Problem 28 - Fundamentals of Physics

Question

A \(2.0 \mathrm{~kg}\) particle-like object moves in a plane with velocity components \(v_{x}=30 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=60 \mathrm{~m} / \mathrm{s}\) as it passes through the point with \((x, y)\) coordinates of \((3.0,-4.0) \mathrm{m} .\) Just then, in unitvector notation, what is its angular momentum relative to (a) the origin and (b) the point located at (-2.0,-2.0) \(\mathrm{m} ?\)

Step-by-Step Solution

Answer

(a) 600 kg⋅m²/s (b) 720 kg⋅m²/s

Step 1: Identify known values

The mass of the particle is given as \( m = 2.0 \, \text{kg} \). The velocity components are \( v_{x} = 30 \, \text{m/s} \) and \( v_{y} = 60 \, \text{m/s} \). The position coordinates of the particle are \( (x, y) = (3.0, -4.0) \).

Step 2: Calculate the velocity vector

The velocity vector \( \mathbf{v} \) of the object in unit-vector notation is \( \mathbf{v} = v_{x} \mathbf{i} + v_{y} \mathbf{j} = 30\mathbf{i} + 60\mathbf{j} \).

Step 3: Calculate angular momentum at the origin

The angular momentum \( \mathbf{L} \) relative to a point is defined as \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \), where \( \mathbf{r} \) is the position vector and \( \mathbf{p} = m\mathbf{v} \) is the momentum. The position vector at the origin \( \mathbf{r} = 3.0\mathbf{i} - 4.0\mathbf{j} \). The momentum vector \( \mathbf{p} = 2.0(30\mathbf{i} + 60\mathbf{j}) = 60\mathbf{i} + 120\mathbf{j} \). The cross product \( \mathbf{r} \times \mathbf{p} \) is evaluated as the determinant:\[\mathbf{r} \times \mathbf{p} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \3.0 & -4.0 & 0 \60 & 120 & 0\end{vmatrix} = \mathbf{k}(3.0 \cdot 120 - (-4.0) \cdot 60) = \mathbf{k}(360 + 240) = 600\mathbf{k}\]Thus, \( \mathbf{L} = 600\mathbf{k} \, \text{kg} \cdot \text{m}^2/\text{s} \).

Step 4: Calculate angular momentum at the point (-2.0, -2.0)

First, determine the position vector \( \mathbf{r} \) relative to the new point, \(-2.0, -2.0\), as \( \mathbf{r} = (3.0 - (-2.0))\mathbf{i} + (-4.0 - (-2.0))\mathbf{j} = 5.0\mathbf{i} - 2.0\mathbf{j} \). Using the same momentum vector \( \mathbf{p} = 60\mathbf{i} + 120\mathbf{j} \), calculate the cross product:\[\mathbf{r} \times \mathbf{p} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \5.0 & -2.0 & 0 \60 & 120 & 0\end{vmatrix} = \mathbf{k}(5.0 \cdot 120 - (-2.0) \cdot 60) = \mathbf{k}(600 + 120) = 720\mathbf{k}\]Therefore, \( \mathbf{L} = 720\mathbf{k} \, \text{kg} \cdot \text{m}^2/\text{s} \).

Key Concepts

Velocity Vector

In the context of physics, a velocity vector indicates both how fast an object is moving and in what direction. It's a combination of both the speed and direction, which makes it a vector quantity. For the exercise in question, the velocity vector is expressed in unit vector notation as \( \mathbf{v} = v_{x} \mathbf{i} + v_{y} \mathbf{j} \). This represents the horizontal and vertical velocity components.
The velocity components given are \( v_{x} = 30 \text{ m/s} \) and \( v_{y} = 60 \text{ m/s} \). When these are combined, the velocity vector becomes \( 30 \mathbf{i} + 60 \mathbf{j} \). This means the object is moving rightward at 30 m/s and upward at 60 m/s.
Understanding the velocity vector is crucial for determining other dynamics of the motion, especially when computing quantities like momentum and angular momentum.

Cross Product

The cross product is an operation between two vectors that results in another vector, which is perpendicular to the plane formed by the two input vectors. In the exercise, the cross product \( \mathbf{r} \times \mathbf{p} \) involves the position vector \( \mathbf{r} \) and the momentum vector \( \mathbf{p} \).
This operation is essential in calculating angular momentum, which is expressed as \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \). The cross product in this context uses the determinant of a 3x3 matrix to find the resulting vector's components. The determinant arranges the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and the components of \( \mathbf{r} \) and \( \mathbf{p} \) into a matrix. Solving the determinant helps us find a vector perpendicular to \( \mathbf{r} \) and \( \mathbf{p} \).
In practice, this process helps find the angular momentum of a particle, critical for understanding rotational dynamics.

Momentum

Momentum is a fundamental concept in physics, representing the quantity of motion of a moving body. It is calculated as the product of an object's mass and its velocity. The formula for momentum \( \mathbf{p} \) is \( \mathbf{p} = m \mathbf{v} \), where \( m \) is mass and \( \mathbf{v} \) is velocity.
In this exercise, given the mass \( m = 2.0 \text{ kg} \) and a velocity vector of \( 30 \mathbf{i} + 60 \mathbf{j} \), the momentum vector is \( \mathbf{p} = 2.0(30 \mathbf{i} + 60 \mathbf{j}) = 60 \mathbf{i} + 120 \mathbf{j} \). Momentum is a vector quantity, meaning it has both magnitude and direction.
Understanding momentum is essential for solving problems related to forces, collisions, and motion, including scenarios that involve angular momentum.

Position Vector

A position vector describes the location of a point in space relative to an origin. In this exercise, the position vector \( \mathbf{r} \) represents the point through which the particle passes. For example, relative to the origin at \((0,0)\), the position vector is \( 3.0 \mathbf{i} - 4.0 \mathbf{j} \).
Changing references, such as considering the position vector relative to a different point (like \((-2.0, -2.0)\)), affects the calculations of angular momentum. By shifting the origin, the new position vector becomes \( 5.0 \mathbf{i} - 2.0 \mathbf{j} \).
The position vector is critical in calculating the angular momentum because it defines the axis about which the rotation is calculated. Knowing how to compute the position vector allows for determining relationships among points in physical space, an indispensable tool in physics.