Four-Bar Parallel Linkage Pendulum

[erobz]

Something is a miss because using the datum at the fulcrum:

$$ PE_{r}( \theta) = -m_r g \frac{r}{2} \cos \theta $$

$$ PE_{b}( \theta) = -M_b g r \cos \theta $$

I don't know how you would be getting the same result...but it doesn't matter we are going to work it out. Do you agree that from the datum at the fulcrum both of those( above ) describe the potential energy of each mass in the pendulum?

The total potential energy as a function of $\theta$ is given by:

$$ PE(\theta) = -rg \left( \frac{m_r}{2}+ M_b \right) \cos \theta $$

Are you in agreement up until there?

 

[erobz]

Let's just write the equation like below and focus on the potential energy which is giving the confusion:

$$ PE_o = PE( \theta) + KE_{total}$$

$$ -rg \left( \frac{m_r}{2}+ M_b \right) \cos \theta_o = -rg \left( \frac{m_r}{2}+ M_b \right) \cos \theta + KE_{total} $$

$$ \begin{align} KE_{total} &= rg \left( \frac{m_r}{2}+ M_b \right) \cos \theta - rg \left( \frac{m_r}{2}+ M_b \right) \cos \theta_o \tag*{} \\ &= rg \left( \frac{m_r}{2}+ M_b \right) \left( \cos \theta - \cos \theta_o \right) \tag*{} \end{align} $$

That is the result for the total kinetic energy when PE = 0 is at the pivot.

 

[erobz]

Now, comparing that when PE = 0 is at a distance $r$ below the pivot:

$$ rg \left[ \left( 1- \cos \theta_o \right) M_b + \left( 1- \frac{1}{2}\cos \theta_o \right) m_r \right] = rg \left[ \left( 1- \cos \theta \right) M_b + \left( 1- \frac{1}{2}\cos \theta \right) m_r \right] + KE_{total} $$

$$ \begin{align} KE_{total} &= rg \left[ \left( 1- \cos \theta_o \right) M_b + \left( 1- \frac{1}{2}\cos \theta_o \right) m_r \right] - \left[ \left( 1- \cos \theta \right) M_b + \left( 1- \frac{1}{2}\cos \theta \right) m_r \right] \tag*{}\\ &= rg \left[ \left( 1- \cos \theta_o \right) M_b + \left( 1- \frac{1}{2}\cos \theta_o \right) m_r - \left( 1- \cos \theta \right) M_b - \left( 1- \frac{1}{2}\cos \theta \right) m_r \right] \tag*{} \\ &= rg \left[ \cancel{M_b} - M_b \cos \theta_o + \cancel{m_r} - \frac{1}{2}m_r \cos \theta_o - \cancel{M_b} + M_b \cos \theta - \cancel{m_r} + \frac{1}{2}m_r \cos \theta \right] \tag*{} \\ &= rg \left[ M_b \left( \cos \theta - \cos \theta_o \right)+ \frac{1}{2} m_r \left( \cos \theta - \cos \theta_o \right) \right] \tag*{} \\ &= rg \left( \frac{m_r}{2}+ M_b \right) \left( \cos \theta - \cos \theta_o \right) \tag*{} \end{align} $$

The results are equivalent... this one is just much less clean and obvious perhaps.

It should be clear that all three of the results are not equivalent, as:

$$ rg \left[ \left( 1- \cos \theta \right) M_b + \left( 1- \frac{1}{2}\cos \theta \right) m_r \right] \neq rg \left[ \left( 1- \cos \theta \right) M_b + \frac{1}{2}\left( 1- \cos \theta \right) m_r \right] = rg \left( M_b + \frac{m_r}{2} \right) \left( 1 - \cos \theta \right) $$