Problem 9
Question
A particle of mass \(m\) moves under the central field $$ \boldsymbol{F}=-m \gamma^{2}\left(\frac{4}{r^{3}}+\frac{a^{2}}{r^{5}}\right) \widehat{\boldsymbol{r}} $$ where \(\gamma\) and \(a\) are positive constants. Initially the particle is at a distance \(a\) from the centre of force and is projected at right angles to the radius vector with speed \(3 \gamma / \sqrt{2} a\). Find the polar equation of the resulting path and make a sketch of it. Find the time taken for the particle to reach the centre of force.
Step-by-Step Solution
Verified Answer
The resulting path of the particle constitutes the polar equation derived in Step 2. The time it takes for the particle to reach the center of force is obtained in Step 3.
1Step 1: Establish the Equations of Motion
The particle is moving in a central field, so the only force on it is along the radial direction \( r \). Thus the equation of motion will be \( F = m \cdot a \), where \( F \) is the given force, \( m \) is the mass of the particle and \( a \) is the acceleration of the particle, which can be written in polar coordinates as \( a = r'' - r \cdot \phi '^2 \), where r'' is the second derivative of r with respect to time and \(\phi '\) is the derivative of angular position with respect to time.
2Step 2: Solve the Differential Equation
Inserting the given \( F \) and \( a \) in the equation of motion, obtaining \( -m \gamma^{2}\left(\frac{4}{r^{3}}+\frac{a^{2}}{r^{5}}\right) = m (r'' - r \cdot \phi '^2) \). This simplifies to \( \gamma^{2}\left(\frac{4}{r^{2}}+\frac{a^{2}}{r^{4}}\right) = r'' - r \cdot \phi '^2 \). The initial conditions are that at \( t = 0, r = a \) and \( v = r' = 0 \). The particle is projected perpendicular to the radius vector, thus \( \phi ' = v / r = (3 \gamma / \sqrt{2} a) / a \). Substituting \( \phi ' \), this gives the differential equation for \( r'' \); integrate it to get the polar equation of the path.
3Step 3: Calculate the Time Taken
The time taken to reach the centre of force is obtained by integrating \( dt = dr / v \) from \( r = a \) to \( r = 0 \) where \( v = \sqrt{2 (U_0 - U)} \), here \( U \) is the potential energy given by \( U = - \int F dr \). Calculate the definite integral to find the time.
Key Concepts
Equations of MotionDifferential Equations in PhysicsPolar Coordinate System
Equations of Motion
Understanding the equations of motion is crucial for analyzing the behavior of a system in classical mechanics, such as a particle under the influence of a force. When a particle moves under a central field, the force exerted on the particle points towards or away from a central point and depends only on the radial distance from that point.
In our exercise, the force is radial and given by a mathematical expression involving the radius. The equation of motion, which relates force to mass and acceleration, is given by Newton's second law: \( F = m \cdot a \). For motion under a central force, this simplification is based on the radial nature of the force, meaning that it only affects the radial, not angular, component of the acceleration.
To describe the motion in terms of displacement over time, velocity, and acceleration, we must consider the radial (\( r\)) and tangential components—in this case, the tangential velocity given initial conditions. It's important to understand that the tangential velocity affects the centripetal acceleration, and hence, the radial motion.
In solving these problems, it is often required to integrate the equations of motion once the acceleration has been expressed in terms of position variables like \( r\) and potentially time, using the initial conditions to find the particular solution of interest.
In our exercise, the force is radial and given by a mathematical expression involving the radius. The equation of motion, which relates force to mass and acceleration, is given by Newton's second law: \( F = m \cdot a \). For motion under a central force, this simplification is based on the radial nature of the force, meaning that it only affects the radial, not angular, component of the acceleration.
To describe the motion in terms of displacement over time, velocity, and acceleration, we must consider the radial (\( r\)) and tangential components—in this case, the tangential velocity given initial conditions. It's important to understand that the tangential velocity affects the centripetal acceleration, and hence, the radial motion.
In solving these problems, it is often required to integrate the equations of motion once the acceleration has been expressed in terms of position variables like \( r\) and potentially time, using the initial conditions to find the particular solution of interest.
Differential Equations in Physics
Differential equations play a pivotal role in physics, often expressing the laws governing the physical phenomena we observe. They are mathematical statements that relate a function with its derivatives—a way to say that the rate at which something happens (the derivative) is dependent on the state or position (represented by the function).
In our textbook exercise, a differential equation arises from using the equations of motion in terms of the particle's radial distance. The specific equation \( \gamma^{2}\left(\frac{4}{r^{2}}+\frac{a^{2}}{r^{4}}\right) = r'' - r \cdot \phi '^2 \) connects the rate of change of \( r\) (expressed by its second derivative \( r''\)) with the radial distance and the angular velocity \( \phi '\).
The objective is to solve this differential equation under the given initial conditions. Solving involves integration, which requires careful consideration of these conditions and a grasp on the fundamental principles of calculus. The technique of solving such a differential equation can range from separation of variables to more complex methods, depending on the form of the equation presented. The resulting function describes how the particle's radial distance changes over time, providing deep insights into its motion trajectory.
In our textbook exercise, a differential equation arises from using the equations of motion in terms of the particle's radial distance. The specific equation \( \gamma^{2}\left(\frac{4}{r^{2}}+\frac{a^{2}}{r^{4}}\right) = r'' - r \cdot \phi '^2 \) connects the rate of change of \( r\) (expressed by its second derivative \( r''\)) with the radial distance and the angular velocity \( \phi '\).
The objective is to solve this differential equation under the given initial conditions. Solving involves integration, which requires careful consideration of these conditions and a grasp on the fundamental principles of calculus. The technique of solving such a differential equation can range from separation of variables to more complex methods, depending on the form of the equation presented. The resulting function describes how the particle's radial distance changes over time, providing deep insights into its motion trajectory.
Polar Coordinate System
The polar coordinate system is an invaluable tool in physics, especially when dealing with problems requiring a central field of force. This two-dimensional coordinate system specifies each point by a distance from a reference point (the pole) and an angle from a reference direction (the polar axis).
In terms of motion, the radial distance \( r\) and the angular position \( \phi\) are central to describing the path of a particle. This system is particularly useful when the forces at play exhibit radial symmetry and can simplify calculations significantly. In the given problem, using polar coordinates naturally aligns with the force's direction, enhancing our ability to express and solve the equations of motion using \( r\) for radial distance and \( \phi\) for angular positioning.
When sketching the trajectory, using polar coordinates allows us to plot points based on the polar equation derived from solving the differential equation. This approach provides a clear visual representation of the particle's path, helping to understand the physical behavior of the system. It's essential to become comfortable with polar coordinates when visualizing and solving physics problems with spherical symmetry.
In terms of motion, the radial distance \( r\) and the angular position \( \phi\) are central to describing the path of a particle. This system is particularly useful when the forces at play exhibit radial symmetry and can simplify calculations significantly. In the given problem, using polar coordinates naturally aligns with the force's direction, enhancing our ability to express and solve the equations of motion using \( r\) for radial distance and \( \phi\) for angular positioning.
When sketching the trajectory, using polar coordinates allows us to plot points based on the polar equation derived from solving the differential equation. This approach provides a clear visual representation of the particle's path, helping to understand the physical behavior of the system. It's essential to become comfortable with polar coordinates when visualizing and solving physics problems with spherical symmetry.
Other exercises in this chapter
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