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 \(r > a\).