Question

A particle of mass \(m\) is free to move in a horizontal plane. It is attached to a fixed point \(O\) by a light elastic string, of natural length \(a\) and modulus \(\lambda\). Show that the tension in the string is conservative and show that the Lagrangian for the motion when \(r>a\) is $$ L=\frac{1}{2} m\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}\right)-\frac{\lambda}{2 a}(r-a)^{2} $$ where \(r\) and \(\theta\) are plane polar coordinates with origin \(O\).

Short answer

A: The Lagrangian expression for the given system when \(r > a\) is \(L = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2) - \frac{\lambda}{2a}(r - a)^2\).

Step-by-step solution

Tension in the string and its conservativeness

The tension in the elastic string follows Hooke's Law: \(T = \frac{\lambda}{a}(r - a)\), where \(\lambda\) is the modulus and \(a\) is the natural length of the string. We can check if it's conservative by calculating the curl of the force that corresponds to the tension or, equivalently, by checking that it can be derived from a potential energy function. Since we have a radial force, \(\mathbf{F} = T\mathbf{e_r}\), and its curl will be in the \(z\) direction only, we can write: $$ \nabla \times \mathbf{F} = \frac{1}{r} \frac{\partial(T)}{\partial \theta}\mathbf{k} $$ Since tension \(T\) does not depend on \(\theta\), the term \(\frac{\partial(T)}{\partial \theta} = 0\), and thus \(\nabla \times \mathbf{F} = 0\). The tension in the string is conservative.

Kinetic energy in plane polar coordinates

In plane polar coordinates, the position vector of the particle is given by \(\mathbf{r} = r\mathbf{e_r}\). The velocity vector can be obtained by taking the time derivative of the position vector: $$ \mathbf{v} = \frac{d \mathbf{r}}{dt} = \dot{r} \mathbf{e_r} + r \dot{\theta} \mathbf{e_\theta} $$ The kinetic energy, \(T\), can be expressed as: $$ T = \frac{1}{2}m\mathbf{v} \cdot \mathbf{v} = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2) $$

Potential energy from tension in the string

Since the tension in the string is conservative, there exists a potential energy function, \(V\), such that \(-\nabla V = \mathbf{F}\). Integrating the radial component of the tension force, we obtain the potential energy function as: $$ V(r) = -\int_{a}^{r} T dr' = -\int_{a}^{r} \frac{\lambda}{a}(r' - a) dr' = \frac{\lambda}{2a}(r - a)^2 $$ Note, we have set \(V(a) = 0\) for deriving the above result.

Constructing the Lagrangian

The Lagrangian, \(L\), can now be constructed using the expressions for kinetic and potential energy: $$ L = T - V = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2) - \frac{\lambda}{2a}(r - a)^2 $$ This Lagrangian describes the motion of the particle in the plane polar coordinates \((r, \theta)\) for the given system when \(r > a\).

Key concepts

Conservative Forces

In mechanics, conservative forces are special because they conserve mechanical energy. A force is deemed conservative if it can be derived from a potential energy function. This means that the work done by the force on an object moving from one point to another only depends on the initial and final positions, not on the trajectory taken. This attribute leads to path independence.
To determine if a force like the tension in an elastic string is conservative, we can check its curl. If the curl is zero, the force is conservative. For our case, the tension is given by Hooke’s Law: \[ T = \frac{\lambda}{a}(r - a) \]Where the tension is a function of the radial distance, but not of the angular position \(\theta\). Calculating the curl, the partial derivative with respect to \(\theta\) yields zero, confirming the conservative nature of the tension in the string.

Plane Polar Coordinates

Plane polar coordinates are a two-dimensional coordinate system, particularly useful for problems with radial symmetry, like our elastic string problem. They consist of two coordinates: \(r\) for the radial distance from a fixed origin, and \(\theta\) for the angular position measured from a reference direction.
When working with plane polar coordinates, kinetics takes a special form. A position vector \(\mathbf{r} = r\mathbf{e_r}\) is described based on these coordinates. To find velocity, we differentiate the position vector with respect to time. This gives:\[ \mathbf{v} = \dot{r} \mathbf{e_r} + r \dot{\theta} \mathbf{e_\theta} \]Where \(\dot{r}\) is the rate of change of the radial distance, and \(r\dot{\theta}\) is related to the angular velocity component. These expressions are used to calculate the kinetic energy of a system in plane polar coordinates.

Potential Energy Function

The potential energy function is associated with conservative forces, representing the energy stored in a system due to its position or configuration. For our string, potential energy \(V(r)\) arises from its tension. Knowing the force's form, we can find \(V(r)\) by integrating the force over distance.
The integration for the radial component of tension gives:\[ V(r) = \int_{a}^{r} -\frac{\lambda}{a}(r' - a) dr' \]Carrying out this integration, we find:\[ V(r) = \frac{\lambda}{2a}(r - a)^2 \]This potential function sets the reference energy at the natural length \(a\) to zero. This makes sense since no tension means no stored energy, aligning with the physical setup. The potential energy function is an integral part of defining a system's Lagrangian, which combines both kinetic and potential energies to describe motion holistically.