Question

\({ }^{\dagger}\) A particle \(A\) of mass \(m\) is suspended from a fixed point \(O\) by a light string of length \(3 a\). A second particle \(B\) of mass \(m\) is suspended from \(A\) by a second light string of length \(2 b\) and a third particle \(C\) of mass \(m\) is suspended from \(B\) by a third light string of length \(c\). The system can move in a vertical plane through \(O\). Let \(x, y, z\) denote the horizontal displacements of \(A, B, C\) from their equilibrium positions. Show that an approximate Lagrangian

Short answer

Answer: The approximate Lagrangian for the system is given by \(L \approx \frac{1}{2}m (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - mg(x^2 + y^2 + z^2)/(6a)\), where m is the mass of particles, g is the gravitational acceleration, and a is a constant related to the equilibrium positions of the particles. The coordinates x, y, and z represent the small displacements of the particles from their equilibrium positions.

Step-by-step solution

Find the kinetic and potential energies of each particle

To find the energies, we first need to express the positions of particles A, B, and C in terms of x, y, z and use these positions to calculate the kinetic and potential energies. For particle A, the position vector is: \(r_A = (x, 3a - \sqrt{3a^2 - x^2})\) For particle B, the position vector is: \(r_B = (x + y, 3a - \sqrt{3a^2 - x^2} - \sqrt{2b^2 - y^2})\) For particle C, the position vector is: \(r_C = (x + y + z, 3a - \sqrt{3a^2 - x^2} - \sqrt{2b^2 - y^2} - \sqrt{c^2 - z^2})\) Now, we calculate the kinetic energy for each particle, T_i = (1/2) m v_i^2 The kinetic energy of particle A is: \(T_A = \frac{1}{2}m\dot{r}_A^2\) The kinetic energy of particle B is: \(T_B = \frac{1}{2}m\dot{r}_B^2\) The kinetic energy of particle C is: \(T_C = \frac{1}{2}m\dot{r}_C^2\) The potential energy for each particle is \(V_i=mgh\). For particle A: \(V_A = mg(3a - \sqrt{3a^2 - x^2})\) For particle B: \(V_B = mg(3a - \sqrt{3a^2 - x^2} - \sqrt{2b^2 - y^2})\) For particle C: \(V_C = mg(3a - \sqrt{3a^2 - x^2} - \sqrt{2b^2 - y^2} - \sqrt{c^2 - z^2})\)

Obtain the Lagrangian

The Lagrangian is given by the expression \(L = T - V\), where T is the total kinetic energy and V is the total potential energy. So, we need to sum the kinetic and potential energies for all the particles: \(T = T_A + T_B + T_C\) \(V = V_A + V_B + V_C\) \(L = T - V\) Now, we assume that the displacements x, y, z are small. This means that terms containing higher powers of the displacements can be neglected. We expand and simplify the expressions for T and V, keeping only the first-order terms: \(L \approx \frac{1}{2}m (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - mg(x^2 + y^2 + z^2)/(6a)\) Now, we have an approximate Lagrangian for the system: \(L \approx \frac{1}{2}m (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - mg(x^2 + y^2 + z^2)/(6a)\)

Key concepts

Kinetic Energy

In the context of Lagrangian mechanics, the kinetic energy of a system is crucial for understanding its dynamics. It represents the energy that the system possesses due to the motion of its parts. For example, in the system given in the problem, we consider the kinetic energy of each moving particle, denoted as particle A, B, and C respectively.

Kinetic energy is mathematically expressed as \( T = \frac{1}{2}mv^2 \), where \( m \) represents the mass of the particle and \( v \) is the velocity. For a particle in motion along multiple axes, the total kinetic energy is the sum of kinetic energies for motion in each direction. Since we're dealing primarily with horizontal motion, the kinetic energy calculations consider only the horizontal components of the velocity. The energy is quadratic in terms of velocity, indicating that it rapidly increases as the speed of the particle increases.

When solving problems, understanding the quadratic nature of kinetic energy aids in the recognition that higher-order terms in velocity will significantly influence the system’s total energy. This knowledge is particularly important when working with the Lagrangian, which involves both kinetic and potential energy.

Potential Energy

Potential energy, often symbolized as \( V \), in a physical system, refers to the energy stored by an object as a result of its position or configuration. Considering the same system in our exercise, the potential energy is related to the height of each particle above a reference level, typically chosen to be where the potential energy is zero. This reference level could be the equilibrium position of the particles when the system is undisturbed.

The formula for gravitational potential energy is \( V = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above the reference level. Since the system under consideration is in a vertical plane, the potential energy depends on the vertical displacements of the particles from their equilibrium positions.

When dealing with Lagrangian mechanics, potential energy is subtracted from the kinetic energy to calculate the Lagrangian function, emphasizing its equal importance. Recognizing changes in potential energy as a result of displacement is important for understanding the system's dynamics, particularly in instances of small oscillations where potential energy and restoring forces are intimately connected.

Small Oscillations

Small oscillations refer to the motion near the equilibrium position where the displacement of the system is small enough to allow approximations that simplify the mathematical analysis. It's essential to identify such conditions because they allow the application of Taylor series expansion and usually result in linear equations of motion, which are much simpler to solve.

The assumption of small oscillations suggests that we can neglect higher-order terms of displacement in our calculations, as performed in the step-by-step solution. This is because these terms would contribute insignificantly to the system's dynamics when the displacements are sufficiently small.

In the study of Lagrangian mechanics, small oscillations are often used to explore the stable equilibrium of a system. The Lagrangian can be simplified significantly under the assumption of small oscillations, leading to a clear understanding of the natural frequencies and modes of the system. These simplifications are critical for designing and analyzing mechanical systems and structures that exhibit vibrational behavior.