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Before solving this system, note that the rigid bars $\mathrm{\overline{OA}}$ and $\mathrm{\overline{OB}}$ are rotating about point $\mathrm{O}$, not their respective CGs. To account for this two methods are available.

  1. Use the mass moments of inertia of the rigid bars $\mathrm{\overline{OA}}$ and $\mathrm{\overline{OB}}$ about hinge $\mathrm{O}$, which can be obtained using parallel axis theorem. We will use this method when deriving equations of motion using Lagrange's equation.

$$I_{O} = I_{CG} + M d^{2}$$

  1. Use the mass moments of inertia of the rigid bars $\mathrm{\overline{OA}}$ and $\mathrm{\overline{OB}}$ about their respective CGs and account for the moments of normal and tangential forces developed due to rotation. Since the line of action of normal force (AKA centripetal force given by $\mathrm{m r \dot{\theta}^{2}}$) passes through axis of rotation, its moment is zero by definition and only the moment of the tangential force $\mathrm{m r \ddot{\theta}}$) needs to be determined. We will use this approach when deriving equation of motion using Newton's second law.

To derive the equation of motion, we will only use moment balance about point $\mathrm{O}$. Since we have already seen that the moments due to normal forces is zero, and the moment of hinge reaction is also zero, we will not show them on the free-body diagram.

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  1. Moment due to $\mathrm{\dfrac{m a \ddot{\theta}}{2}}$ acting on CG of $\mathrm{\overline{OA}}$: $\mathrm{\dfrac{m a \ddot{\theta}}{2} \times \dfrac{a}{2} = \dfrac{m a^{2} \ddot{\theta}}{4}}$

  2. Moment due to $\mathrm{\dfrac{m a \ddot{\theta}}{2}}$ acting on CG of $\mathrm{\overline{OB}}$: $\mathrm{\dfrac{m a \ddot{\theta}}{2} \times \dfrac{a}{2} = \dfrac{m a^{2} \ddot{\theta}}{4}}$

  3. Moment due to $\mathrm{m a \ddot{\theta}}$ acting on point mass $\mathrm{M}$: $\mathrm{M a \ddot{\theta}\times a = M a^{2} \ddot{\theta}}$

  4. Moment due to weight $\mathrm{mg}$ of $\mathrm{\overline{OA}}$: $\mathrm{\dfrac{a}{2} \cos \theta \times m g}$

  5. Moment due to weight $\mathrm{mg}$ of $\mathrm{\overline{OB}}$: $\mathrm{\dfrac{a}{2} \sin \theta \times m g}$

  6. Moment due to weight $\mathrm{Mg}$ of point mass $\mathrm{M}$: $\mathrm{a \sin \theta \times M g}$

  7. Moment due to spring force $\mathrm{K a \sin \theta}$: $\mathrm{-K a \theta \times a}$

Once the moment of all applied and inertial forces have been determined, the final step is to apply Newton's second law for rotation about hinge $\mathrm{O}$.

$$ \left( J_{CG} \right)_{OA} \ddot{\theta} + \left( J_{CG} \right)_{OB} \ddot{\theta} + \left( J_{CG} \right)_{M} \ddot{\theta} + \dfrac{m a^{2} \ddot{\theta}}{4} + \dfrac{m a^{2} \ddot{\theta}}{4} + M a^{2} \ddot{\theta} = \dfrac{a}{2} \cos \theta \times m g + \dfrac{a}{2} \sin \theta \times m g + a \sin \theta \times M g - K a \theta \times a $$

Substituting $\mathrm{\left( J_{CG} \right)_{OA} = \left( J_{CG} \right)_{OB} = \dfrac{ma^2}{12}}$ and $\mathrm{\left( J_{CG} \right)_{M} = 0}$ and simplifying gives

$$ \left(\dfrac{2}{3} m a^{2} + M a^{2} \right) \ddot{\theta} + K a^{2} \theta = \dfrac{m g a}{2} \left( \sin \theta + \cos \theta \right) + M g a \sin \theta $$

Next we will derive the same equation of motion using Lagrange's equation.

