A uniform plank of length \(2 a\) is placed with one end on a smooth horizontal floor and the other against a smooth vertical wall. Write down the Lagrangian function, using two generalized co-ordinates, the distance \(x\) of the foot of the plank from the wall, and its angle \(\theta\) of inclination to the horizontal, with a suitable constraint between the two. Given that the plank is initially at rest at an inclination of \(60^{\circ}\), find the angle at which it loses contact with the wall. (Hint: First write the co-ordinates of the centre of mass in terms of \(x\) and \(\theta\). Note that the reaction at the wall is related to the Lagrange multiplier.)

In order to find the Lagrangian, we need to determine the position of the center of mass of the plank in terms of x and θ. We can do this using vector notation and trigonometry. The position vector of the center of mass, r, can be written as: \(r = x\hat{i} + y\hat{j}\) For the plank, the center of mass is located at its midpoint, which is a distance a from both ends. Using trigonometry, we can express x and y in terms of a and θ: \(x = a\sin\theta\) \(y = a\cos\theta\)

The kinetic energy, T, can be written as: \(T = \frac{1}{2}m\left(\frac{d r}{d t}\right)^2\) Where m is the mass of the plank and r is the position. The potential energy, V, of the system is due to the gravitational force acting on the center of mass of the plank and can be written as: \(V = mgy\) Where g is the acceleration due to gravity. Now we can substitute the expressions for x and y, and their time derivatives, into the equations for the kinetic and potential energies: \(T = \frac{1}{2}m\left(\frac{d}{dt}(a\sin\theta)\hat{i} + \frac{d}{dt}(a\cos\theta)\hat{j}\right)^2\) \(V = mga\cos\theta\)

Now that we have expressions for the kinetic and potential energies, we can write down the Lagrangian function L, which is defined as the difference between the kinetic and potential energies: \(L = T - V\) Substituting and simplifying the expressions from Step 2, we get: \(L = \frac{1}{2}ma^2(\dot\theta^2) - mga\cos\theta\)

Now, we can apply the Euler-Lagrange equations to find the equations of motion for the plank. For this problem, the Euler-Lagrange equation can be written as: \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot\theta}\right) - \frac{\partial L}{\partial \theta} = 0\) Applying this to the Lagrangian from step 3 and solving for the acceleration, \(\ddot\theta\), we get: \((ma^2)\ddot\theta - (-mga\sin\theta) = 0\) \(\ddot\theta = \frac{g}{a}\sin\theta\)

In the last step, we have derived the equation of motion for the plank: \(\ddot\theta = \frac{g}{a}\sin\theta\) Now we just need to find the angle, θ, at which the plank loses contact with the wall. At this moment, there is no reaction force at the wall, and the plank is moving freely under the effect of gravity. In this case, the angle is given by the condition: \(\frac{d\theta}{dt} = 0\) Using the initial condition that the plank is at rest (notated as \(\dot{\theta} = 0\)) when the angle is at 60 degrees or \(\pi/3\) radians, we integrate the equations of motion to find θ: \(\ddot\theta = \frac{g}{a}\sin\theta\) Integration gives: \(\dot\theta^2 = \frac{2g}{a}\cos\theta\) Since the plank is initially at rest when the angle is \(\pi/3\), and loses contact with the wall when \(\dot\theta = 0\), we obtain: \(\cos\theta = \frac{1}{2}\cos\frac{\pi}{3}\) This gives us the angle at which the plank loses contact with the wall: \(\theta = \cos^{-1}\left(\frac{1}{2}\cos\frac{\pi}{3}\right) \approx 41.41^\circ\)