Problem 27 - Fundamentals of Physics

Question

At one instant, force \(\vec{F}=4.0 \mathrm{j} \mathrm{N}\) acts on a \(0.25 \mathrm{~kg}\) object that has position vector \(\vec{r}=(2.0 \mathrm{i}-2.0 \mathrm{k}) \mathrm{m}\) and velocity vector \(\vec{v}=(-5.0 \hat{i}+5.0 \mathrm{k}) \mathrm{m} / \mathrm{s}\). About the origin and in unit-vector notation, what are (a) the object's angular momentum and (b) the torque acting on the object?

Step-by-Step Solution

Answer

(a) \( \vec{L} = \vec{0} \), (b) \( \vec{\tau} = 8.0 \hat{i} + 8.0 \hat{k} \).\)

Step 1: Find Angular Momentum

Angular momentum \( \vec{L} \) is given by \( \vec{L} = \vec{r} \times \vec{p} \), where \( \vec{p} \) is the linear momentum. First, calculate \( \vec{p} \) using \( \vec{p} = m \vec{v} \). Thus, \( \vec{p} = 0.25 \times (-5.0 \hat{i} + 5.0 \hat{k}) = -1.25 \hat{i} + 1.25 \hat{k} \). Now, compute \( \vec{L} \): \[ \vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2.0 & 0 & -2.0 \ -1.25 & 0 & 1.25 \end{vmatrix} \] Evaluating the determinant gives: \[ \vec{L} = \hat{j} (2.0 \times 1.25 - (-2.0) \times (-1.25)) = \hat{j} (2.5 - 2.5) = \vec{0} \]. Therefore, the angular momentum is \( \vec{L} = \vec{0} \).

Step 2: Find the Torque

Torque \( \vec{\tau} \) is given by \( \vec{\tau} = \vec{r} \times \vec{F} \). Given \( \vec{F} = 4.0 \hat{j} \) and \( \vec{r} = 2.0 \hat{i} - 2.0 \hat{k} \), calculate \( \vec{\tau} \): \[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2.0 & 0 & -2.0 \ 0 & 4.0 & 0 \end{vmatrix} \] Evaluating the determinant gives: \[ \vec{\tau} = \hat{i} (0 \times 0 - 4 \times (-2.0)) + \hat{j} (-2.0 \times 0 - 2.0 \times 0) + \hat{k} (2.0 \times 4.0 - 0 \times 0) \]\[ \vec{\tau} = \hat{i} (8.0) + \hat{k} (8.0) = 8.0 \hat{i} + 8.0 \hat{k} \] Thus, the torque is \( \vec{\tau} = 8.0 \hat{i} + 8.0 \hat{k} \).

Key Concepts

Torque

Torque is a fundamental concept in physics, often symbolized by \( \vec{\tau} \). It describes the tendency of a force to rotate an object around an axis. You can think of it as a twist or a rotational equivalent of force.
The equation for torque is \( \vec{\tau} = \vec{r} \times \vec{F} \), where \( \vec{r} \) is the position vector (the distance and direction from the pivot point to the point of force application) and \( \vec{F} \) is the force vector.

• The greater the distance (magnitude of \( \vec{r} \)), the larger the torque if the force is applied perpendicularly.
• If the force is aligned with \( \vec{r} \), the torque is zero.
• The direction of \( \vec{\tau} \) is given by the right-hand rule, which is useful for determining the sense of rotation.

When calculating torque, using the cross product calculations helps find the precise direction and magnitude of this rotational force.

Cross Product

The cross product is a mathematical operation, often used in physics, that takes two vectors and produces another vector that is perpendicular to both. In terms of two vectors \( \vec{A} \) and \( \vec{B} \), the cross product \( \vec{A} \times \vec{B} \) is calculated as:
\[\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]
Here, \( a_1, a_2, a_3 \) and \( b_1, b_2, b_3 \) are the components of \( \vec{A} \) and \( \vec{B} \). The determinant is used to calculate the resulting vector perpendicular to \( \vec{A} \) and \( \vec{B} \).

• The magnitude of the cross product is given by \( |\vec{A}| |\vec{B}| \sin(\theta) \), where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \).
• Vector direction: follow the right-hand rule to determine the direction of the result.

Understanding the cross product is essential for solving problems related to rotational forces and motions, like calculating torque and angular momentum.

Linear Momentum

Linear momentum is a measure of an object's motion, denoted by \( \vec{p} \). It is the product of an object's mass and its velocity expressed as \( \vec{p} = m \vec{v} \). This means:

• Mass \( (m) \): how much matter is in the object.
• Velocity \( (\vec{v}) \): the speed and direction of the object's motion.

The direction of linear momentum is the same as velocity's. It is a vector quantity, meaning it has both magnitude and direction.
In physics, linear momentum is conserved in isolated systems, meaning \( \vec{p} \) remains the same if no external forces act on the system.
In the exercise, calculating linear momentum was used to find angular momentum, showcasing its role in linking linear and rotational dynamics.

Determinant Evaluation

Determinant evaluation is a crucial mathematical technique used in this context to compute the cross product of vectors. This method involves calculating the determinant of a 3x3 matrix comprised of the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of the vectors involved.
For example, for two vectors \( \vec{r} = (r_1, r_2, r_3) \) and \( \vec{F} = (f_1, f_2, f_3) \):
\[\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ r_1 & r_2 & r_3 \ f_1 & f_2 & f_3 \end{vmatrix}\]

• Determine the product for each part by using the formula \( \text{Det} = \hat{i}(b_2c_3 - b_3c_2) - \hat{j}(a_1c_3 - a_3c_1) + \hat{k}(a_1b_2 - a_2b_1) \).
• Each term involves computations between components alongside applying sign changes.

Determinant evaluation not only aids in finding cross products but also appears in various mathematical applications in physics and engineering, making it an essential tool for students.