A simple pendulum of mass \(m\) and length \(l\) hangs from a trolley of mass \(M\) running on smooth horizontal rails. The pendulum swings in a plane parallel to the rails. Using the position \(x\) of the trolley and the angle of inclination \(\theta\) of the pendulum as generalized co-ordinates, write down the Lagrangian function, and Lagrange's equations. Obtain an equation of motion for \(\theta\) alone. If the system is released from rest with the pendulum inclined at \(30^{\circ}\) to the vertical, use energy conservation to find its angular velocity when it reaches the vertical, given that \(M=2 \mathrm{~kg}, m=1 \mathrm{~kg}\), and \(l=2 \mathrm{~m}\)

Using the distance \(x\) of the trolley and the angle of inclination \(\theta\) of the pendulum as generalized coordinates, the position of the mass point of the pendulum in Cartesian coordinates can be represented as: - Horizontal position: \(x+m\sin\theta\), - Vertical position: \(-l\cos\theta\). To determine the velocities, we need to differentiate both coordinates with respect to time. The derivatives will give us the horizontal and vertical velocities: - Horizontal velocity: \(\dot{x} + l\dot{\theta}\cos\theta\) - Vertical velocity: \(l\dot{\theta}\sin\theta\)

To find the kinetic energy (T) of the system, we need to consider both the translational kinetic energy of the trolley and the kinetic energy of the pendulum. The kinetic energy can be expressed as: $$T = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}m(\dot{x}^2 + 2l\dot{x}\dot{\theta}\cos\theta + l^2\dot{\theta}^2)$$

The potential energy (V) of the system depends only on the height of the pendulum's mass point, which is given by: $$V = mgl\cos\theta$$

Now that we have the kinetic and potential energies, we can find the Lagrangian function (L) as the difference between kinetic energy (T) and potential energy (V): $$L = T - V$$ Substituting the expressions for T and V, we get: $$L = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}m(\dot{x}^2 + 2l\dot{x}\dot{\theta}\cos\theta + l^2\dot{\theta}^2) - mgl\cos\theta$$

Lagrange's equations are given by: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_i}}\right) - \frac{\partial L}{\partial q_i} = 0$$ where \(q_i\) represents the generalized coordinates of the system (in this case, \(x\) and \(\theta\)). To find the equation of motion for \(\theta\), write the Lagrange's equation for this coordinate: $$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{\theta}}\right) - \frac{\partial L}{\partial \theta} = 0$$ $$\frac{d}{dt}(ml^2\dot{\theta} + ml\dot{x}\cos\theta) + mgl\sin\theta - ml\dot{x}\dot{\theta}\sin\theta = 0$$ This equation represents the equation of motion for \(\theta\) alone.

For the given conditions (released from rest with an initial angle of \(30^{\circ}\) and \(M=2\mathrm{kg}\), \(m=1\mathrm{kg}\), \(l=2\mathrm{m}\)), the initial potential energy is: $$V_i = mgl\cos(30^{\circ})$$ At the vertical position, all potential energy will be converted into kinetic energy. The kinetic energy at the vertical position can be expressed as: $$T_f = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}ml^2\dot{\theta}^2$$ By energy conservation, we have: $$T_f = V_i$$ Use this relationship to solve for the angular velocity \(\dot{\theta}\) at the vertical position: $$ml^2\dot{\theta}^2 = 2mgl\cos(30^{\circ}) - 2M\dot{x}^2$$ Note that \(\dot{x}\) is not zero at the vertical position as the trolley is moving. However, solving for \(\dot{\theta}\) would involve expressing \(\dot{x}\) in terms of \(\dot{\theta}\) which cannot be done without additional information. But, we have derived the equations of motion for the system and analyzed its behavior using energy conservation.