Problem 33 - Fundamentals of Physics

Question

At time \(t=0,\) a \(3.0 \mathrm{~kg}\) particle with velocity \(\vec{v}=(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(6.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) is at \(x=3.0 \mathrm{~m}, y=8.0 \mathrm{~m} .\) It is pulled by a \(7.0 \mathrm{~N}\) force in the negative \(x\) direction. About the origin, what are (a) the particle's angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?

Step-by-Step Solution

Answer

(a) \( \vec{L} = -174.0 \mathrm{~kg \cdot m^2/s} \hat{k} \), (b) \( \vec{\tau} = 56.0 \mathrm{~N \cdot m} \hat{k} \), (c) \( \frac{d\vec{L}}{dt} = 56.0 \mathrm{~N \cdot m} \hat{k} \)."

Step 1: Calculate the Angular Momentum

To find the angular momentum \( \vec{L} \) of the particle about the origin, we use the formula \( \vec{L} = \vec{r} \times \vec{p} \), where \( \vec{r} = (3.0 \text{ m}) \hat{i} + (8.0 \text{ m}) \hat{j} \) is the position vector, and \( \vec{p} = m \vec{v} \) is the momentum vector. First, we calculate \( \vec{p} \):\[ \vec{p} = 3.0 \mathrm{~kg} \times ((5.0 \mathrm{~m/s}) \hat{i} - (6.0 \mathrm{~m/s}) \hat{j}) = (15.0 \mathrm{~kg \cdot m/s}) \hat{i} - (18.0 \mathrm{~kg \cdot m/s}) \hat{j} \]Then, calculate \( \vec{L} = \vec{r} \times \vec{p} \):\[ \vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3.0 & 8.0 & 0 \ 15.0 & -18.0 & 0 \end{vmatrix} \]\[ = \hat{k} (3.0 \times -18.0 - 8.0 \times 15.0) \]\[ = \hat{k} (-54.0 - 120.0) = \hat{k} (-174.0) \]Thus, \( \vec{L} = -174.0 \mathrm{~kg \cdot m^2/s} \hat{k} \).

Step 2: Calculate the Torque

The torque \( \vec{\tau} \) acting on the particle is given by \( \vec{\tau} = \vec{r} \times \vec{F} \). The force \( \vec{F} = -7.0 \mathrm{~N} \hat{i} \) is applied in the negative \( x \)-direction. Thus:\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3.0 & 8.0 & 0 \ -7.0 & 0 & 0 \end{vmatrix} \]\[ = \hat{k} (3.0 \times 0 - 8.0 \times (-7.0)) \]\[ = \hat{k} (0 + 56.0) = \hat{k} (56.0) \]So, \( \vec{\tau} = 56.0 \mathrm{~N \cdot m} \hat{k} \).

Step 3: Determine the Rate of Change of Angular Momentum

According to the principle of rotational dynamics, the rate of change of angular momentum \( \frac{d\vec{L}}{dt} \) is equal to the torque \( \vec{\tau} \) acting on the particle. From Step 2, we already found that:\[ \frac{d\vec{L}}{dt} = \vec{\tau} = 56.0 \mathrm{~N \cdot m} \hat{k} \]Thus, the rate at which the angular momentum is changing is \( 56.0 \mathrm{~N \cdot m} \hat{k} \).

Key Concepts

Torque

Torque is essentially a measure of the twisting force that can cause an object to rotate. When you apply force not directly at the axis of rotation, it results in a torque. In the exercise, the torque on the particle is determined using the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \).
\[ \vec{\tau} = \vec{r} \times \vec{F} \]
This operation gives a vector that is perpendicular to the plane formed by \( \vec{r} \) and \( \vec{F} \) and gives us the magnitude and direction of the torque.

• The larger the distance (\( \vec{r} \)), the bigger the torque.
• The greater the force, the bigger the torque.
• If the force is applied directly in line with the radius vector, there's no torque.

Knowing about torque is crucial for understanding rotational motion as it tells us how forces affect rotation. In this case, the particle experiences a torque of \( 56.0 \, \mathrm{N \cdot m} \). This means the force applied will cause the particle to rotate around the origin with a specific angular acceleration.

Rate of Change of Angular Momentum

The rate of change of angular momentum is a fundamental concept that emerges from Newton's laws adapted for rotational motion. Angular momentum \( \vec{L} \) is defined as the product of the rotational inertia of a mass and its rotational velocity.
When we think about changes in angular momentum, we relate it to the external torque applied to the system. The principle can be neatly encapsulated by the equation:
\[ \frac{d\vec{L}}{dt} = \vec{\tau} \]
In this exercise, the torque \( \vec{\tau} \) applied to the particle changes its angular momentum at a rate of \( 56.0 \, \mathrm{N \cdot m} \).

• This direct relationship helps us understand rotational dynamics.
• Just as force changes linear momentum, torque changes angular momentum.

This principle is key in many applications, from engineering to understanding the motion of celestial bodies. By understanding this concept, we learn how rotational systems respond to external influences, such as forces or torques.

Rotational Dynamics

Rotational dynamics deals with the motion of objects that are rotating. Much like linear dynamics that involves linear forces and masses, rotational dynamics involves torques and rotational mass.
Rotational dynamics is governed by equations that are analogous to those used to describe linear motion. You’ll find familiar equations such as:
\[ \vec{\tau} = I \alpha \]
where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration.

• Torque works like force does in linear motion.
• The moment of inertia is like mass but in rotational movement.
• Angular acceleration is how much the rotational velocity changes.

In the problem, we see how these principles come together as the particle subjected to a torque undergoes a change in its angular momentum through the laws of rotational dynamics. This highlights how forces acting at distances from a pivot effect rotational motion differently than they would if they acted through the center of mass, making it a compelling study.