/*! 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 33 At time \(t=0,\) a \(3.0 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At time \(t=0,\) a \(3.0 \mathrm{~kg}\) particle with velocity \(\vec{v}=(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(6.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) is at \(x=3.0 \mathrm{~m}, y=8.0 \mathrm{~m} .\) It is pulled by a \(7.0 \mathrm{~N}\) force in the negative \(x\) direction. About the origin, what are (a) the particle's angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?

Short Answer

Expert verified
(a) \( \vec{L} = -174.0 \mathrm{~kg \cdot m^2/s} \hat{k} \), (b) \( \vec{\tau} = 56.0 \mathrm{~N \cdot m} \hat{k} \), (c) \( \frac{d\vec{L}}{dt} = 56.0 \mathrm{~N \cdot m} \hat{k} \)."

Step by step solution

01

Calculate the Angular Momentum

To find the angular momentum \( \vec{L} \) of the particle about the origin, we use the formula \( \vec{L} = \vec{r} \times \vec{p} \), where \( \vec{r} = (3.0 \text{ m}) \hat{i} + (8.0 \text{ m}) \hat{j} \) is the position vector, and \( \vec{p} = m \vec{v} \) is the momentum vector. First, we calculate \( \vec{p} \):\[ \vec{p} = 3.0 \mathrm{~kg} \times ((5.0 \mathrm{~m/s}) \hat{i} - (6.0 \mathrm{~m/s}) \hat{j}) = (15.0 \mathrm{~kg \cdot m/s}) \hat{i} - (18.0 \mathrm{~kg \cdot m/s}) \hat{j} \]Then, calculate \( \vec{L} = \vec{r} \times \vec{p} \):\[ \vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3.0 & 8.0 & 0 \ 15.0 & -18.0 & 0 \end{vmatrix} \]\[ = \hat{k} (3.0 \times -18.0 - 8.0 \times 15.0) \]\[ = \hat{k} (-54.0 - 120.0) = \hat{k} (-174.0) \]Thus, \( \vec{L} = -174.0 \mathrm{~kg \cdot m^2/s} \hat{k} \).
02

Calculate the Torque

The torque \( \vec{\tau} \) acting on the particle is given by \( \vec{\tau} = \vec{r} \times \vec{F} \). The force \( \vec{F} = -7.0 \mathrm{~N} \hat{i} \) is applied in the negative \( x \)-direction. Thus:\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3.0 & 8.0 & 0 \ -7.0 & 0 & 0 \end{vmatrix} \]\[ = \hat{k} (3.0 \times 0 - 8.0 \times (-7.0)) \]\[ = \hat{k} (0 + 56.0) = \hat{k} (56.0) \]So, \( \vec{\tau} = 56.0 \mathrm{~N \cdot m} \hat{k} \).
03

Determine the Rate of Change of Angular Momentum

According to the principle of rotational dynamics, the rate of change of angular momentum \( \frac{d\vec{L}}{dt} \) is equal to the torque \( \vec{\tau} \) acting on the particle. From Step 2, we already found that:\[ \frac{d\vec{L}}{dt} = \vec{\tau} = 56.0 \mathrm{~N \cdot m} \hat{k} \]Thus, the rate at which the angular momentum is changing is \( 56.0 \mathrm{~N \cdot m} \hat{k} \).

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.

