/*! 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 \(\vec{T}_{2}\) has a magnitude of \(3.0 \mathrm{~N} \cdot \mathrm{m}\) 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
\(\frac{d \vec{\ell}}{dt} = 2.0 \, \hat{i} - 3.0 \, \hat{j} \).

Step by step solution

01

Understanding the Problem

We need to find the rate of change of angular momentum \(\frac{d \vec{\ell}}{dt}\) when the particle is acted upon by two torques. The total torque \(\vec{\tau}_{\text{net}}\) is responsible for this rate of change according to the equation: \(\frac{d \vec{\ell}}{dt} = \vec{\tau}_{\text{net}}\). We are given two torques \(\vec{\tau}_{1} = 2.0 \, \mathrm{N} \cdot \mathrm{m}\) in the \(\hat{i}\) direction and \(\vec{\tau}_{2} = 3.0 \, \mathrm{N} \cdot \mathrm{m}\) in the \(-\hat{j}\) direction.
02

Calculate Net Torque

Add the two given torques to get the net torque \(\vec{\tau}_{\text{net}}\). In unit vector notation, \(\vec{\tau}_{\text{net}} = \vec{\tau}_{1} + \vec{\tau}_{2} = 2.0 \, \hat{i} + 0 \, \hat{j} + 0 \, \hat{k} + 0 \, \hat{i} - 3.0 \, \hat{j} + 0 \, \hat{k}\). This simplifies to \(\vec{\tau}_{\text{net}} = 2.0 \, \hat{i} - 3.0 \, \hat{j}\).
03

Compute Rate of Change of Angular Momentum

Since \(\frac{d \vec{\ell}}{dt} = \vec{\tau}_{\text{net}}\), substitute the net torque into the equation: \(\frac{d \vec{\ell}}{dt} = 2.0 \, \hat{i} - 3.0 \, \hat{j}\).
04

Express the Solution in Unit Vector Notation

The rate of change of angular momentum \(\frac{d \vec{\ell}}{dt}\), in unit vector notation, is \(2.0 \, \hat{i} - 3.0 \, \hat{j} \).

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Key Concepts

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

Torque
Torque is a fundamental concept in physics and engineering that describes the rotational effect of a force applied at a distance from a pivot or fulcrum. It's analogous to force in linear dynamics but acts on rotations. You can think of torque as the twist that causes an object to rotate.
Torque is calculated using the formula:
  • The formula for torque (\( \tau \)) is \( \tau = r \times F \), where \( r \) is the lever arm (distance between the pivot and the point where the force is applied) and \( F \) is the force applied.
Torque can cause an object to start rotating, stop rotating, or change its rotational speed. The direction of the torque vector is determined by the right-hand rule, which helps us visualize the direction of rotational effects based on the orientation of the force.
In the problem we discussed, two torques are acting on a particle: one in the direction of the x-axis and the other in the negative y-axis. Their combined effect determines the net torque responsible for changing the particle's angular momentum.
Unit Vector Notation
Unit vector notation is a convenient way to express vector quantities in a coordinate system. Vectors represent various physical quantities like displacement, velocity, and torque, which have both magnitude and direction. Using unit vectors standardizes the way we express these quantities.
The most common unit vectors are:
  • \( \hat{i} \): Represents the unit vector in the x-direction.
  • \( \hat{j} \): Represents the unit vector in the y-direction.
  • \( \hat{k} \): Represents the unit vector in the z-direction.
When solving problems in mechanics, especially those involving torque and angular momentum, unit vector notation allows us to clearly and concisely express vector components.
In the exercise, unit vector notation helps us express the torques \( \tau_{1} \) and \( \tau_{2} \) as \( 2.0 \hat{i} \) and \( -3.0 \hat{j} \), respectively. This notation then simplifies computations, such as finding the net torque by simply adding or subtracting the unit vector components.
Rate of Change
In physics, the rate of change is a measure of how a quantity varies with respect to time. It is a pivotal concept in calculus and physics, helping to describe how quickly or slowly changes occur. When applied to motion, it can describe changes in position (velocity), changes in speed (acceleration), and, as in this case, changes in angular momentum.
Angular momentum, denoted by \( \vec{\ell} \), characterizes an object's rotational motion. The change of angular momentum over time, \( \frac{d\vec{\ell}}{dt} \), is directly linked to torque. This relationship is foundational in dynamics, stated simply as:
  • \( \frac{d\vec{\ell}}{dt} = \vec{\tau}_{\text{net}} \)
This equation tells us that the net torque acting on an object causes a change in that object's angular momentum. In our specific problem, the combined torques on the particle give rise to a rate of change expressed in unit vector notation as \( 2.0 \hat{i} - 3.0 \hat{j} \), indicating how the angular momentum's direction and magnitude evolve over time.

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Most popular questions from this chapter

Figure 11-45 shows an overhead view of a ring that can rotate about its center like a merry-goround. Its outer radius \(R_{2}\) is \(0.800 \mathrm{~m}\), its inner radius \(R_{1}\) is \(R_{2} / 2.00\), its mass \(M\) is \(6.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2}\). By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1}\) ?

A particle of mass \(m\) is shot from ground level at initial speed \(u\) and initial angle \(\theta\) relative to the horizontal. When it reaches its highest point in its flight over the level ground, what are the magnitudes of (a) the torque acting on it from the gravitational force and (b) its angular momentum, both measured about the launch point?

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 \(F_{2 x}=0, F_{2 y}=2.0 \mathrm{~N}, F_{2 z}=4.0 \mathrm{~N} ?\)

Two \(2.00 \mathrm{~kg}\) balls are attached to the ends of a thin rod of length \(50.0 \mathrm{~cm}\) and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (Fig. 11-52), a \(50.0 \mathrm{~g}\) wad of wet putty drops onto one of the balls, hitting it with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) and then sticking to it. (a) What is the angular speed of the system just after the putty wad hits? (b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before? (c) Through what angle will the system rotate before it momentarily stops?

A top spins at \(25 \mathrm{rev} / \mathrm{s}\) about an axis that makes an angle of \(30^{\circ}\) with the vertical. The mass of the top is \(0.50 \mathrm{~kg}\), its rotational inertia about its central axis is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and its center of mass is \(4.2 \mathrm{~cm}\) from the pivot point. If the spin is clockwise from an overhead view, what are the (a) precession rate and (b) direction of the precession as viewed from overhead?

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