Problem 30 - Fundamentals of Physics

Question

At the instant the displacement of a \(2.00 \mathrm{~kg}\) object relative to the origin is \(\vec{d}=(2.00 \mathrm{~m}) \hat{i}+(4.00 \mathrm{~m}) \mathrm{j}-(3.00 \mathrm{~m}) \mathrm{k},\) its velocity is \(\vec{v}=-(6.00 \mathrm{~m} / \mathrm{s}) \mathrm{i}+(3.00 \mathrm{~m} / \mathrm{s}) \mathrm{j}+(3.00 \mathrm{~m} / \mathrm{s}) \mathrm{k}\) and it is subject to a force \(\vec{F}=(6.00 \mathrm{~N}) \hat{\mathrm{i}}-(8.00 \mathrm{~N}) \mathrm{j}+(4.00 \mathrm{~N}) \mathrm{k} .\) Find (a) the acceleration of the object, (b) the angular momentum of the object about the origin, (c) the torque about the origin acting on the object, and (d) the angle between the velocity of the object and the force acting on the object.

Step-by-Step Solution

Answer

(a) \(3.00\hat{i} - 4.00\hat{j} + 2.00\hat{k}\) m/s², (b) \(66.00\hat{i} + 60.00\hat{j} + 60.00\hat{k}\) kg·m²/s, (c) \(-8.00\hat{i} + 26.00\hat{j} - 40.00\hat{k}\) N·m, (d) 128.74°.

Step 1: Calculate Acceleration

To find the acceleration of the object, use Newton's second law of motion, which states that \( \vec{F} = m \cdot \vec{a} \). Rearranging for acceleration gives \( \vec{a} = \frac{\vec{F}}{m} \). Here, \( \vec{F} = (6.00 \text{ N}) \hat{i} - (8.00 \text{ N}) \hat{j} + (4.00 \text{ N}) \hat{k} \) and mass \( m = 2.00 \text{ kg} \). Thus, \[ \vec{a} = \frac{1}{2.00} \left( 6.00 \hat{i} - 8.00 \hat{j} + 4.00 \hat{k} \right) \]Calculating each component, we get\[ \vec{a} = (3.00 \text{ m/s}^2) \hat{i} - (4.00 \text{ m/s}^2) \hat{j} + (2.00 \text{ m/s}^2) \hat{k} \].

Step 2: Calculate Angular Momentum

The angular momentum \( \vec{L} \) of an object about the origin is given by \( \vec{L} = \vec{r} \times \vec{p} \), where \( \vec{r} \) is the position vector and \( \vec{p} = m \cdot \vec{v} \) is the momentum. Here, \( \vec{r} = (2.00 \text{ m}) \hat{i} + (4.00 \text{ m}) \hat{j} - (3.00 \text{ m}) \hat{k} \). Momentum \( \vec{p} = m \cdot \vec{v} = 2.00 \cdot \left( -6.00 \hat{i} + 3.00 \hat{j} + 3.00 \hat{k} \right) = (-12.00 \hat{i} + 6.00 \hat{j} + 6.00 \hat{k}) \text{ kg m/s} \).To calculate the cross product, use the determinant:\[ \vec{L} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 2.00 & 4.00 & -3.00 \ -12.00 & 6.00 & 6.00 \end{array} \right| \] This simplifies to \[ \vec{L} = \hat{i}(4.00 \times 6.00 - (-3.00) \times 6.00) - \hat{j}(2.00 \times 6.00 - (-3.00) \times (-12.00))+ \hat{k}(2.00 \times 6.00 - 4.00 \times (-12.00)) \]\[ = (48.00 + 18.00)\hat{i} + (24.00 + 36.00)\hat{j} + (12.00 + 48.00)\hat{k} \]\[ = (66.00 \text{ kg m}^2/\text{s}) \hat{i} + (60.00 \text{ kg m}^2/\text{s}) \hat{j} + (60.00 \text{ kg m}^2/\text{s}) \hat{k} \].

Step 3: Calculate Torque

The torque \( \vec{\tau} \) about the origin is given by \( \vec{\tau} = \vec{r} \times \vec{F} \). Using the same position vector \( \vec{r} = (2.00 \text{ m}) \hat{i} + (4.00 \text{ m}) \hat{j} - (3.00 \text{ m}) \hat{k} \) and force \( \vec{F} = (6.00 \text{ N}) \hat{i} - (8.00 \text{ N}) \hat{j} + (4.00 \text{ N}) \hat{k} \), we calculate:\[ \vec{\tau} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 2.00 & 4.00 & -3.00 \ 6.00 & -8.00 & 4.00 \end{array} \right| \]This simplifies to:\[ \vec{\tau} = \hat{i}(4.00 \times 4.00 - (-3.00)\times(-8.00)) - \hat{j}(2.00 \times 4.00 - (-3.00)\times6.00) + \hat{k}(2.00\times(-8.00) - 4.00\times6.00) \]\[ = (16.00 - 24.00)\hat{i} + (8.00 + 18.00)\hat{j} + (-16.00 - 24.00)\hat{k} \]\[ = (-8.00 \text{ N m}) \hat{i} + (26.00 \text{ N m}) \hat{j} - (40.00 \text{ N m}) \hat{k} \].

