A uniform cylindrical drum of mass \(M\) and radius \(a\) is free to rotate about its axis, which is horizontal. A cable of negligible mass and length \(l\) is wound on the drum, and carries on its free end a mass \(m\). Write down the Lagrangian function in terms of an appropriate generalized co-ordinate, assuming no slipping or stretching of the cable. If the cable is initially fully wound up, and the system is released from rest, find the angular velocity of the drum when it is fully unwound.

We will use the angle \(\theta\) as the generalized coordinate. As the drum rotates, the angle increases and the mass m descends. We can define \(x = a\theta\) as the displacement of the mass m. The drum's moment of inertia about its axis is given by \(I = \frac{1}{2}Ma^2\). The kinetic energy of the system is the sum of the kinetic energies of the drum and the mass m, which can be expressed as \(T = \frac{1}{2}I\dot{\theta}^2 + \frac{1}{2}m(\dot{x})^2\). The potential energy of the system is given by the height of mass m above the ground, which is \(U = mgh\), where \(h = l - x = l - a\theta\).

The Lagrangian function is defined as \(L = T - U\). From the previous step, we substitute the expressions for T and U into the Lagrangian function: \(L = \frac{1}{2}I\dot{\theta}^2 + \frac{1}{2}m(\dot{x})^2 - mg(l-x)\) In order to write the Lagrangian in terms of the generalized coordinate \(\theta\), we use the relationship \(x = a\theta\). Differentiating this with respect to time gives \(\dot{x} = a\dot{\theta}\). Substitute this into the Lagrangian function: \(L = \frac{1}{2}I\dot{\theta}^2 + \frac{1}{2}m(a\dot{\theta})^2 - mg(l-a\theta)\)

The Euler-Lagrange equation for the generalized coordinate \(\theta\) is given by: \(\frac{d}{dt}(\frac{\partial L}{\partial \dot{\theta}}) - \frac{\partial L}{\partial \theta} = 0\) With the expressions for the Lagrangian, we differentiate with respect to \(\dot{\theta}\) and \(\theta\), and plug them into the equation: \(I\ddot{\theta} + ma^2\ddot{\theta} - mga\frac{\partial}{\partial \theta} (l-a\theta) = 0\) \(I\ddot{\theta} + ma^2\ddot{\theta} + mga^2 = 0\) Apply Euler-Lagrange equation and solve for \(\ddot{\theta}\): \(\ddot{\theta} = \frac{-mga^2}{I+ma^2}\)

We will solve the equation of motion to find \(\theta(t)\), then find the final angular velocity \(\omega_f = \dot{\theta}(t_f)\). We know that at time \(t = t_0\), the drum is released from rest, which means \(\dot{\theta}(t_0) = 0\). Also, the cable is initially fully wound up, so \(\theta(t_0) = 0\). We can find \(t_f\) as the time when the drum is fully unwound by solving \(l - a\theta(t_f) = 0\). Now, integrate the equation of motion with respect to time twice: \(\theta(t) = \int \int \ddot{\theta} dt dt\) \(\theta(t) = -\frac{mga^2}{I+ma^2}t^2 + C_1 t + C_2\) Using the initial conditions, we can find \(C_1\) and \(C_2\): \(C_2 = 0\), \(C_1 = \frac{lg}{2t_f}\) Substitute them back into the expression for \(\theta(t)\): \(\theta(t) = -\frac{mga^2}{I+ma^2}t^2 + \frac{lg}{2t_f}t\) Now, obtain the angular velocity by differentiating \(\theta(t)\) with respect to time: \(\dot{\theta}(t) = -\frac{2mga^2}{I+ma^2}t + \frac{lg}{2t_f}\) Finally, substitute \(t = t_f\) into the expression of \(\dot{\theta}(t)\) to find the final angular velocity: \(\omega_f = \dot{\theta}(t_f) = -\frac{2mga^2}{I+ma^2}t_f + \frac{lg}{2}\)