91Ó°ÊÓ

Torque is a concept that describes a force's ability to cause an object to rotate around an axis. It is a vector quantity, which means it has both magnitude and direction.
To compute torque, we use the formula \( \overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F} \), where - \( \overrightarrow{r} \) is the position vector, indicating the point of application of the force relative to the axis of rotation.- \( \overrightarrow{F} \) is the applied force vector.
This formula utilizes the cross product, a mathematical operation that helps calculate the perpendicular force component relative to the radius, producing torque.
In our example, the torque calculation showed that the applied force of 4.0 N along the x-axis, at a given position, resulted in a torque vector of \( 0 \hat{\mathbf{i}} + 16.0 \hat{\mathbf{j}} - 8.0 \hat{\mathbf{k}} \ text{N} \cdot \text{m} \). This means the force causes rotation mainly in the direction of the y-axis and negatively along the z-axis.

The cross product is essential in vector mathematics, especially when dealing with rotational forces like torque or angular momentum.
The cross product of two vectors, say \( \overrightarrow{A} \) and \( \overrightarrow{B} \), is a vector that is perpendicular to the plane containing \( \overrightarrow{A} \) and \( \overrightarrow{B} \). It has a magnitude equal to the area of the parallelogram formed by the vectors. The formula is:\[\overrightarrow{A} \times \overrightarrow{B} = (a_2b_3 - a_3b_2) \hat{\mathbf{i}} - (a_1b_3 - a_3b_1) \hat{\mathbf{j}} + (a_1b_2 - a_2b_1) \hat{\mathbf{k}} \]This operation helps determine the new vector's orientation and magnitude, crucial in computing angular momentum and torque.
In our problem, both torque and angular momentum are derived using the cross product to examine rotational effects relative to a certain axis.
The careful application of the cross product gives us the resultant vectors indicating the precise rotational influence of the forces and momentum involved.

Linear momentum is the product of an object's mass and its velocity, represented as a vector pointing in the direction of velocity. It is a fundamental concept in mechanics and is conserved in isolated systems, meaning that the total linear momentum remains constant unless acted upon by external forces.
The formula for linear momentum is given as:- \( \overrightarrow{p} = m \overrightarrow{v} \)where - \( m \) is the mass of the particle,- \( \overrightarrow{v} \) is its velocity vector.
Linear momentum is key in determining the angular momentum of a particle when combined with the position vector through the cross product.
In our case, with a mass of 1.00 kg and velocity of \( 7.0 \hat{\mathbf{i}} + 6.0 \hat{\mathbf{j}} \ \text{m/s} \), we calculated the linear momentum to be \( 7.0 \hat{\mathbf{i}} + 6.0 \hat{\mathbf{j}} \ \text{kg} \cdot \text{m/s} \). This momentum helps predict how our particle would contribute to the overall rotational dynamics around the origin when computing angular momentum.