/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 A particle is acted on by two to... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 direction of the \(y\) axis. In unit-vector notation, find \(d \vec{\ell} / d t\), where \(\vec{\ell}\) is the angular momentum of the particle about the origin.

Short Answer

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

Step by step solution

01

Understanding the Relation Between Torque and Angular Momentum

Torque (\(\vec{\tau}\)) is related to the rate of change of angular momentum (\(\vec{\ell}\)) by the equation: \(\vec{\tau} = \frac{d \vec{\ell}}{dt}\). In this problem, we have two torques acting on a particle. We are asked to find the rate of change of the angular momentum, \(\frac{d \vec{\ell}}{dt}\).
02

Express Torque in Vector Form

We have two torques: \(\vec{\tau}_{1}\) and \(\vec{\tau}_{2}\). The vector form of \(\vec{\tau}_{1}\) is \(2.0 \hat{i}\) N·m because it's directed along the positive \(x\)-axis. The vector form of \(\vec{\tau}_{2}\) is \(-4.0 \hat{j}\) N·m because it's directed along the negative \(y\)-axis.
03

Sum the Torques

To find the total torque, add the individual torques vectorially: \(\vec{\tau} = \vec{\tau}_{1} + \vec{\tau}_{2} = 2.0 \hat{i} + (-4.0) \hat{j}\). Thus, \(\vec{\tau} = 2.0 \hat{i} - 4.0 \hat{j}\) N·m.
04

Relate Total Torque to Change in Angular Momentum

Since \(\vec{\tau} = \frac{d \vec{\ell}}{dt}\), the total torque we calculated represents the rate of change of angular momentum. Thus, \(\frac{d \vec{\ell}}{dt} = 2.0 \hat{i} - 4.0 \hat{j}\) N·m.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Torque
Torque is a fundamental concept in physics that describes the rotational effect of a force applied at a distance from a pivot point or axis. Picture it as the twist you feel when you turn a doorknob. This rotational effect can be calculated using the formula:
  • \( \tau = r \times F \)
  • where \( \tau \) is the torque, \( r \) is the distance from the pivot point, and \( F \) is the applied force.
If you imagine a wrench turning a bolt, the longer the wrench (distance \( r \)), or the stronger you push/pull (force \( F \)), the greater the torque.
In our problem, two torques are applied, each in different directions, which is crucial because direction affects the overall torque.
Torque is directional, meaning it has both magnitude and direction. It is a vector quantity.
Angular Momentum Rate of Change
Angular momentum, denoted by \( \vec{\ell} \), is a measure of the quantity of rotation a particle has. When the angular momentum of an object changes over time, it is due to the presence of torque. This relationship is described by the equation:
  • \( \vec{\tau} = \frac{d \vec{\ell}}{dt} \)
This expression means that the torque \( \vec{\tau} \) exerted on a particle results in the particle's angular momentum \( \vec{\ell} \) changing at a rate given by \( \frac{d \vec{\ell}}{dt} \).
In simpler terms, the bigger the torque, the faster the change in angular momentum.
In the exercise, two torques are applied, totaling \(2.0 \hat{i} - 4.0 \hat{j}\) N·m. This combined torque tells us how the particle's angular momentum is changing every second.
Understanding this relationship aids in predicting how objects will move when forces are applied.
Vector Addition in Torque Calculation
Vector addition is essential in calculating the total effect when multiple forces or torques act on a point. Each torque in the problem is represented as a vector because it has both direction and magnitude.
In vector notation, each component (e.g., \( \hat{i} \) and \( \hat{j} \)) represents a directional axis, like \( x \) and \( y \), and helps to understand how forces or torques overlap or counteract each other.
For instance, in the original exercise, \( \vec{\tau}_{1} = 2.0 \hat{i} \) N·m and \( \vec{\tau}_{2} = -4.0 \hat{j} \) N·m. Adding these vectors gives the total torque:
  • \( \vec{\tau} = 2.0 \hat{i} + (-4.0) \hat{j} = 2.0 \hat{i} - 4.0 \hat{j} \)
This highlights how each torque's directional influence is combined to find the net effect. Properly summing these vectors is key in determining how the angular momentum of the system is changing.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two disks are mounted (like a merry-go-round) on lowfriction bearings on the same axle and can be brought together so that they couple and rotate as one unit. The first disk, with rotational inertia \(3.30 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis, is set spinning counterclockwise at 450 rev/min. The second disk, with rotational inertia \(6.60 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis, is set spinning counterclockwise at 900 rev/min. They then couple together. (a) What is their angular speed after coupling? If instead the second disk is set spinning clockwise at \(900 \mathrm{rev} / \mathrm{min}\), what are their (b) angular speed and (c) direction of rotation after they couple together?

A uniform solid ball rolls smoothly along a floor, then up a ramp inclined at \(15.0^{\circ} .\) It momentarily stops when it has rolled \(1.50 \mathrm{~m}\) along the ramp. What was its initial speed?

During a jump to his partner, an aerialist is to make a quadruple somersault lasting a time \(t=1.87 \mathrm{~s}\). For the first and last quarter- revolution, he is in the extended orientation shown in Fig. \(11-55\), with rotational inertia \(I_{1}=19.9 \mathrm{~kg} \cdot \mathrm{m}^{2}\) around his center of mass (the dot). During the rest of the flight he is in a tight tuck, with rotational inertia \(I_{2}=3.93 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) What must be his angular speed \(\omega_{2}\) around his center of mass during the tuck?

In a long jump, an athlete leaves the ground with an initial angular momentum that tends to rotate her body forward, threatening to ruin her landing. To counter this tendency, she rotates her outstretched arms to "take up" the angular momentum (Fig. 11- 18). In \(0.700 \mathrm{~s}\), one arm sweeps through \(0.500\) rev and the other arm sweeps through \(1.000\) rev. Treat each arm as a thin rod of mass \(4.0 \mathrm{~kg}\) and length \(0.60 \mathrm{~m}\), rotating around one end. In the athlete's reference frame, what is the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?

A wheel rotates clockwise about its central axis with an angular momentum of \(600 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). At time \(t=0\), a torque of magnitude \(50 \mathrm{~N} \cdot \mathrm{m}\) is applied to the wheel to reverse the rotation. At what time \(t\) is the angular speed zero?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.