91Ó°ÊÓ

Angular momentum is a fundamental concept in rotational mechanics. It is similar to linear momentum but for rotational motion. If you compare it to how linear momentum is the product of mass and velocity, angular momentum involves the object's mass distribution (moment of inertia) and its rotational velocity (angular velocity). The formula for angular momentum \(\vec{L}\) is: \[ \vec{L} = \vec{r} \times \vec{p} \] Here, \(\vec{r}\) is the position vector, and \(\vec{p}\) is the linear momentum.
In the context of the given exercise, we had a particle at a specific position with a given velocity. By calculating the position and velocity vectors, we determined the linear momentum. Using the cross product, we obtained the angular momentum about the origin.

The cross product, also known as the vector product, is an operation on two vectors in three-dimensional space. It results in a vector perpendicular to both original vectors.
Mathematically, for two vectors \(\vec{A}=(A_x\hat{\mathtt{i}}, A_y\hat{\mathtt{j}}, A_z\hat{\mathtt{k}})\) and \(\vec{B}=(B_x\hat{\mathtt{i}}, B_y\hat{\mathtt{j}}, B_z\hat{\mathtt{k}})\), the cross product is:
\[\begin{vmatrix} \hat{\mathtt{i}} & \hat{\mathtt{j}} & \hat{\mathtt{k}} \ A_x & A_y & A_z \ B_x & B_y & B_z \end{vmatrix}\]
This determinant results in a new vector: \[\vec{C} = (A_y B_z - A_z B_y)\hat{\mathtt{i}} + (A_z B_x - A_x B_z)\hat{\mathtt{j}} + (A_x B_y - A_y B_x)\hat{\mathtt{k}} \] In the problem, we computed the cross product of the position vector \(\vec{r}\) and the linear momentum \(\vec{p}\) to find the angular momentum \(\vec{L}\) and later \(\vec{r}\) with force \(\vec{F}\) to find the torque \(\vec{\tau}\).

Torque is essentially the rotational equivalent of force. It measures how much a force acting on an object causes it to rotate.
The formula for torque \(\vec{\tau}\) is given by: \[\vec{\tau} = \vec{r} \times \vec{F} \] Here, \(\vec{r}\) represents the position vector from the axis of rotation to the point of force application, and \(\vec{F}\) is the force vector.
In our problem, the particle experienced a force in the negative \(\hat{\mathtt{i}}\) direction. By applying the cross product between the position vector \(\vec{r}\) and the force vector \(\vec{F}\), we found that the torque about the origin was \(\vec{\tau} = -56 \hat{\mathtt{k}} \ \mathrm{N \cdot m}\). This negative sign suggests that the torque caused a clockwise rotation around the origin.

Newton's second law for rotational motion is analogous to his second law for translational motion: it states that the net torque \(\vec{\tau}\) acting on an object is equal to the rate of change of its angular momentum \(\vec{L}\). Mathematically: \[ \vec{\tau} = \frac{d \vec{L}}{dt} \] This principle helps us understand how external torques affect the rotational state of an object.
In our exercise, we found the torque acting on the particle to be \( -56 \ \mathrm{N \cdot m} \ \hat{\mathtt{k}}\). According to Newton's second law for rotation, this torque equals the rate of change of the particle's angular momentum, i.e., \. \frac{d \vec{L}}{dt} = -56 \ \mathrm{N \cdot m} \ \hat{\mathtt{k}} \.
This information indicates that the particle's rotational momentum is changing at a rate of \ -56 \ \mathrm{N \cdot m} \ \hat{\mathtt{k}}.