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Chapter 3: Problem 26

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 \( O \).

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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.

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Most popular questions from this chapter

In the early stages of the Saturn V rocket's launch, mass was ejected at about \(15,000 \mathrm{kg} / \mathrm{s}\), with a speed \(v_{\mathrm{ex}} \approx 2500 \mathrm{m} / \mathrm{s}\) relative to the rocket. What was the thrust on the rocket? Convert this to tons (1 ton \(\approx 9000\) newtons) and compare with the rocket's initial weight (about 3000 tons).

Many applications of conservation of momentum involve conservation of energy as well, and we haven't yet begun our discussion of energy. Nevertheless, you know enough about energy from your introductory physics course to handle some problems of this type. Here is one elegant example: An elastic collision between two bodies is defined as a collision in which the total kinetic energy of the two bodies after the collision is the same as that before. (A familiar example is the collision between two billiard balls, which generally lose extremely little of their total kinetic energy.) Consider an elastic collision between two equal mass bodies, one of which is initially at rest. Let their velocities be \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}=0\) before the collision, and \(\mathbf{v}_{1}^{\prime}\) and \(\mathbf{v}_{2}^{\prime}\) after. Write down the vector equation representing conservation of momentum and the scalar equation which expresses that the collision is elastic. Use these to prove that the angle between \(\mathbf{v}_{1}^{\prime}\) and \(\mathbf{v}_{2}^{\prime}\) is \(90^{\circ} .\) This result was important in the history of atomic and nuclear physics: That two bodies emerged from a collision traveling on perpendicular paths was strongly suggestive that they had equal mass and had undergone an elastic collision.

The first couple of minutes of the launch of a space shuttle can be described very roughly as follows: The initial mass is \(2 \times 10^{6} \mathrm{kg}\), the final mass (after 2 minutes) is about \(1 \times 10^{6} \mathrm{kg}\), the average exhaust speed \(v_{\mathrm{ex}}\) is about \(3000 \mathrm{m} / \mathrm{s},\) and the initial velocity is, of course, zero. If all this were taking place in outer space, with negligible gravity, what would be the shuttle's speed at the end of this stage? What is the thrust during the same period and how does it compare with the initial total weight of the shuttle (on earth)?

Use spherical polar coordinates \(r, \theta, \phi\) to find the CM of a uniform solid hemisphere of radius R, whose flat face lies in the \(x y\) plane with its center at the origin. Before you do this, you will need to convince yourself that the element of volume in spherical polars is \(d V=r^{2} d r \sin \theta d \theta d \phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)

Consider a uniform solid disk of mass \(M\) and radius \(R\), rolling without slipping down an incline which is at angle \(\gamma\) to the horizontal. The instantaneous point of contact between the disk and the incline is called \(P\). (a) Draw a free-body diagram, showing all forces on the disk. (b) Find the linear acceleration \(\dot{v}\) of the disk by applying the result \(\dot{\mathbf{L}}=\Gamma^{\text {ext }}\) for rotation about \(P .\) (Remember that \(L=I \omega\) and the moment of inertia for rotation about a point on the circumference is \(\frac{3}{2} M R^{2} .\) The condition that the disk not slip is that \(v=R \omega \text { and hence } \dot{v}=R \dot{\omega} .)\) (c) Derive the same result by applying \(\dot{\mathbf{L}}=\mathbf{\Gamma}^{\mathrm{ext}}\) to the rotation about the CM. (In this case you will find there is an extra unknown, the force of friction. You can eliminate this by applying Newton's second law to the motion of the CM. The moment of inertia for rotation about the \(\mathrm{CM}\) is \(\frac{1}{2} M R^{2}\).)
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