Derive the equations of motion for the crankshaft system shown in Figure P6.6 using Lagrange's Equations. The spring is undeformed when the connecting pin \(A\) is directly above or below the hub of the wheel.

First, let's describe the position of pin \(A\) relative to the hub of the wheel. Using polar coordinates, assign an angle of \(\theta\) to represent the orientation of the crankshaft with respect to the horizontal axis. The length of the crankshaft is denoted by \(l\). The vertical displacement of pin \(A\) is \(y\), and the horizontal displacement is \(x\). We can write the position of pin A as: $$x = l\cos\theta$$ $$y = l\sin\theta$$

Now, we'll find the velocity of pin \(A\), which is necessary for determining the kinetic energy of the system. We start by differentiating the position equations with respect to time \(t\): $$\frac{dx}{dt} = -l\sin\theta \frac{d\theta}{dt}$$ $$\frac{dy}{dt} = l\cos\theta \frac{d\theta}{dt}$$ The velocities are the time derivatives of the position coordinates (\(v_x\) for horizontal velocity and \(v_y\) for vertical velocity). Therefore, $$v_x = -l\sin\theta \dot{\theta}$$ $$v_y = l\cos\theta \dot{\theta}$$

The kinetic energy of the system is given by the sum of the translational and rotational kinetic energies: $$T = \frac{1}{2}m(v_x^2 + v_y^2) + \frac{1}{2}I\omega^2$$ Here, \(m\) is the mass of pin \(A\), \(I\) is the moment of inertia of the wheel, and \(\omega\) is the angular velocity of the wheel. Substituting the expressions for \(v_x\) and \(v_y\) from Step 2, we get: $$T = \frac{1}{2}ml^2\dot{\theta}^2 + \frac{1}{2}I\dot{\theta}^2 = \frac{1}{2}(ml^2+I)\dot{\theta}^2$$ For potential energy, we have the gravitational potential energy (\(mgy\)) and the potential energy of the spring (\(\frac{1}{2}kx^2\)). When the spring is undeformed, \(l_a = l\cos\theta\), and the spring constant is \(k\). Thus, the potential energy can be written as: $$V = mgy + \frac{1}{2}k(l-l_a)^2 = mg(l\sin\theta) + \frac{1}{2}k(l\sin\theta)^2$$

The Lagrangian \(L\) is defined as the difference between kinetic and potential energy (\(L=T-V\)): $$L = \frac{1}{2}(ml^2+I)\dot{\theta}^2 - mg(l\sin\theta) - \frac{1}{2}k(l\sin\theta)^2$$ Now, we apply Lagrange's equation: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) - \frac{\partial L}{\partial \theta} = 0$$ Taking the time derivative of \(\frac{\partial L}{\partial \dot{\theta}}\) and partial derivative of \(\frac{\partial L}{\partial \theta}\), we get: $$\frac{d}{dt}((ml^2+I)\dot{\theta}) - (mgl\cos\theta + kl^2\sin\theta\cos\theta) = 0$$ After simplifying the equation, we have the equation of motion for the crankshaft system: $$(ml^2+I)\ddot{\theta} - mgl\cos\theta - kl^2\sin\theta\cos\theta = 0$$