Problem 5 - Classical Mechanics - An Undergraduate Text

Question

A particle \(P\) moves under the attractive inverse square field \(\boldsymbol{F}=-\left(m \gamma / r^{2}\right) \widehat{\boldsymbol{r}}\). Initially \(P\) is at the point \(C\), a distance \(c\) from \(O\), and is projected with speed \((3 \gamma / c)^{1 / 2}\) perpendicular to \(O C\). Find the polar equation of the path make a sketch of it. Deduce the angle between \(O C\) and the final direction of departure of \(P\).

Step-by-Step Solution

Answer

The polar equation of the path is obtained by integrating the radial velocity and combining it with angular information. The final angle is obtained by considering the limits of \(r(t)\) and \(\theta\) as \(t\) approaches infinity.

Step 1: Identify the Radial Force on the Particle

First, identify the radial force acting on the particle: it is given by \(F=-\left(m \gamma / r^{2}\right) \widehat{r}\), where \(\widehat{r}\) indicates the directed radial coordinate from the origin \(O\) to the location of the particle.

Step 2: Apply Conservation of Angular Momentum

Applying the concept of angular momentum conservation based on the radial force \(F\), the angular momentum is constant throughout the particle's motion. Thus, write it in polar coordinates: \(m r^{2} d\theta / dt = mrc\), where \(c = (3 \gamma / c)^{1 / 2}\) is the initial speed of the particle and \(m\) is the mass of the particle.

Step 3: Solve for Particle's Radial Velocity

Now, solve the equality from Step 2 for the velocity in terms of the radial position: \(dr/dt = \sqrt{3 \gamma / r}\).

Step 4: Integrate the Radial Velocity

To obtain \(r\) as a function of time \(t\), we integrate the radial velocity: \(\int dr/\sqrt{3 \gamma / r} = \int dt\). This results in a radical function of time \(r(t)\) which represents the radial distance at any given time.

Step 5: Obtain Polar Equation of Path

From the above equation and the radial linearity of \(\theta\), write the polar equation of the path by eliminating \(t\) between \(r(t)\) and \(\theta = wt + \theta_0\), where \(w = mrc\) is the initial angular velocity, and \( \theta_0\) is the original angle.

Step 6: Calculate the Final Angle

Finally, to find the angle between \(OC\) and the final direction, consider the bounds of the radial function \(r(t)\) and \(\theta\) as \(t\) approaches infinity. The difference between the final and initial angles is the desired measure.

Key Concepts

Conservation of Angular Momentum

Understanding the conservation of angular momentum is essential when studying the movement of particles under a central force. In classical mechanics, this principle states that if no external torque acts upon a particle or a system of particles, its angular momentum remains constant. For the particle described in our exercise, moving under the inverse square law force, the absence of external torque ensures that the particle’s angular momentum does not change over time.

This constancy can be mathematically represented using polar coordinates, with the relationship: \( m r^2 \frac{d\theta}{dt} = constant \), where \( m \) is the mass of the particle, \( r \) is the radial distance from the origin, and \( \theta \) is the angular position. In the exercise, the initial projection velocity of the particle allows us to specify this constant, leading to the understanding that regardless of the radial distance changes, the angular velocity will adjust to maintain angular momentum.

Polar Coordinates

Polar coordinates provide a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point (usually the origin) and an angle from a reference direction (commonly the positive x-axis). This system is particularly useful for problems involving radial symmetry or central forces, such as the motion of our particle \( P \) in the exercise.

In polar coordinates, the position of a particle is given by two values: the radial distance \( r \), and the angle \( \theta \). The attractive inverse square law force which the particle experiences is directed along the radial coordinate, making polar coordinates a natural choice for analyzing the particle's motion and simplifying the determination of the path equation.

Particle Dynamics

Particle dynamics refers to the study of how forces affect the motion of a particle. When a particle moves under the influence of a force, its acceleration is determined by Newton's second law, \( F = ma \). In the context of an inverse square law field, the force acting on the particle is radial and varies with the inverse square of the distance from the force center.

For our particle \( P \), projected perpendicular to \( OC \), particle dynamics will involve solving for the velocity and trajectory of the particle using the given initial conditions. By integrating the expressions derived from the forces, we acquire a complete description of the particle's motion over time, allowing us to predict its future position and velocity.

Radial Force

A radial force is a force that acts along the radius connecting the center of a force field to the particle within it. It is central to the dynamics of a particle moving under central forces, such as gravitational or electrostatic fields. In the case of our exercise, the particle experiences a radial force given by \( F = -\left(m \gamma / r^{2}\right) \hat{r} \), where \( m \) is the mass of the particle, \( \gamma \) is a constant characteristic of the force field, \( r \) is the radius, and \( \hat{r} \) is the unit vector in the direction of the radius.

The negative sign in the force indicates that it is attractive, pulling the particle towards the origin. Knowing the nature of this force is key to solving for the particle's motion. It governs how the particle's speed and direction change as it moves, leading to the particle's eventual trajectory, which can be sketched out following the obtained polar path equation.