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Chapter 11: Problem 32

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 dire莽tion 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.

Short Answer

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

Step by step solution

01

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.
02

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 \( \vec{\tau}_{2} = -4.0 \mathrm{~N} \cdot \mathrm{m} \hat{j} \).
03

Calculating the Net Torque

Add the vectors \( \vec{\tau}_{1} \) and \( \vec{\tau}_{2} \) to find the net torque: \[ \vec{\tau}_{\text{net}} = \vec{\tau}_{1} + \vec{\tau}_{2} = 2.0 \hat{i} \mathrm{~N} \cdot \mathrm{m} + (-4.0 \hat{j} \mathrm{~N} \cdot \mathrm{m}) = 2.0 \hat{i} - 4.0 \hat{j} \mathrm{~N} \cdot \mathrm{m}. \]
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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.

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Most popular questions from this chapter

A bowler throws a bowling ball of radius \(R=11 \mathrm{~cm}\) along a lane. The ball (Fig. \(11-38\) ) slides on the lane with initial speed \(v_{\mathrm{com}}=8.5 \mathrm{~m} / \mathrm{s}\) and initial angular speed \(\omega_{0}=0 .\) The coefficient of kinetic friction between the ball and the lane is \(0.21 .\) The kinetic frictional force \(\vec{f}_{k}\) acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed \(v_{\text {com }}\) has decreased enough and angular speed \(\omega\) has increased enough, the ball stops sliding and then rolls smoothly. (a) What then is \(v_{\operatorname{com}}\) in terms of \(\omega ?\) During the sliding, what are the ball's (b) linear acceleration and (c) angular acceleration? (d) How long does the ball slide? (e) How far does the ball slide? (f) What is the linear speed of the ball when smooth rolling begins?

A wheel of radius \(0.250 \mathrm{~m},\) moving initially at \(43.0 \mathrm{~m} / \mathrm{s},\) rolls to a stop in \(225 \mathrm{~m}\). Calculate the magnitudes of its (a) linear acceleration and (b) angular acceleration. (c) Its rotational inertia is \(0.155 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis. Find the magnitude of the torque about the central axis due to friction on the wheel.

In unit-vector notation, what is the torque about the origin on a particle located at coordinates \((0,-4.0 \mathrm{~m}, 3.0 \mathrm{~m})\) if that torque is due to (a) force \(\vec{F}_{1}\) with components \(F_{1 x}=2.0 \mathrm{~N}, F_{1 y}=F_{1 z}=0,\) and (b) force \(\vec{F}_{2}\) with components \(F_{2 x}=0, F_{2 y}=2.0 \mathrm{~N}, F_{2 t}=4.0 \mathrm{~N} ?\)

A ballerina begins a tour jet茅 (Fig. \(11-19 a\) ) with angular speed \(\omega_{i}\) and a rotational inertia consisting of two parts:\(I_{\mathrm{leg}}=1.44 \mathrm{~kg} \cdot \mathrm{m}^{2}\) for her leg extended outward at angle \(\theta=90.0^{\circ}\) to her body and \(I_{\text {trunk }}=0.660 \mathrm{~kg} \cdot \mathrm{m}^{2}\) for the rest of her body (primarily her trunk ). Near her maximum height she holds both legs at angle \(\theta=30.0^{\circ}\) to her body and has angular speed \(\omega_{f}(\) Fig. \(11-19 b)\). Assuming that \(I_{\text {trunk }}\) has not changed, what is the ratio \(\omega_{f} / \omega_{i} ?\)

A horizontal vinyl record of mass \(0.10 \mathrm{~kg}\) and radius \(0.10 \mathrm{~m}\) rotates freely about a vertical axis through its center with an angular speed of \(4.7 \mathrm{rad} / \mathrm{s}\) and a rotational inertia of \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\). Putty of mass \(0.020 \mathrm{~kg}\) drops vertically onto the record from above and sticks to the edge of the record. What is the angular speed of the record immediately afterwards?
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