Problem 32 - Fundamentals of Physics

Question

A particle is acted on by two torques about the origin: \(\vec{\tau}_{1}\) has a magnitude of \(2.0 \mathrm{~N} \cdot \mathrm{m}\) and is directed in the positive direction of the \(x\) axis, and \(\vec{\tau}_{2}\) has a magnitude of \(4.0 \mathrm{~N} \cdot \mathrm{m}\) and is directed in the negative direçtion of the \(y\) axis. In unit-vector notation, find \(d \ell / d t,\) where \(\vec{\ell}\) is the angular momentum of the particle about the origin.

Step-by-Step Solution

Answer

\( \frac{d \vec{\ell}}{dt} = 2.0 \hat{i} - 4.0 \hat{j} \mathrm{~N} \cdot \mathrm{m} \).

Step 1: Understanding the Concept

The rate of change of angular momentum \( \frac{d \vec{\ell}}{dt} \) of a particle is equal to the net external torque acting on it. Therefore, we need to find the vector sum of the two given torques.

Step 2: Express the Torques in Vector Form

We are given two torques, \( \vec{\tau}_{1} \) and \( \vec{\tau}_{2} \). The first torque, \( \vec{\tau}_{1} \), acts in the positive \( x \)-direction and is \( 2.0 \mathrm{~N} \cdot \mathrm{m} \). Thus, it is represented as \( \vec{\tau}_{1} = 2.0 \mathrm{~N} \cdot \mathrm{m} \hat{i} \). The second torque, \( \vec{\tau}_{2} \), acts in the negative \( y \)-direction with magnitude \( 4.0 \mathrm{~N} \cdot \mathrm{m} \), represented as \( \vec{\tau}_{2} = -4.0 \mathrm{~N} \cdot \mathrm{m} \hat{j} \).

Step 3: Calculating the Net Torque

Add the vectors \( \vec{\tau}_{1} \) and \( \vec{\tau}_{2} \) to find the net torque: \[ \vec{\tau}_{\text{net}} = \vec{\tau}_{1} + \vec{\tau}_{2} = 2.0 \hat{i} \mathrm{~N} \cdot \mathrm{m} + (-4.0 \hat{j} \mathrm{~N} \cdot \mathrm{m}) = 2.0 \hat{i} - 4.0 \hat{j} \mathrm{~N} \cdot \mathrm{m}. \]

Step 4: Determine \( \frac{d \vec{\ell}}{dt} \)

Since \( \frac{d \vec{\ell}}{dt} = \vec{\tau}_{\text{net}} \), we have \( \frac{d \vec{\ell}}{dt} = 2.0 \hat{i} - 4.0 \hat{j} \mathrm{~N} \cdot \mathrm{m} \). This is the rate of change of angular momentum in unit-vector notation.

Key Concepts

Torque

Torque is a measure of how much a force acting on an object causes that object to rotate. Imagine pushing a door to open it; the harder you push and the farther from the hinge you push, the easier the door will open.
The formula for torque is given by:\[ \vec{\tau} = \vec{r} \times \vec{F} \]Where:

• \( \vec{\tau} \) is the torque vector.
• \( \vec{r} \) is the position vector (distance from the pivot point).
• \( \vec{F} \) is the force vector.
• The symbol \( \times \) indicates a cross product, which means the direction of \( \vec{\tau} \) is perpendicular to both \( \vec{r} \) and \( \vec{F} \).

In this exercise, torque is given directly in terms of its magnitude and direction. Knowing how torque works helps you understand the rotational effects of forces.

Vector Addition

To find the net torque, it’s essential to understand vector addition. Vectors have both magnitude and direction. Adding them isn't straightforward like adding numbers. Instead, you align them like arrows and connect them to find the resultant vector.
In this exercise, you were given two torque vectors:

• \( \vec{\tau}_{1} = 2.0 \ \mathrm{N} \cdot \mathrm{m} \ \hat{i} \)
• \( \vec{\tau}_{2} = -4.0 \ \mathrm{N} \cdot \mathrm{m} \ \hat{j} \)

To find the net torque \( \vec{\tau}_{\text{net}} \), you simply add them:\[ \vec{\tau}_{\text{net}} = 2.0 \hat{i} - 4.0 \hat{j} \ \mathrm{N} \cdot \mathrm{m} \]This combined vector represents the overall effect of the two torques acting together.

Unit-Vector Notation

Unit-vector notation is a way to express vectors using unit vectors. These are vectors with a magnitude of one, pointing in the direction of the axes. They are usually represented as:

• \( \hat{i} \) for the x-direction
• \( \hat{j} \) for the y-direction
• \( \hat{k} \) for the z-direction

Using unit-vector notation, you can easily write down any vector using its components along the axes.
For example, the torque \( \vec{\tau}_{\text{net}} = 2.0 \hat{i} - 4.0 \hat{j} \ \mathrm{N} \cdot \mathrm{m} \) is in unit-vector notation. This tells us:

• There is a torque of \( 2.0 \ \mathrm{N} \cdot \mathrm{m} \) acting in the positive x-direction.
• There is a torque of \( 4.0 \ \mathrm{N} \cdot \mathrm{m} \) acting in the negative y-direction.

This notation makes handling vectors in physics problems much simpler.

Rate of Change

The rate of change is a concept that tells us how a quantity changes over time. In physics, it is often represented using derivatives. In this problem, you're interested in the rate of change of angular momentum, expressed as \( \frac{d \vec{\ell}}{dt} \).
The rate of change of angular momentum is directly related to the net torque applied to the system. This relationship is expressed by:\[ \frac{d \vec{\ell}}{dt} = \vec{\tau}_{\text{net}} \]For this exercise:

• We found the net torque to be \( 2.0 \hat{i} - 4.0 \hat{j} \ \mathrm{N} \cdot \mathrm{m} \).
• This means the angular momentum is changing in the positive x-direction and the negative y-direction.

Understanding rate of change helps you predict how a system evolves over time when subjected to certain forces.