Problem 19 Find Lagrange's equations in pol... [FREE SOLUTION]

Chapter 9: Problem 19

Find Lagrange's equations in polar coordinates for a particle moving in a plane if the potential energy is \(V=\frac{1}{2} k r^{2}\).

Short Answer

Expert verified

The Lagrange's equations in polar coordinates are \[ \frac{d^{2} r}{d t^{2}} = r \left( \frac{d \theta}{d t} \right)^{2} - \frac{k}{m} r \] and \[ \frac{d}{d t} \left( r^2 \frac{d \theta}{d t} \right) = 0 \].

Step by step solution

Define the Lagrangian

The Lagrangian is given by the kinetic energy minus the potential energy. In polar coordinates, the kinetic energy for a particle is given by \[ T = \frac{1}{2} m (\frac{d r}{d t})^2 + \frac{1}{2} m r^2 (\frac{d \theta}{d t})^2 \]and the potential energy is \[ V = \frac{1}{2} k r^2 \].Thus the Lagrangian is \[ L = T - V = \frac{1}{2} m (\frac{d r}{d t})^2 + \frac{1}{2} m r^2 (\frac{d \theta}{d t})^2 - \frac{1}{2} k r^2 \].

Write the Euler-Lagrange Equations

The Euler-Lagrange equation for a coordinate q is given by \[ \frac{d}{dt} \left( \frac{\frac{\theta}{\theta}}{\frac{\frac{Halg}{dHalg}} \sum^{3.2}{i}} \right) - \frac{\frac{\theta}{\theta}}{frac{\partial L}{\partial q}} = 0 \].

Derive the Equation for r

For the radial coordinate r, applying the Euler-Lagrange equation we get \[ \frac{d}{dt} \left( m \frac{d r}{d t} \right) - \frac{\partial L}{\partial r} = 0 \].This results in \[ m \frac{d^{2} r}{d t^{2}} - m r \left( \frac{d \theta}{d t} \right)^{2} + k r = 0 \].Therefore, the equation of motion for r is \[ \frac{d^{2} r}{d t^{2}} = r \left( \frac{d \theta}{ d t} \right)^{2} - \frac{k}{m} r \].

Derive the Equation for \( \theta \)

For the angular coordinate \( \theta \), the Euler-Lagrange equation is \[ \frac{d}{dt} \left( m r^2 \frac{d \theta}{d t} \right) = 0 \].This simplifies to \[ m r^2 \frac{d \theta}{d t} = c \], where c is a constant.Hence, the equation of motion for \( \theta \) is \[ \frac{d}{d t} \left( r^2 \frac{d \theta}{d t} \right) = 0 \].

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Key Concepts

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

Lagrangian mechanics

Lagrangian mechanics is a reformulation of classical mechanics. It was developed by Joseph-Louis Lagrange in the 18th century. The core idea revolves around the Lagrangian function, denoted as \(L\), which is the difference between the kinetic energy (\(T\)) and potential energy (\(V\)) of a system: \(L = T - V\).
Here鈥檚 how it works:

Identify the system's kinetic and potential energies.
Construct the Lagrangian function using these energies.
Apply the Euler-Lagrange equation to derive the equations of motion.

This approach can simplify complex mechanics problems especially when dealing with constraints and non-Cartesian coordinates, such as polar coordinates.

Euler-Lagrange equation

The Euler-Lagrange equation is a fundamental equation in the Lagrangian formulation of mechanics. It can be written as:
\ \frac{d}{dt} \left( \frac{\rott L}{\rott \dot{q}} \right) - \frac{\rott L}{\rott q} = 0 \,
where \(q\) represents a generalized coordinate and \(\dot{q}\) its time derivative. To apply it:

Compute the partial derivatives of the Lagrangian with respect to the coordinate (\(q\)) and its derivative (\(\frac{\rott \dot{q}}{\rott t}\)).
Formulate the equation by setting the time derivative of the partial derivative equal to the partial derivative with respect to the coordinate.
Solve this to find the equations of motion.

This equation encapsulates how the system's behavior can be predicted through variations in the system's path.

Polar coordinates

Polar coordinates offer a way to describe two-dimensional motion using a radius (\(r\)) and an angle (\(\theta\)). This is particularly useful when dealing with circular or orbital motion.
Key points to remember:

The position of a point is determined by \(r\) (distance from the origin) and \(\theta\) (angle from some fixed direction).
Kinetic energy must include both radial and angular components: \(T = \frac{1}{2}m \left( \frac{dr}{dt} \right)^2 + \frac{1}{2} m r^2 \left( \frac{d \theta}{dt} \right)^2\).
Potential energy often depends only on \(r\), like \(V = \frac{1}{2}k r^2\) in our example.

Using polar coordinates can simplify the analysis of systems that have radial symmetry.

Kinetic energy

Kinetic energy is the energy a particle has because of its motion. In polar coordinates, kinetic energy is split into two parts:

Radial kinetic energy: associated with the movement along the radius, \(T_r = \frac{1}{2}m \left( \frac{dr}{dt} \right)^2\).
Angular kinetic energy: associated with the rotational movement, \(T_\theta = \frac{1}{2}m r^2 \left( \frac{d \theta}{dt} \right)^2\).

Thus, the total kinetic energy in polar coordinates combines these two contributions: \(T = \frac{1}{2}m \left( \frac{dr}{dt} \right)^2 + \frac{1}{2} m r^2 \left( \frac{d \theta}{dt} \right)^2\).

Potential energy

Potential energy is the energy stored in a system due to its position in a force field, such as gravity or a spring force. In our problem, the potential energy given is \(V = \frac{1}{2} k r^2\).
Important points:

Potential energy typically depends on positional coordinates (e.g., \(r\) in polar coordinates).
It represents the energy that can be converted into kinetic energy or work.
In the context of the Lagrangian, it is subtracted from the kinetic energy to form the Lagrangian: \(L = T - V\).

Recognizing the potential energy form helps in constructing the equations of motion using the Lagrangian method.

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