Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A particle moves under the influence of a central force directed toward a fixed origin \(O\). (a) Explain why the particle's angular momentum about \(O\) is constant. (b) Give in detail the argument that the particle's orbit must lie in a single plane containing \(O\).

Short Answer

Expert verified
Angular momentum is constant due to zero torque from the central force, ensuring planar motion.

Step by step solution

01

Define Angular Momentum

The angular momentum \( L \) of a particle about a point \( O \) is defined as \( L = r \times p \), where \( r \) is the position vector of the particle relative to \( O \), and \( p \) is the linear momentum \( p = m \cdot v \), with \( m \) being the mass and \( v \) the velocity of the particle.
02

Understand Central Force Concept

Central forces act along the line joining the particle to a fixed point (the origin \( O \)). This means that the force vector \( F \) is of the form \( F = f(r) \cdot \hat{r} \), where \( \hat{r} \) is the unit vector along \( r \), implying that the force is radial.
03

Apply Torque and Angular Momentum Relationship

The torque \( \tau \) acting on the particle is given by \( \tau = r \times F \). For a central force, since \( F \) is along \( r \), the torque is \( \tau = r \times (f(r) \cdot \hat{r}) = 0 \). Since torque is zero, angular momentum \( L \) is conserved (\( \frac{dL}{dt} = 0 \)).
04

Conclude Constant Angular Momentum

With the torque being zero, the derivative of angular momentum with respect to time is zero. Thus, the angular momentum \( L \) of the particle about \( O \) is constant over time.
05

Preservation of Plane Motion

The motion being influenced by a central force and the constancy of angular momentum imply that the particle remains in a plane perpendicular to the angular momentum vector (which is unchanging in direction).
06

Argue Single Plane Orbit

Any deviation from this plane would necessitate either a change in the magnitude of angular momentum or result from an out-of-plane force, neither of which occurs here. Therefore, the particle's orbit lies entirely in the plane containing \( O \).

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 Vaia!

Key Concepts

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

Angular Momentum
Angular momentum is a fundamental concept in physics, especially when dealing with central forces. It is represented by the symbol \( L \) and is calculated using the formula \( L = r \times p \), where \( r \) is the position vector of a particle relative to a fixed point \( O \), and \( p \) is the linear momentum defined as \( p = m \cdot v \). - Here, \( m \) is the mass of the particle and \( v \) is its velocity.
- The angular momentum depends on both the motion of the particle and its position. For particles moving under a central force, one key property of angular momentum is its conservation. This occurs because the torque acting on the particle is zero, meaning there are no external influences causing a change in its rotational motion. Thus, once set, the angular momentum of a particle remains constant over time since \( \frac{dL}{dt} = 0 \).
Torque
Torque, often referred to as the rotational equivalent of force, plays a vital role in determining the motion of particles affected by forces. For a particle under a central force, torque \( \tau \) is calculated as \( \tau = r \times F \), where \( F \) is the force vector acting on the particle.- In a central force scenario, such forces act along the line connecting the particle to a fixed origin.
- This indicates that \( F \) is always radial, pointing directly toward or away from the center.Because \( F = f(r) \cdot \hat{r} \) and due to the property of cross product \( r \times (f(r) \cdot \hat{r}) = 0 \), we deduce that torque vanishes.With zero torque, the rotational motion of the system is conserved, reinforcing the constancy of angular momentum.
Radial Force
A radial force is distinctive as it always acts along the direction between two points. In central force dynamics, this means the force points directly toward or away from the origin, following a line created by the position vector \( r \).- This form of force is characterized as \( F = f(r) \cdot \hat{r} \), where \( \hat{r} \) is the unit vector along \( r \).
- The central force doesn't have any components perpendicular to the position vector. Such forces contribute to the motion being confined to a plane. Since the force does not introduce any additional components to alter the path's flatness, it maintains the integrity of motion in a plane perpendicular to the angular momentum vector.
Plane of Motion
The plane of motion is an important concept to grasp when discussing particles in central force fields. Due to the constant nature of the angular momentum, the motion of a particle is constrained to a single plane. - This plane includes the fixed origin and is perpendicular to the constant angular momentum vector.
- Any deviation from this plane would indicate forces that don't align with the central force's radial nature or angular momentum being altered. The characteristics of central forces ensure that movements off this plane do not naturally occur. Therefore, the trajectory of the particle aligns perfectly within this fixed plane, supporting both the force and momentum conservation principles.

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

Consider a gun of mass \(M\) (when unloaded) that fires a shell of mass \(m\) with muzzle speed \(v\). (That is, the shell's speed relative to the gun is \(v\).) Assuming that the gun is completely free to recoil (no external forces on gun or shell), use conservation of momentum to show that the shell's speed relative to the ground is \(v /(1+m / M).\)

Find the position of the center of mass of three particles lying in the \(x y\) plane at \(\mathbf{r}_{1}=(1,1,0)\) \(\mathbf{r}_{2}=(1,-1,0),\) and \(\mathbf{r}_{3}=(0,0,0),\) if \(m_{1}=m_{2}\) and \(m_{3}=10 m_{1} .\) Illustrate your answer with a sketch and comment.

A particle of mass \(m\) is moving on a frictionless horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where I am holding it. Initially the particle is moving in a circle of radius \(r_{\mathrm{o}}\) with angular velocity \(\omega_{\mathrm{o}},\) but I now pull the string down through the hole until a length \(r\) remains between the hole and the particle. What is the particle's angular velocity now?

A uniform thin sheet of metal is cut in the shape of a semicircle of radius \(R\) and lies in the \(x y\) plane with its center at the origin and diameter lying along the \(x\) axis. Find the position of the CM using polar coordinates. [In this case the sum (3.9) that defines the CM position becomes a two- dimensional integral of the form \(\int \mathbf{r} \sigma d A\) where \(\sigma\) denotes the surface mass density (mass/area) of the sheet and \(d A \text { is the element of area } d A=r d r d \phi .]\)

A rocket (initial mass \(m_{\mathrm{o}}\) ) needs to use its engines to hover stationary, just above the ground. (a) If it can afford to burn no more than a mass \(\lambda m_{\mathrm{o}}\) of its fuel, for how long can it hover? [Hint: Write down the condition that the thrust just balance the force of gravity. You can integrate the resulting equation by separating the variables \(t\) and \(m .\) Take \(v_{\text {ex }}\) to be constant.] (b) If \(v_{\mathrm{ex}} \approx 3000 \mathrm{m} / \mathrm{s}\) and \(\lambda \approx 10 \%,\) for how long could the rocket hover just above the earth's surface?

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.

Sign-up for free