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.

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: \[T = T_{rod} + T_{base}\]

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: \[V = V_{spring} + V_{support}\]

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: \[ L = \left(\frac{1}{2} m_a\left(\frac{L}{2}\right)^2 \dot{\theta}^2 + \frac{1}{2} m_b \left(\frac{L}{2}\dot{\theta}\right)^2\right) - \left(\frac{1}{2} k_{T} \theta^2 + \frac{1}{2} k\left(\frac{L}{2}\theta\right)^2\right) \]

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.