91Ó°ÊÓ

Torque is a measure of how much a force acting on an object causes that object to rotate. It is not just about pushing or pulling something but twisting it. The strength of the torque depends on three factors. First, the amount of force applied. Second, the distance from the point of rotation. Finally, the angle at which the force is applied. The formula for torque is given by\[\mathbf{\tau} = \mathbf{r} \times \mathbf{F}\]where \(\mathbf{\tau}\) is the torque, \(\mathbf{r}\) is the position vector from the point of rotation to the point where the force is applied, and \(\mathbf{F}\) is the force vector. In our specific exercise, the particle had a constant velocity. This means there was no net force acting on it, resulting in zero torque being needed to maintain its motion. When no external forces or twists act on an object, the torque remains zero since torque is directly related to a change in angular momentum.

Linear momentum is a fundamental concept in physics that describes the quantity of motion an object possesses. It connects mass and velocity in a single measurement. Essentially, it signifies how much "oomph" an object has while moving.The mathematical expression is:\[ p = m \cdot v \]where \(p\) is the linear momentum, \(m\) is the mass, and \(v\) is the velocity. This expression highlights that a more massive object or a faster-moving one will have greater momentum. In the given problem, the momentum of the particle is calculated as the mass of the particle, \(3\text{ kg}\), multiplied by its speed, \(2\text{ ms}^{-1}\). This gives us a linear momentum value of \(6 \hat{j} \text{ kg ms}^{-1}\), indicating it's moving in the positive \(y\)-direction at \(2\text{ ms}^{-1}\). Since the particle's speed and direction remain unchanged, its linear momentum remains constant.

The cross product is a mathematical operation used in physics to determine a vector's direction and magnitude perpendicular to two other vectors. It is fundamental in calculating both torque and angular momentum.The cross product formula is:\[ \mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}| \sin(\theta) \hat{n} \]where \(\mathbf{a}\) and \(\mathbf{b}\) are vectors, \(\theta\) is the angle between them, and \(\hat{n}\) is the unit vector perpendicular to the plane containing \(\mathbf{a}\) and \(\mathbf{b}\). For the exercise, the cross product \(\mathbf{r} \times \mathbf{p}\) is calculated to find the angular momentum. Here, the position vector \(\mathbf{r} = 4 \hat{i} + y \hat{j}\) and linear momentum \(\mathbf{p} = 6 \hat{j}\) are used, resulting in \(4 \hat{i} \times 6 \hat{j} = 24 \hat{k}\). This illustrates that a calculated cross product results in a new vector pointing in the \(z\)-direction, perpendicular to both \(\mathbf{r}\) and \(\mathbf{p}\). Understanding the cross product is essential to navigating problems involving vectors in rotational dynamics.