Torque
Torque is essentially a measure of the twisting force that can cause an object to rotate. When you apply force not directly at the axis of rotation, it results in a torque. In the exercise, the torque on the particle is determined using the cross product of the position vector \( \vec{r} \) and the force vector \( \vec{F} \).
\[ \vec{\tau} = \vec{r} \times \vec{F} \]
This operation gives a vector that is perpendicular to the plane formed by \( \vec{r} \) and \( \vec{F} \) and gives us the magnitude and direction of the torque.
  • The larger the distance (\( \vec{r} \)), the bigger the torque.
  • The greater the force, the bigger the torque.
  • If the force is applied directly in line with the radius vector, there's no torque.
Knowing about torque is crucial for understanding rotational motion as it tells us how forces affect rotation. In this case, the particle experiences a torque of \( 56.0 \, \mathrm{N \cdot m} \). This means the force applied will cause the particle to rotate around the origin with a specific angular acceleration.
Rate of Change of Angular Momentum
The rate of change of angular momentum is a fundamental concept that emerges from Newton's laws adapted for rotational motion. Angular momentum \( \vec{L} \) is defined as the product of the rotational inertia of a mass and its rotational velocity.
When we think about changes in angular momentum, we relate it to the external torque applied to the system. The principle can be neatly encapsulated by the equation:
\[ \frac{d\vec{L}}{dt} = \vec{\tau} \]
In this exercise, the torque \( \vec{\tau} \) applied to the particle changes its angular momentum at a rate of \( 56.0 \, \mathrm{N \cdot m} \).
  • This direct relationship helps us understand rotational dynamics.
  • Just as force changes linear momentum, torque changes angular momentum.
This principle is key in many applications, from engineering to understanding the motion of celestial bodies. By understanding this concept, we learn how rotational systems respond to external influences, such as forces or torques.
Rotational Dynamics
Rotational dynamics deals with the motion of objects that are rotating. Much like linear dynamics that involves linear forces and masses, rotational dynamics involves torques and rotational mass.
Rotational dynamics is governed by equations that are analogous to those used to describe linear motion. You’ll find familiar equations such as:
\[ \vec{\tau} = I \alpha \]
where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration.
  • Torque works like force does in linear motion.
  • The moment of inertia is like mass but in rotational movement.
  • Angular acceleration is how much the rotational velocity changes.
In the problem, we see how these principles come together as the particle subjected to a torque undergoes a change in its angular momentum through the laws of rotational dynamics. This highlights how forces acting at distances from a pivot effect rotational motion differently than they would if they acted through the center of mass, making it a compelling study.

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

At the instant the displacement of a \(2.00 \mathrm{~kg}\) object relative to the origin is \(\vec{d}=(2.00 \mathrm{~m}) \hat{i}+(4.00 \mathrm{~m}) \mathrm{j}-(3.00 \mathrm{~m}) \mathrm{k},\) its velocity is \(\vec{v}=-(6.00 \mathrm{~m} / \mathrm{s}) \mathrm{i}+(3.00 \mathrm{~m} / \mathrm{s}) \mathrm{j}+(3.00 \mathrm{~m} / \mathrm{s}) \mathrm{k}\) and it is subject to a force \(\vec{F}=(6.00 \mathrm{~N}) \hat{\mathrm{i}}-(8.00 \mathrm{~N}) \mathrm{j}+(4.00 \mathrm{~N}) \mathrm{k} .\) Find (a) the acceleration of the object, (b) the angular momentum of the object about the origin, (c) the torque about the origin acting on the object, and (d) the angle between the velocity of the object and the force acting on the object.

A \(2.0 \mathrm{~kg}\) particle-like object moves in a plane with velocity components \(v_{x}=30 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=60 \mathrm{~m} / \mathrm{s}\) as it passes through the point with \((x, y)\) coordinates of \((3.0,-4.0) \mathrm{m} .\) Just then, in unitvector notation, what is its angular momentum relative to (a) the origin and (b) the point located at (-2.0,-2.0) \(\mathrm{m} ?\)

An overhead view of a thin uniform rod of length \(0.800 \mathrm{~m}\) and mass \(M\) rotating horizontally at angular speed \(20.0 \mathrm{rad} / \mathrm{s}\) about an axis through its center. A particle of mass \(M / 3,00\) initially attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particle's speed \(v_{p}\) is \(6.00 \mathrm{~m} / \mathrm{s}\) greater than the speed of the rod end just after ejection, what is the value of \(v_{p} ?\)

A cockroach of mass \(m\) lies on the rim of a uniform disk of mass \(4.00 \mathrm{~m}\) that can rotate freely about its center like a merrygo-round. Initially the cockroach and disk rotate together with an angular velocity of \(0.260 \mathrm{rad} / \mathrm{s}\). Then the cockroach walks halfway to the center of the disk. (a) What then is the angular velocity of the cockroach-disk system? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the system to its initial kinetic energy? (c) What accounts for the change in the kinetic energy?

A thin-walled pipe rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic energy about the central axis parallel to its length?

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.