As discussed earlier, in this section, we will use mass moments of inertia about hinge $\mathrm{O}$. Using parallel axis theorem, the mass moments of inertia of rigid bars $\mathrm{\overline{OA}}$ and $\mathrm{\overline{OB}}$ about hinge $\mathrm{O}$ are equal and given by $\mathrm{\dfrac{m a^{2}}{12} + \dfrac{m a^2}{4} = \dfrac{m a^2}{3}}$. The mass moment of inertia of point mass $\mathrm{M}$ about hinge $\mathrm{O}$ is given by $\mathrm{M a^{2}}$.

$$ T = \dfrac{1}{3} m a^{2} \dot{\theta}^{2} + \dfrac{1}{2} M a^{2} \dot{\theta}^{2} $$

The centre line of bar $\mathrm{\overline{OA}}$ marks the datum for zero potential energy. As a result of positive rotation of the system, all three components fall down in gravity field. So, their potential energy will be negative.

$$ U_{g} = -m g \dfrac{a \sin \theta}{2} - m g \dfrac{a \left( 1 - \cos \theta \right)}{2} - M g a \left( 1 - \cos \theta \right) $$

The elastic potential energy of the system is given by

$$ U = -m g \dfrac{a \sin \theta}{2} - m g \dfrac{a \left( 1 - \cos \theta \right)}{2} - M g a \left( 1 - \cos \theta \right) + \dfrac{1}{2} K \left( a \theta \right)^2 $$

Evaluating the terms in Lagrange equation:

$$ \dfrac{d}{d t} \left( \dfrac{\partial T}{\partial \dot{q}_{1}} \right) = \dfrac{2}{3} m a^{2} \ddot{\theta} + M a^2 \ddot{\theta} \\\\$$

$$ \dfrac{\partial T}{\partial q_{1}} = 0 \\ $$

$$ \dfrac{\partial U}{\partial q_{1}} = -\dfrac{m g a}{2} \cos \theta - \dfrac{m g a}{2} \sin \theta - M g a \sin \theta + K a^{2} \theta $$

with $\mathrm{q_{1} = \theta}$. Substituting values in to the Lagrange's equation gives

$$ \dfrac{2}{3} m a^{2} \ddot{\theta} + M a^2 \ddot{\theta} - \dfrac{m g a}{2} \cos \theta - \dfrac{m g a}{2} \sin \theta - M g a \sin \theta + K a^{2} \theta = 0 $$

Re-arranging the above equation gives the same equation of motion already derived earlier.

$$ \left( \dfrac{2}{3} m a^{2} + M a^2 \right) \ddot{\theta} + K a^{2} \theta = \dfrac{m g a}{2} \left( \cos \theta + \sin \theta \right) + M g a \sin \theta $$

Assuming large rotations, are these derivations correct?

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  • $\begingroup$ To be honest, I din't follow through all of your calculations, I just checked that the two results with the independent methods of derivation were the same. To me that usually, in this type of problem is strong indication that you are doing it right. Normally if you mess up, you don't get the same results. $\endgroup$ – NMech Sep 29 '20 at 17:49
  • $\begingroup$ @NMech Thank you for reviewing it. I am not quite sure whether this statement is true or false: "The centre line of bar $\mathrm{\overline{OA}}$ marks the datum for zero potential energy. As a result of positive rotation of the system, all three components fall down in gravity field. So, their potential energy will be negative." $\endgroup$ – Ali Baig Sep 29 '20 at 17:52
  • $\begingroup$ Also, assuming positive $\mathrm{\theta}$ rotation of rigid bars, what will be resulting deformation in the spring? $\mathrm{a \times \theta}$ or $\mathrm{a \times \sin \theta}$. $\endgroup$ – Ali Baig Sep 29 '20 at 17:53
  • $\begingroup$ Regarding the potential energy that statement seems ok. The only gray point is the vertical bar. However, even that has a energy change which is negative. $\endgroup$ – NMech Sep 29 '20 at 18:26
  • $\begingroup$ Regardign the spring deformation, that is the only tricky thing about the large displacements. Namely, for small deformations $\theta = \sin\theta$, however, when you have large deformations, you will have also a horizontal displacement. From what I gather you are not accounting for that. $\endgroup$ – NMech Sep 29 '20 at 18:28
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Isn't the equation of motion more like this:

\begin{align} \left( \frac{2\,ma^2}{3} + Ma^2\right)\,\, \frac{d^2 \theta}{dt^2}\, =\, - \, \frac{amg}{2} \, \cos(\theta) \, + \, \left(\frac{amg}{2} + aMg \right)\, \sin(\theta) \, - \, aK\left(2a\, \sin\Big(\frac{\theta}{2}\Big) \, - \, l_0\right) \cos\Big(\frac{\theta}{2}\Big) \end{align} The easiest way is to use Lagrangian mechanics framework.