Step 4: Calculate Angle Between Velocity and Force

To find the angle \( \theta \) between two vectors \( \vec{v} \) and \( \vec{F} \), use the formula \( \cos\theta = \frac{\vec{v} \cdot \vec{F}}{||\vec{v}|| ||\vec{F}||} \). First, find the dot product: \( \vec{v} \cdot \vec{F} = (-6.00)(6.00) + (3.00)(-8.00) + (3.00)(4.00) = -36.00 - 24.00 + 12.00 = -48.00 \).Next, calculate the magnitudes:\( ||\vec{v}|| = \sqrt{(-6.00)^2 + (3.00)^2 + (3.00)^2} = \sqrt{36.00 + 9.00 + 9.00} = \sqrt{54.00} \approx 7.35 \).\( ||\vec{F}|| = \sqrt{(6.00)^2 + (-8.00)^2 + (4.00)^2} = \sqrt{36.00 + 64.00 + 16.00} = \sqrt{116.00} \approx 10.77 \).Now, substitute these into the formula:\( \cos\theta = \frac{-48.00}{7.35 \times 10.77} \approx \frac{-48.00}{79.11} \approx -0.61 \).Finally, calculate \( \theta \) using \( \cos^{-1}(-0.61) \approx 128.74^\circ \).

Key Concepts

Newton's Laws of Motion

Newton's laws of motion are fundamental in understanding how objects behave under various forces. The second of these laws, which states \( \vec{F} = m \cdot \vec{a} \), is particularly important in mechanics. This law tells us that the force \( \vec{F} \) acting on an object is equal to the mass \( m \) of the object multiplied by its acceleration \( \vec{a} \). Simply put, if you know the force acting on an object and its mass, you can determine its acceleration.

In the problem provided, we used this law to find the acceleration of the object. We took the force vector \( (6.00 \text{ N}) \hat{i} - (8.00 \text{ N}) \hat{j} + (4.00 \text{ N}) \hat{k} \) and divided each of its components by the mass, 2.00 kg. This gives us an acceleration of \( 3.00 \text{ m/s}^2 \) in the \( \hat{i} \) direction, \( -4.00 \text{ m/s}^2 \) in the \( \hat{j} \) direction, and \( 2.00 \text{ m/s}^2 \) in the \( \hat{k} \) direction.
This demonstrates how the force applied to an object influences its rate of change of motion based on its mass.

Angular Momentum

Angular momentum is a key concept that explains the rotational motion of objects. It is the product of an object's momentum and its distance from a point, typically calculated using the cross product: \( \vec{L} = \vec{r} \times \vec{p} \). Here, \( \vec{r} \) is the position vector and \( \vec{p} \) is the linear momentum, calculated as the object's mass times its velocity \( m \cdot \vec{v} \).

In our problem, we computed the angular momentum using the provided position and velocity vectors. The position vector is \( (2.00 \text{ m}) \hat{i} + (4.00 \text{ m}) \hat{j} - (3.00 \text{ m}) \hat{k} \), and the momentum is \( ( -12.00 \hat{i} + 6.00 \hat{j} + 6.00 \hat{k} ) \text{ kg m/s} \), after multiplying by the mass, 2.00 kg. We found the angular momentum through determinant calculations of the cross product. The result is \( (66.00 \text{ kg m}^2/ ext{s}) \hat{i} + (60.00 \text{ kg m}^2/ ext{s}) \hat{j} + (60.00 \text{ kg m}^2/ ext{s}) \hat{k} \).
Understanding angular momentum is crucial for analyzing phenomena where objects rotate or revolve around a point.

Torque

Torque is like rotational force that causes objects to rotate. It can be thought of as the rotational equivalent of linear force, determined by the vector cross product \( \vec{\tau} = \vec{r} \times \vec{F} \). \( \vec{r} \) is the position vector and \( \vec{F} \) is the force applied. Torque is vital for understanding how effectively a force causes an angular acceleration in an object.

In the exercise provided, we calculated torque using the given force and position vectors. The position vector is \( (2.00 \text{ m}) \hat{i} + (4.00 \text{ m}) \hat{j} - (3.00 \text{ m}) \hat{k} \) and the applied force is \( (6.00 \text{ N}) \hat{i} - (8.00 \text{ N}) \hat{j} + (4.00 \text{ N}) \hat{k} \). By calculating the determinant of the cross product, the torque results in \( (-8.00 \text{ N m}) \hat{i} + (26.00 \text{ N m}) \hat{j} - (40.00 \text{ N m}) \hat{k} \).
This demonstrates the torque's effectiveness in causing the object to rotate, with each component reflecting the influence in its respective axis.

Vector Calculations

Vectors are fundamental in physics, representing quantities that have both magnitude and direction. Operations with vectors, such as addition, dot products, and cross products, are crucial in mechanics to describe real-world phenomena.

The dot product gives a scalar and represents how much one vector goes in the direction of another. It's used to find angles between vectors using the formula \( \cos\theta = \frac{\vec{v} \cdot \vec{F}}{||\vec{v}|| ||\vec{F}||} \). In our problem, the angle \( \theta \) between velocity and force vectors is found using this calculation. The dot product of the velocity \( (-6.00 \hat{i} + 3.00 \hat{j} + 3.00 \hat{k}) \) and force \( (6.00 \hat{i} - 8.00 \hat{j} + 4.00 \hat{k}) \) is \(-48.00\), leading to \( \cos^{-1}(-0.61) \approx 128.74^\circ \).

Cross products give a vector perpendicular to the plane formed by two vectors and are used for torque and angular momentum. Grasping these vector operations helps solve complex physics problems and gives insights into the relationships between dynamic quantities.