Problem 9.1.
a. Choose coordinate axes \(Oxy\) with \(O\) at the initial position of \(A\). See Figure 1 for a definition of basis \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}})\). The position is defined by angle \(\theta\) alone by noting that \(G\) moves along axis \(Oy\), that is, \(x_G =0\) at all time. This comes from the application of Euler’s first principle to the system \(\{rod, A, B\}\): \[ 3m \boldsymbol{a}_G \cdot \boldsymbol{\hat{\imath}}=0 \Longrightarrow \dot{x}_G = Cst = 0 \Longrightarrow x_G = 0 \] This is a consequence of \({\bf F}_{ext\to \Sigma} \cdot \boldsymbol{\hat{\imath}}=0\).
We can find the coordinates of \(A\) and \(B\): \[ x_A = -2l \sin\theta, \qquad y_A = 0 \] \[ x_B = l \sin\theta, \qquad y_B = 3l \cos\theta \]
b. Time-differentiation gives the component of \(\boldsymbol{v}_A\) and \(\boldsymbol{v}_B\): \[ \dot{x}_A = -2l\dot{\theta}\cos\theta, \qquad \dot{y}_A =0 \] \[ \dot{x}_B = l\dot{\theta}\cos\theta, \qquad \dot{y}_B = -3l \dot{\theta}\sin\theta \] The kinetic energy of the system is then \[ K_\Sigma= \frac{1}{2}m v_A^2 + m v_B^2 = \frac{1}{2}m (2l\dot{\theta}\cos\theta)^2 + m (l\dot{\theta}\cos\theta)^2 + m (3l \dot{\theta}\sin\theta)^2 \] \[ K_\Sigma=ml^2 \dot{\theta}^2 (2 \cos^2\theta+ \cos^2\theta+ 9\sin^2\theta)= 3ml^2 \dot{\theta}^2 (1+2\sin^2\theta) \] We can apply the Work-Energy Principle to the system:
the work of the internal force is zero since \(A\) and \(B\) stay at a constant distance from one another.
the work of the contact force is also zero: \({\bf N}_A \cdot \boldsymbol{v}_A =0\).
The energy of the system is conserved: \[ E_\Sigma= K_\Sigma+ 2mg y_B = 3ml^2 \dot{\theta}^2 (1+2\sin^2\theta) + 6mlg \cos\theta= Cst = 6mlg \] using \(\dot{\theta}=0\) at \(\theta=0\). This gives \[ \boxed{ \dot{\theta}^2 = \frac{2g}{l} \frac{1-\cos\theta}{1+2\sin^2\theta} } \]
c. At \(\theta=\pi/2\), we find \(\dot{\theta}^2 = 2g/3l\) which gives \[ \boxed{ v_A =0, \qquad v_B = \sqrt{6gl} } \]
d If the system is now replaced by a slender rod of length \(L=3l\) and mass \(M= 3m\), the mechanical energy takes the expression \[ E = \frac{1}{2}M \dot{y}_G^2 + \frac{1}{2}I_{Gzz} \dot{\theta}^2 + Mg y_G = \frac{1}{2}ML^2 \dot{\theta}^2 (\frac{1}{12}+\frac{1}{4}\sin^2\theta) + \frac{1}{2}ML g \cos\theta= Cst = \frac{1}{2}ML g \] using \(y_G= \frac{1}{2}L\cos\theta\) and \(\dot{y}_G= - \frac{1}{2}L\dot{\theta}\sin\theta\) (again \(G\) moves along a vertical axis \(Oy\)). This gives \[ \boxed{ \dot{\theta}^2 = \frac{4g}{l} \frac{1-\cos\theta}{1+3\sin^2\theta} } \] At \(\theta=\pi/2\), we find \(\dot{\theta}^2 = g/l\).
Problem 9.2.
a. Assuming sufficiently large friction coefficient, the body will roll without slipping during the time interval \(0\leq t \leq t_1\), where time \(t=t_1\) corresponds to the vertical position of rod \(AB\). To find angular velocity \(\omega_1\boldsymbol{\hat{k}}\) at \(t=t_1\), the simplest method consists of applying the Work-Energy Principle in the interval \(0\leq t \leq t_1\) realizing that the reaction force does not work: \(W^R_{0\to 1} = \int_0^{t_1} {\bf R}\cdot \boldsymbol{v}_{I\in 1/0} dt = 0\) since \(\boldsymbol{v}_{I\in 1/0} =\boldsymbol{0}\). Then the total work done by the external forces acting on the body is given by \[ W_{0\to 1}^{ext} = W^R_{0\to 1}+ W^{2mg}_{0\to 1} = 2mg \Delta y_G = 2mg \frac{r}{4} \] We have used the fact that the mass center \(G\) of the body is located on \(AB\) at a distance \(r/4\) from \(A\).