Let me start with a picture of how the system should behave, according to me: enter image description here I have assumed that $OA = OB = OC = a$, as well as $\angle \, AOB = 90^{\circ}$ and that the lower end of the spring is attached to the ground at the point $C$ on the picture and its upper end is attached to the frame $OAB$ at point $A$, since you have not clearly specified the exact position of the spring. Since $OA = OC$ in any configuration, the triangle $AOC$ is alsways isosceles and we can calculate the distance $|AC| = 2a \, \sin\left(\frac{\theta}{2}\right)$.

The moving object $OAB$ with extra mass-point $B$ of mass $M$ should have moment of inertia the sum of the moments of inertia of the three components $OA, \, OB$ and $B$ relative to the stationary point of rotation $O$. By assumption, the masses of $OA, \, OB$ and $B$ are $m, \, m$ and $M$ respectively. So, we should add together the moments of inertia of two rigid rods $OA$ and $OB$ and a mass point $B$ relative to $O$. Therefore $$I = I_{OAB,B} = \frac{ma^2}{3} + \frac{ma^2}{3} + Ma^2 = \frac{2\,ma^2}{3} + Ma^2 $$
Consequently, the kinetic energy of this system is $$T = \frac{1}{2}\left(\frac{2\,ma^2}{3} + Ma^2 \right)\, \left(\frac{d\theta}{dt}\right)^2$$ All the forces acting on the system are potential forces:

  1. Gravity force acting on rod $OA$ applied to its center of mass, the midpoint $M_A$. Its potential energy is $$U_A = mg\, y_{M_A} = \frac{amg}{2} \, \sin(\theta)$$

  2. Gravity force acting on rod $OB$ applied to its center of mass, the midpoint $M_B$. Its potential energy is $$U_B = mg\, y_{M_B} = \frac{amg}{2} \, \cos(\theta)$$

  3. Gravity force acting on mass-point $B$. Its potential energy is $$U_M = Mg\, y_{B} = aMg \, \cos(\theta)$$

  4. Elastic force, which possibly obeys Hooke's law, so it is a force with a quadratic potential: $$U_S = \frac{K}{2} \Big(|AC| - l_0 \Big)^2 = \frac{K}{2} \left(2a \, \sin\left(\frac{\theta}{2} \right)- l_0\right)^2$$ where $l_0$ is the length of the spring when it is in equilibrium.

Consequently, the Lagrangian of the system is \begin{align} L \, =& \, T - U_A - U_B - U_M - U_S \\ =& \, \frac{1}{2}\left(\frac{2\,ma^2}{3} + Ma^2 \right)\, \left(\frac{d\theta}{dt}\right)^2 \, - \, \frac{amg}{2} \, \sin(\theta)\\ & \, - \, \frac{amg}{2} \, \cos(\theta) \, - \, aMg \, \cos(\theta) \, - \,\frac{K}{2} \left(2a \, \sin\left(\frac{\theta}{2} \right)- l_0\right)^2 \end{align} If we calculate explicitly the Euler-Lagrange equation $$\frac{d}{dt}\left(\,\frac{\partial L}{\partial \dot{\theta}}\left(\,\theta, \, \frac{d\theta}{dt}\,\right)\,\right) = \frac{\partial L}{\partial {\theta}}\left(\,\theta, \, \frac{d\theta}{dt}\,\right)$$ we arrive at the differential equation \begin{align} \left( \frac{2\,ma^2}{3} + Ma^2\right)\,\, \frac{d^2 \theta}{dt^2}\, =& \, - \, \frac{amg}{2} \, \cos(\theta) \, + \, \left(\frac{amg}{2} + aMg \right)\, \sin(\theta)\\ &\, - \, aK\left(2a\, \sin\Big(\frac{\theta}{2}\Big) \, - \, l_0\right) \cos\Big(\frac{\theta}{2}\Big) \end{align}

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