Question

You are provided with a line of length \(\pi a / 2\) and negligible mass and some lead shot of total mass \(M .\) Use a variational method to determine how the lead shot must be distributed along the line if the loaded line is to hang in a circular arc of radius \(a\) when its ends are attached to two points at the same height. (Measure the distance \(s\) along the line from its centre.)

Short answer

The lead shot must be uniformly distributed along the line to hang in a circular arc of radius a.

Step-by-step solution

- Understanding the Problem

The goal is to determine the distribution of the lead shot mass along a line so that it hangs in a circular arc of radius a. The length of the line is \(\frac{\ra}{2}\) and its total mass is M. The ends of the line are attached to two points at the same height. We measure the distance s along the line from its center.

- Set Up the Variational Problem

To use the variational method, define the potential energy of the system. Assuming the line hangs under gravity, the potential energy is given by the integral of the weight distribution times height. \(U = \int _{-L/2}^{L/2} μ(s) y(s) ds \), where \( μ(s) \) is the mass per unit length and \( y(s) \) is the height.

- Parametrize the Arc

Assume the line forms a circular arc. The height of a point on the arc of radius \( a \) and central angle \( θ \) is given by \( y = a \sin (θ) \). Also, the arc length from the center can be written as \( s = a \theta \).

- Express Potential Energy

Substitute these into the potential energy integral. Since \( ds = a d \theta \), we have: \( U = \int_{-\frac{\rL}{2a}}^{\frac{\rL}{2a}} \rho (a \theta) \cdot a \sin \theta \cdot a d \theta = a^2 \int_{-\frac{\rL}{2a}}^{\frac{\rL}{2a}} \rho(aθ) \cdot \sin \theta dθ \).

- Apply Variational Method

Since the mass is uniformly distributed, \( \int_{-\frac{\rL}{2a}}^{\frac{\rL}{2a}} \rho(a \theta) a d \theta = M \). Given the constraints, use the Lagrange multiplier method to find \( \rho(a \theta)\). Solve the Euler-Lagrange equation (stationary action) to get the distribution.

- Determine the Distribution

The result is that \( \rho = \frac{M}{a L} = \frac{2a}{\pi a} \), giving a constant mass distribution. This makes sense because a circular arc with uniform mass distribution balances itself when hanging.

Key concepts

Potential Energy

In physics, potential energy is the energy held by an object because of its position relative to other objects. For the exercise, the main focus is on gravitational potential energy.
We define the potential energy, U, of the system with a given formula:
\( U = \int _{-L/2}^{L/2} μ(s) y(s) ds \),
where μ(s) represents the mass per unit length and y(s) is the height function.
The potential energy tells us how the mass distribution of the lead shots along the line influences the system's energy state under gravity.

Gravity

Gravity is the force that pulls objects towards each other. On Earth, this force gives weight to physical objects and influences their potential energy.
In this problem, the line with lead shots is subject to gravity, meaning its potential energy depends on the height of each mass element along the arc.
Each segment’s contribution to the total potential energy is its weight (mass times gravity) multiplied by its height relative to the reference point at the bottom of the arc. Gravity is fundamental to understanding how the potential energy is calculated for each part of the arc.

Euler-Lagrange Equation

The Euler-Lagrange equation is a key tool in the variational method. It's used to find the function that minimizes or maximizes a given integral.
For our purpose, this involves finding the mass distribution that minimizes the potential energy, U.
The general form of the Euler-Lagrange equation is:
\[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0 \]
where L is the Lagrangian, q represents the coordinates, and \dot{q} denotes the time derivatives of those coordinates.
For the mass distribution problem, you adapt this concept to work with the spatial distribution rather than time.
After setting up the variational integral, solving the Euler-Lagrange equation helps us find ρ(s), the mass per unit length, that ensures the system’s stationary action.

Mass Distribution

Mass distribution refers to how mass is spread along the line. In this problem, the aim was to determine this distribution so the line forms a circular arc under gravity.
It's essential to understand that the mass distribution directly affects the potential energy of the system.
In our solved example:
- Start by assuming a uniform distribution initially.
- Use the variational method and Euler-Lagrange equation to account for the constraints.
- Perform integration and solve for the mass per unit length, ρ(s).
Finally, it was found that the mass is uniformly distributed because this results in a circular arc hanging in a balanced manner. Therefore, the mass density becomes constant and can be calculated using the total mass M and the arc length.