Then the Work-Energy Principle can be stated as \[ \Delta K = K(t_1) -K (t_0) = \frac{1}{2} I_{Bzz} \omega_1 ^2 = mg \frac{r}{2} \] This is equivalent to stating conservation of mechanical energy. The kinetic energy at time \(t=t_1\) is determined as that of a body (instantaneously) in rotation about \(B\) (since \(B\) takes the position of the contact point): \(K (t_1) = \frac{1}{2}I_{Bzz} \omega_1^2\) (see Figure 2). To find \(I_{Bzz}\), we first find \(I_{Azz}\): \[ I_{Azz}= \underbrace{mr^2}_{\text{hoop}} + \underbrace{m\frac{r^2}{4}+ m \frac{r^2}{12}}_{\text{bar}} =\frac{4}{3} mr^2 \] Then we use the parallel axis theorem: \[ I_{Azz}= I_{Gzz}+ 2m\frac{r^2}{16} \] \[ I_{Bzz}= I_{Gzz} + 2m \frac{9r^2}{16} \] giving \[ I_{Bzz}= I_{Azz} + 2m \frac{r^2}{2}=\frac{7}{3} m r^2 \] This gives the expression of angular velocity \(\omega_1\) at time \(t_1\): \[ \boxed{\omega_1 = \sqrt{\frac{3g}{7r}}} \]
b. To find the equation of motion, we introduce angle \(\theta\) which defines the orientation of bar \(AB\): see Figure 3. To find the equation of motion, i.e. the ODE governing angle \(\theta\), we follow the same method as in a): we apply the Work-Energy Principle from \(t=0\) to some arbitrary time \(t\). First we find the KE at time \(t\) by using the fact that the contact point \(I\) is the IC of the body: \[ K= \frac{1}{2} I_{Izz} \dot{\theta}^2 \] with \[ I_{Izz} = I_{Gzz}+ 2m |IG|^2 =I_{Gzz}+ 2m(r^2 + \frac{r^2}{16}- \frac{r^2}{2}\cos\theta) = mr^2(\frac{10}{3}-\cos\theta) \] This gives \[ K = \frac{1}{2} m r^2 \dot{\theta}^2 \left( \frac{10}{3} -\cos\theta\right) \] Since the contact force does not work, the mechanical energy is conserved: \[ E= K+ U = \frac{1}{2} m r^2 \dot{\theta}^2 \left( \frac{10}{3} -\cos\theta\right) - \frac{1}{4} (2m) g r \cos\theta= Cst \] To find the ODE, we set \(dE/dt =0\) to obtain \[ \boxed{ \left( \frac{10}{3} -\cos\theta\right) \ddot{\theta}+ \frac{1}{2} \dot{\theta}^2 \sin\theta+ \frac{g}{2r} \sin\theta=0 } \] After linearization (\(\theta\approx 0\)) we obtain the ODE \[ \ddot{\theta}+ \frac{3g}{14 r} \theta=0 \] The period of the small-amplitude oscillations is then given by \[ \boxed{ T = 2\pi \sqrt{ \frac{14r}{3g} } } \]
Problem 9.3.
Step 1: Kinematics: from homework 8 we found the following results \[ v_A=v_B = R \omega, \quad \omega_1 = \omega_2 =\omega= \dot{\phi} \] where \(\phi\) is an angle defining the orientation of the disks, \[ v_G = v_C = v_D = v_A \] and \[ v_G^2 = v_A^2 + 2r v_A \dot{\theta}\cos\theta+ r^2 \dot{\theta}^2 \]
Step 2: We can find the kinetic energy of the system by finding \[ K_1 = \frac{1}{2} Mv_A^2 + \frac{1}{4} M R^2 \dot{\phi}^2 = \frac{3}{4} M R^2 \dot{\phi}^2 = K_2 \] \[ K_3 = \frac{1}{2} m v_G^2 = \frac{1}{2} m (R^2 \dot{\phi}^2 + 2rR \dot{\phi}\dot{\theta}\cos\theta + r^2 \dot{\theta}^2) \]
Step 3: External work due to contact forces at \(I\) and \(J\) (no-slip), and gravity \[ W^{ext}_{0\to t} = - \Delta U^g, \quad U^g= -m g r \cos\theta \]
Step 4: Internal Work due to \(F_A\) and \(F_B\): \[ W^{int}_{0\to t} =0 \] Step 5: Conclusion: the mechanical energy of the system is conserved: \[ E= K_1 + K_2 + K_3 - mgr \cos\theta= Cst \] The equation \(\frac{d}{dt} E =0\) does give an ODE, but this ODE is not the ODE governing angle \(\theta\) (equation 10 in homework 10): the system has 2 DOF (\(\phi\) and \(\theta\)) and the application of the WEP is insufficient to find the evolution of the system.
