1. To find \(I_2\), we find the directions of \(\boldsymbol{v}_{A\in 2}\) and \(\boldsymbol{v}_{B\in 2}\) (2 points of body 2): Figure 1 shows that \(I_2\) is at the intersection of lines \(OA\) and \(AB\). Hence, IC \(I_2\) of body 2 is at the location of point \(A\). This implies that, at time \(t_1\), \(v_A= 0\) and that \[ \boxed{ \omega_1 =0, \qquad v_B = 2l \omega_2 } \] This will be used in question 2.
2. To apply the Work-Energy Principle between \(t_0\) and \(t_1\), we need to find
The variation of KE of the system: \(K_\Sigma(t_0) =0\) (the system is released from rest). We find the final KE \(K_\Sigma(t_1)\) as follows \[ K_\Sigma(t_1) =K_1 + K_2 = \underbrace{\frac{1}{2}I_{Ozz} \, \omega_1^2}_{\text{body 1}} + \underbrace{\frac{1}{2}I_{Azz} \, \omega_2^2}_{\text{body 2}} = \frac{2}{3} m l^2 \omega_2^2 = \frac{1}{6} m v_B^2 \] where we used the fact that \(\omega_1=0\) and that body 2 is instantaneously in rotation about \(A\).
The work done by external forces, i.e. the reaction at \(O\), \(B\) and gravity: \[ W^{ext}_{0\to 1} =W^{R_O}_{0\to 1} + W^{R_B}_{0\to 1} + W^{mg}_{0\to 1} = - \Delta U^g \] where we set \(W^{R_O}_{0\to 1}= W^{R_B}_{0\to 1} =0\) (the joints are frictionless), and where \[ \Delta U^g = -mg \Delta z_{G_1} -mg \Delta z_{G_2}= - mg \frac{l}{\sqrt{2}} - mg \frac{3l}{\sqrt{2}}= -2\sqrt{2} mgl \] (with coordinates \(z_{G_1}\) and \(z_{G_2}\) defined on a vertical axis directed downward). See Figure 1.
- The work done by internal forces,i.e. the reaction at \(A\), and the elastic interaction: \[ W^{int}_{0\to 1} =W^{R_A}_{0\to 1} + W^{el}_{0\to 1} = - \Delta U_{1\leftrightarrow 2}^{el} =0 \] since the internal joint at \(A\) is frictionless, and the spring is unstretched at both \(t=t_0\) and \(t=t_1\).
Conclusion: the mechanical energy of the system is conserved \[ \Delta( K_\Sigma+ U^g + U_{1\leftrightarrow 2}^{el})=0 \] which gives \[ \frac{1}{6} m v_B^2- 2\sqrt{2} mgl =0 \] or \[ \boxed{v_B^2 = 12 \sqrt{2} gl} \]
3. The position/kinematics of ths system is entirely defined in terms of the two angles \(\theta_1\) and \(\theta_2\) (and their derivatives) shown in Figure 2. In fact, these two angles are not independent: they satisfy \[ \boxed{ 2l \cos\theta_1 + 2l \cos\theta_2 = Cst } \qquad(1) \] Hence the system has only 1 DOF. Before we apply the WEP to the system, we need to derive some basic kinematics (defined at arbitrary time \(t\)): \[ \boldsymbol{\omega}_1 = \dot{\theta}_1 \boldsymbol{\hat{k}}, \qquad \boldsymbol{\omega}_2 = - \dot{\theta}_2 \boldsymbol{\hat{k}},\qquad \boldsymbol{v}_{G_2} = -l(2\dot{\theta}_1\sin\theta_1 +\dot{\theta}_2\sin\theta_2) \boldsymbol{\hat{\imath}} + l (2\dot{\theta}_1\cos\theta_1 -\dot{\theta}_2\cos\theta_2)\boldsymbol{\hat{\jmath}} \]
Now we can easily find the KE of the system \[ K_\Sigma= \underbrace{\frac{2}{3} m l^2 \dot{\theta}_1^2}_{\text{body 1}} +\underbrace{ \frac{1}{2} ml^2 (4\dot{\theta}_1^2 -4 \dot{\theta}_1\dot{\theta}_2 \cos(\theta_1+\theta_2) +\dot{\theta}_2^2) +\frac{1}{6} m l^2 \dot{\theta}_2^2}_{\text{body 2}} \] As in question 2, the work done by the non-conservative exterior and interior forces is zero: the mechanical energy of the system is conserved. \[ \boxed{ E_\Sigma= K_\Sigma+ U^g + U^{el}_{1\leftrightarrow 2} = Cst = \frac{1}{\sqrt{2}} mgl } \qquad (2) \] with \[ U^g = -3mgl \sin\theta_1 + mgl \sin\theta_2 \] (the gravitational potential energy of the system) \[ U^{el}_{1\leftrightarrow 2} =\frac{k}{2} (l_{12}-l_0)^2 \] (the elastic potential of interaction) with \[ l_{12}^2 = |G_1 G_2 |^2 = l^2 (2+ 2\cos(\theta_1+\theta_2)), \qquad l_0^2 = l^2 (2 + \sqrt{2}) \] Equations (1-2) are sufficient to predict the dynamic behavior of the system, that is, the values of \(\theta_1\), \(\theta_2\), \(\dot{\theta}_1\) and \(\dot{\theta}_2\) can be computed at any time \(t\), \(t_0 \leq t \leq t_1\) from the knowledge of equations (1-2).
Note 1: The results of Question 1 allow us to relate the KE of body 2 to the speed \(v_B\) of endpoint \(B\) by using the notion of instantaneous center of rotation. The standard way to find the KE of body 2 is to use \(\frac{1}{2}mv^2_{G_2}+ \frac{1}{6} ml^2 \omega_2^2\). We would then have to relate \(v_{G_2}\) and \(\omega_2\) to \(v_B\).
Note 2: In Question 3, we could also use the notion of IC \(I_2\) to find the KE of body 2. This would require us to find the location of \(I_2\).
Note 3: It may be tempting to apply the WEP to body 1, then body 2 separately. However, the reaction at joint \(A\) now needs to be taken into account: \[ W^{R_A}_{0\to 1}= \int_{t_0}^{t_1} {\bf R}_A\cdot \boldsymbol{v}_{A\in 1/0} \, dt \qquad (\text{for body 1}) \] \[ W^{R_A}_{0\to 1}= \int_{t_0}^{t_1} (-{\bf R}_A)\cdot \boldsymbol{v}_{A\in 2/0} \, dt \qquad (\text{for body 2}) \] These works do not vanish, and the WEP is of no use. However, if we sum these 2 expression, we obtain \[ \int_{t_0}^{t_1} {\bf R}_A\cdot (\boldsymbol{v}_{A\in 1/0} -\boldsymbol{v}_{A\in 2/0} ) dt = \int_{t_0}^{t_1} {\bf R}_A\cdot \boldsymbol{v}_{A\in 1/2} \, dt =0 \] This is equivalent to applying the WEP to the system, thereby making the works internal.