration denoted by \(\frac{\text{Ï€}}{6} \text{rad} \cdot \text{s}^{-2}\).
Think of angular acceleration as the rotational counterpart to linear acceleration in straight-line motion.
When the particle moves in its vertical circular path, its rotational speed changes over time due to this constant rate of acceleration.
To find the angular velocity after a certain time, you use the formula \(\text{ω} = \text{ω}_0 + \text{α} t\), where \(\text{ω}_0\) is the initial angular velocity, \(\text{α}\) is the angular acceleration, and \(\text{t}\) is time.
In this problem, substituting \(\text{ω}_0 = 0\), \(\text{α} = \frac{\text{π}}{6}\), and \(\text{t} = 2 \text{s}\) gives us an angular velocity of \(\frac{\text{π}}{3} \text{rad} \cdot \text{s}^{-1}\).

Angular velocity describes how fast an object rotates or revolves relative to another point, in this case, the center of the circle. It is measured in radians per second (rad/s).
For our particle in the vertical circular motion, we calculated the angular velocity as \(\frac{\text{Ï€}}{3} \text{rad} \cdot \text{s}^{-1}\) after 2 seconds.
The formula used was \(\text{ω} = \text{ω}_0 + \text{α} t\), where each variable represents the same values as detailed above.
This angular velocity can be converted to linear speed to understand how fast the particle moves along the circular path using the relationship \(\text{v} = r \text{ω}\), where \(\text{r}\) is the radius of the circle.
In this problem, substituting \(\text{r} = 2 \text{m}\) and \(\text{ω} = \frac{\text{π}}{3} \text{rad} \cdot \text{s}^{-1}\) gives us a linear speed of \(\frac{2\text{π}}{3} \text{m} \cdot \text{s}^{-1}\).

Linear displacement refers to the change in position of a particle along the path of motion. For circular motion, it’s the distance along the circumference.
Given the angular displacement \(\text{θ}\), the linear displacement \(\text{s}\) can be found using the formula \(\text{s} = r \text{θ}\).
In our problem, we initially calculated the angular displacement using \(\text{θ} = \text{ω}_0 t + \frac{1}{2} \text{α} t^2\), which came out to be \(\frac{\text{π}}{3} \text{radians}\).
Using the radius \(\text{r} = 2 \text{m}\), we find the linear displacement as \(\text{s} = 2 \times \frac{\text{Ï€}}{3} = \frac{2\text{Ï€}}{3} \text{m}\).
This helps us understand the exact distance the particle has traveled along its circular path after 2 seconds.

Kinematics in physics describes the motion of objects without considering the forces that cause the motion.
When dealing with circular motion, kinematic equations help determine various aspects like displacement, velocity, and acceleration.
For our particle, we used the equation \(\text{ω} = \text{ω}_0 + \text{α} t\) to find angular velocity and \(\text{θ} = \text{ω}_0 t + \frac{1}{2} \text{α} t^2\) for angular displacement.
These equations are analogous to those used in linear kinematics but are adapted for rotational motion.
Understanding these relationships is crucial for solving problems related to rotational motion efficiently.

Circular motion equations are essential tools for analyzing objects moving along circular paths.
Key formulas include:

• Angular Velocity: \(\text{ω} = \text{ω}_0 + \text{α} t\)
• Angular Displacement: \(\text{θ} = \text{ω}_0 t + \frac{1}{2} \text{α} t^2\)
• Linear Speed: \(\text{v} = r \text{ω}\)
• Linear Displacement: \(\text{s} = r \text{θ}\)

Each formula helps transform between angular and linear measurements.
For our specific problem, these equations allowed us to understand the particle's speed and displacement in its circular path.
With the grasp of these core equations, analyzing any circular motion scenario becomes much more manageable.