/*! 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

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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 relationship between torque and angular momentum

According to the rotational form of Newton's Second Law, the rate of change of angular momentum, \( \frac{d \vec{\ell}}{dt} \), is equal to the net torque acting on the particle. This can be expressed as: \[ \frac{d \vec{\ell}}{dt} = \vec{\tau}_{\text{net}} \].
02

Calculating the net torque

The net torque \( \vec{\tau}_{\text{net}} \) is the vector sum of \( \vec{\tau}_1 \) and \( \vec{\tau}_2 \). Since \( \vec{\tau}_1 \) has a magnitude of \( 2.0 \; \mathrm{N \cdot m} \) in the positive \( x \)-direction, it is represented as \( 2.0 \hat{i} \). Meanwhile, \( \vec{\tau}_2 \) has a magnitude of \( 4.0 \; \mathrm{N \cdot m} \) in the negative \( y \)-direction, represented as \(-4.0 \hat{j} \). Hence, the net torque is:\[ \vec{\tau}_{\text{net}} = 2.0 \hat{i} - 4.0 \hat{j} \].
03

Writing the expression for dâ„“/dt in unit vector notation

Having found the net torque vector, we can express \( \frac{d \vec{\ell}}{dt} \) as:\[ \frac{d \vec{\ell}}{dt} = 2.0 \hat{i} - 4.0 \hat{j} \] N·m.

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Key Concepts

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

Torque
Torque is essentially a measure of how much force acting on an object causes it to rotate around a particular point or axis. When considering rotational motion, torque can be thought of as the equivalent of force in linear motion.
In this problem, we are dealing with two different torques. One torque is in the positive direction of the x-axis, with a magnitude of 2.0 N·m, and the other is in the negative direction of the y-axis, with a magnitude of 4.0 N·m. Therefore, torque can be seen as the tendency of a force to rotate an object about an axis.
To calculate the net effect, you simply take the vector sum of these torques, which in this case is done by adding them in unit vector notation. This gives us \( \vec{\tau}_{\text{net}} = 2.0 \hat{i} - 4.0 \hat{j} \).
Understanding torque is crucial because it directly influences how the angular momentum changes over time. In essence, torque is like a shove, steering the course of this rotational story.
Unit Vector Notation
Unit vector notation is a neat way to express vectors, which are quantities having both a magnitude and a direction. In physics, this method helps us easily handle and visualize the components of vectors such as torque and angular momentum.
In unit vector notation, vectors are broken down into their corresponding axis components. For instance, in the given exercise, \( \vec{\tau}_1 \) is directed along the positive x-axis and is expressed as \( 2.0 \hat{i} \), where \( \hat{i} \) is the unit vector in the x direction. Similarly, \( \vec{\tau}_2 \) is represented as \(-4.0 \hat{j} \), where \( \hat{j} \) corresponds to the y-axis.
By expressing these vectors in this form, we not only simplify calculations but also clearly understand their directions. The net torque \( \vec{\tau}_{\text{net}} \), therefore, becomes a straightforward vector sum, making it easier to proceed with further calculations.
Newton's Second Law
Newton's Second Law in rotational motion is a clever twist on its original form, where it ties torque to the rate of change of angular momentum. The law can be captured by the equation: \( \frac{d \vec{\ell}}{dt} = \vec{\tau}_{\text{net}} \).
This equation serves as the backbone for understanding how forces causing rotational motion affect the angular momentum of a system. When a net torque acts on a system, it brings about a change in its angular momentum at a rate proportionate to the magnitude of the torque.
In the current problem, we use this principle to understand how the torques \( \vec{\tau}_1 \) and \( \vec{\tau}_2 \) combine to create a net torque \( \vec{\tau}_{\text{net}} \), which in turn gives us the rate of change of angular momentum \( \frac{d \vec{\ell}}{dt} \).
By connecting torque with angular momentum in this way, Newton's Second Law provides a powerful tool to study rotational kinematics and dynamics, mirroring its indispensable role in linear motion analysis. This approach leads us to the expression \( \frac{d \vec{\ell}}{dt} = 2.0 \hat{i} - 4.0 \hat{j} \), emphasizing how this relationship elegantly describes rotational behavior.

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Most popular questions from this chapter

A sanding disk with rotational inertia \(1.2 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) is attached to an electric drill whose motor delivers a torque of magnitude \(16 \mathrm{~N} \cdot \mathrm{m}\) about the central axis of the disk. About that axis and with the torque applied for \(33 \mathrm{~ms}\), what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

A hollow sphere of radius \(0.15 \mathrm{~m}\), with rotational inertia \(I=0.040 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about a line through its center of mass, rolls without slipping up a surface inclined at \(30^{\circ}\) to the horizontal. At a certain initial position, the sphere's total kinetic energy is \(20 \mathrm{~J}\). (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved \(1.0 \mathrm{~m}\) up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

A \(1000 \mathrm{~kg}\) car has four \(10 \mathrm{~kg}\) wheels. When the car is moving, what fraction of its total kinetic energy is due to rotation of the wheels about their axles? Assume that the wheels have the same rotational inertia as uniform disks of the same mass and size. Why do you not need to know the radius of the wheels?

Two skaters, each of mass \(50 \mathrm{~kg}\), approach each other along parallel paths separated by \(3.0\) \(\mathrm{m}\). They have opposite velocities of \(1.4 \mathrm{~m} / \mathrm{s}\) each. One skater carries one end of a long pole of negligible mass, and the other skater grabs the other end as she passes. The skaters then rotate around the center of the pole. Assume that the friction between skates and ice is negligible. What are (a) the radius of the circle, (b) the angular speed of the skaters, and (c) the kinetic energy of the two-skater system? Next, the skaters pull along the pole until they are separated by \(1.0 \mathrm{~m}\). What then are (d) their angular speed and (e) the kinetic energy of the system? (f) What provided the energy for the increased kinetic energy?

A girl of mass \(M\) stands on the rim of a frictionless merry-goround of radius \(R\) and rotational inertia \(I\) that is not moving. She throws a rock of mass \(m\) horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is \(v .\) Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the girl?

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