Question

A rigid rod of length \(L\), and mass \(m_{a}\) is connected to a rigid base of mass \(m_{b}\) through a torsional spring of stiffness \(k_{T}\) as shown. The base sits on an elastic support of stiffness \(k\) as indicated. Derive the equations of motion of the system using Lagrange's equations.

Short answer

Answer: The equation of motion for the given system is \[\left(m_a + m_b\right)\ddot{\theta} = \left(k_T + \frac{kL^2}{4}\right)\theta\].

Step-by-step solution

1. Determine the Kinetic Energy (T) of the System

We start by finding the kinetic energy of the system, which consists of two parts: kinetic energy of the rod and kinetic energy of the base. For the rod: Half of the rod's mass, \(m_a\), is concentrated at a distance of L/2 from the axis of rotation. Its angular velocity is \(\dot{\theta}\). Therefore, the kinetic energy of the rod can be written as: \[T_{rod} = \frac{1}{2} m_a\left(\frac{L}{2}\right)^2 \dot{\theta}^2\] For the base: The linear velocity of the center of mass of the base, \(v_{base}\), can be written as: \[v_{base} = \frac{L}{2} \dot{\theta}\] Hence, the kinetic energy of the base is: \[T_{base} = \frac{1}{2} m_b v_{base}^2 = \frac{1}{2} m_b \left(\frac{L}{2}\dot{\theta}\right)^2\] The total kinetic energy of the system is the sum of these two parts: \[T = T_{rod} + T_{base}\]

2. Determine the Potential Energy (V) of the System

The potential energy of the system comes from the torsional spring and the elastic support. For the torsional spring: The potential energy stored in the torsional spring with stiffness \(k_T\) is given by: \[V_{spring} = \frac{1}{2} k_{T} \theta^2\] For the elastic support: The displacement of the base with respect to the support is \(\frac{L}{2}\theta\), and the potential energy stored in the elastic support with stiffness \(k\) is given by: \[V_{support} = \frac{1}{2} k\left(\frac{L}{2}\theta\right)^2\] The total potential energy of the system is the sum of these two parts: \[V = V_{spring} + V_{support}\]

3. Write the Lagrangian of the System (L)

The Lagrangian of the system is the difference between the kinetic energy and potential energy: \[ L = T - V\] Substitute the expressions for the kinetic and potential energies into the Lagrangian equation: \[ L = \left(\frac{1}{2} m_a\left(\frac{L}{2}\right)^2 \dot{\theta}^2 + \frac{1}{2} m_b \left(\frac{L}{2}\dot{\theta}\right)^2\right) - \left(\frac{1}{2} k_{T} \theta^2 + \frac{1}{2} k\left(\frac{L}{2}\theta\right)^2\right) \]

4. Derive the Equations of Motion using Lagrange's Equations

For a single degree-of-freedom system like ours, where the generalized coordinate is \(\theta\), the Lagrange's equation is given by: \[\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}} - \frac{\partial L}{\partial \theta} = 0\] Now, compute the required partial derivatives, substitute the expressions, and solve the equation to obtain the equations of motion: From the Lagrangian expression, we have: \[\frac{\partial L}{\partial \dot{\theta}} = 2 \cdot \frac{1}{2} m_a\left(\frac{L}{2}\right)^2 \dot{\theta} + 2 \cdot \frac{1}{2} m_b \left(\frac{L}{2}\dot{\theta}\right)\left(\frac{L}{2}\right)\] And, \[\frac{\partial L}{\partial \theta} = -k_{T}\theta - k\left(\frac{L}{2}\theta\right)\left(\frac{L}{2}\right)\] Now compute the time derivative of \(\frac{\partial L}{\partial \dot{\theta}}\): \[\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}} = m_a\left(\frac{L}{2}\right)^2 \ddot{\theta} + m_b \left(\frac{L}{2}\right)^2 \ddot{\theta}\] Substitute the expressions for the partial derivatives and their time derivatives into Lagrange's equation: \[(m_a\left(\frac{L}{2}\right)^2 + m_b\left(\frac{L}{2}\right)^2) \ddot{\theta} = k_{T}\theta + k\left(\frac{L}{2}\right)^2\theta\] Finally, the equation of motion for the given problem can be written as: \[\left(m_a + m_b\right)\ddot{\theta} = \left(k_T + \frac{kL^2}{4}\right)\theta\] This is the derived equation of motion for the given system using Lagrange's equations.

Key concepts

Kinetic Energy

The kinetic energy of a system is a measure of the energy possessed by the system due to its motion. For a system involving a rod and a base, each component contributes to the overall kinetic energy.
The rod, being a part of a rotating system, has its kinetic energy calculated considering its angular motion. This results in the equation:

• For the rod: \[T_{rod} = \frac{1}{2} m_a\left(\frac{L}{2}\right)^2 \dot{\theta}^2\]

The base, on the other hand, moves linearly due to the movement of the rod. Its kinetic energy is accounted for by the linear velocity of its center of mass:

• For the base: \[T_{base} = \frac{1}{2} m_b \left(\frac{L}{2}\dot{\theta}\right)^2\]

In Lagrange's equations, the total kinetic energy is the sum of the contributions from the rod and the base, offering insights into how motions of different parts impact the system.

Potential Energy

Potential energy in mechanical systems is associated with forces such as gravity and elastic forces that store energy. In this exercise, potential energy comes from the torsional spring and elastic support.
The torsional spring, which twists about a point, has potential energy stored due to its stiffness when twisted:

• Torsional spring potential energy: \[V_{spring} = \frac{1}{2} k_{T} \theta^2\]

The elastic support contributes potential energy due to its deformation as the system moves:

• Elastic support potential energy:\[V_{support} = \frac{1}{2} k\left(\frac{L}{2}\theta\right)^2\]

Combining these, the total potential energy provides a complete view of stored energy within the system based on its configuration.

Equations of Motion

The equations of motion describe how a system evolves over time under the influence of forces. Using Lagrange's equations helps derive these equations by considering both kinetic and potential energies.
Lagrange's equation for motion is formulated as:

• \[\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}} - \frac{\partial L}{\partial \theta} = 0\]

Where \( L = T - V \) represents the Lagrangian —the difference between kinetic and potential energies. Through differentiation, we derive the system's response to dynamic changes:

• \[\left(m_a + m_b\right)\ddot{\theta} = \left(k_T + \frac{kL^2}{4}\right)\theta\]

This derived equation highlights the interplay between mass and spring forces affecting rotational motion.

Torsional Spring

A torsional spring is a mechanical device that resists rotational movement by applying a torque proportional to the object's angle of twist. In this system, the torsional spring connects the rod to the base.
The spring's stiffness, denoted as \( k_T \), plays a crucial role in determining the potential energy stored:

• Potential energy formula: \[V_{spring} = \frac{1}{2} k_{T} \theta^2\]

By storing energy when twisted, it affects the motion of the system. This mechanics ensure that changes in angular displacement directly impact its contribution to the system through potential energy and subsequent equations of motion.

Elastic Support

Elastic support systems store potential energy through deformation, generally via compression or stretch along their axis. In this exercise, the base is on elastic support that deforms as the system shifts.
Providing energy storage properties:

• Stored energy: \[V_{support} = \frac{1}{2} k\left(\frac{L}{2}\theta\right)^2\]

Here, \( k \)represents the stiffness of the elastic support, which influences how much energy it can store for a specific displacement. This energy storage impacts the equations dictating motion, broadening understanding of the support's role in the overall system dynamics.