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When studying the movement of particles in a system, angular momentum is a key concept. Angular momentum, often denoted as \( \vec{L} \), is the rotational equivalent of linear momentum. In simple terms, it describes how much "spin" an object has based on its motion about some point, often the origin.
For a single particle, angular momentum is calculated as \( \vec{L} = \vec{r} \times \vec{p} \), where \( \vec{r} \) is the position vector and \( \vec{p} \) is the linear momentum vector of the particle. The linear momentum can be expressed as \( \vec{p} = m \vec{v} \) where \( m \) is the mass and \( \vec{v} \) is the velocity.
Analyzing how angular momentum changes with time is critical. This change is directly related to torque, much like how force relates to linear momentum. Here, positive torque indicates that angular momentum increases over time. In the given problem, the torque was confirmed to be positive; thus, the angular momentum of the particle is indeed increasing.

Understanding the cross product is vital when dealing with angular momentum and torque. The cross product is a mathematical operation applied to two vectors resulting in a third vector that is perpendicular to the plane formed by the initial vectors.
For two vectors \( \vec{A} \) and \( \vec{B} \), the cross product \( \vec{A} \times \vec{B} \) can be computed using a determinant of a matrix:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ A_x & A_y & A_z \ B_x & B_y & B_z \end{vmatrix} \]Where \( \hat{\mathrm{i}} \), \( \hat{\mathrm{j}} \), and \( \hat{\mathrm{k}} \) are the unit vectors in the Cartesian coordinate system.
In the context of the problem, the position vector \( \vec{r} \) and force vector \( \vec{F} \) are crossed to find the torque \( \vec{\tau} \). This results in a vector that indicates rotational motion relative to the origin, which is directed along the \( \hat{\mathrm{k}} \)-axis in this exercise.

Newton's Second Law is a cornerstone of classical mechanics. It fundamentally explains how the velocity of an object changes when it is subjected to an external force.
This law is succinctly expressed as \( \vec{F} = m \vec{a} \), where \( \vec{F} \) is the force applied to an object, \( m \) is the mass of the object, and \( \vec{a} \) is the resulting acceleration.
In its usual linear form, it tells us that the change in momentum (which relates directly to linear motion) is proportional to the applied force. However, if we translate this concept into rotational motion, it forms the basis for how torque is linked to angular acceleration.
In our exercise, we first found the acceleration of the particle and then used Newton's Second Law to find the force. This force was later used to calculate the torque via the cross product with the position vector.

The derivative of position with respect to time is a fundamental concept leading to an understanding of a particle’s motion. From the position function, we can determine velocity and further acceleration.
Velocity \( \vec{v}(t) \) is the first derivative of the position vector \( \vec{r}(t) \). It describes how fast and in which direction the particle is moving at any given time. The formula for this derivative is:
\[ \vec{v} = \frac{d\vec{r}}{dt} \]Continuing this logic, the second derivative \( \frac{d\vec{v}}{dt} \) gives us the acceleration \( \vec{a} \). This concept allows us to analyze how the velocity of the particle changes with time and is essential for understanding dynamics problems.
In the provided solution, we see how the position's derivative was vital in determining both velocity and acceleration, which were then used to move forward with analyzing forces and torque acting on the particle.