Four-Bar Parallel Linkage Pendulum

[grvlfun]

Why is θc in your expression for ω(θ) ? It should be a limit of the integration, not in the integration itself.

So I would have to find an equation where ω(θ) is just solved by giving the starting angle?
I'm having trouble to get my head around this, as different starting angles result in different velocities, because of different starting energies. E.g. starting at angle θ = 80° and ending at θ = 20° results in a different velocity compared to starting at θ = 60° and ending at θ = 0°, although both times the traveled distance/angle is 60°.

Let's use this and leave everything as variables for the time being so it's easy to check.

Thank you for the diagram. I've done my calculations with the virtual rod being pinned to the CoM of the bob and not above it.

Also, your potential energy expressions need revised to account for the CoM of the rod

The CoM of the rod is included with r = 0.85 so r/2 = 0.425

$1.72*4*9.81*0.425*(1-cos\theta) = 28.68(1-cos\theta)$

I see. you are using θc as the variable of integration. I wouldn't do this personally because it is meant to signify a specific angle when the block contacts the ground ( not a range of angles ), but as long as you know what you are doing.

Any idea on how to signify a range of angles?

 

[erobz]

So I would have to find an equation where ω(θ) is just solved by giving the starting angle?
I'm having trouble to get my head around this, as different starting angles result in different velocities, because of different starting energies. E.g. starting at angle θ = 80° and ending at θ = 20° results in a different velocity compared to starting at θ = 60° and ending at θ = 0°, although both times the traveled distance/angle is 60°.

Yeah, the starting angle $\theta_o$ determines $\omega( \theta )$. Say you pick some arbitrary angle $\theta_a$ to "instantaneously" measure the angular velocity of the bob. If you start the bob at $\theta_o$ very near $\theta_a$ with no initial angular velocity, when it crosses $\theta_a$ it will still have very little angular velocity. If you choose $\theta_o$ further away, it will hve larger angular velocity when it gets to $\theta_a$

- PE rods = $1.72*4*9.81*0.425*(1-cos\theta) = 28.68(1-cos\theta)$

$$ PE_{rod} = m_r r g\left( 1- \frac{1}{2}\cos \theta \right) $$

Any idea on how to signify a range of angles?

$\theta$ specifies a range of angles "Theta sub whatever" usually signifies a specific angle or point of interest in the range.

 

[grvlfun]

$PE_{rod} = m_r r g\left( 1- \frac{1}{2}\cos \theta \right)$

Shouldn't it be $PE_{rod} = m_r r g ( \frac {1} {2} - \frac {1}{2} \cos \theta )$ ?

$\displaystyle t = \sqrt{ \beta + \varphi} \int_{\theta_o}^{\theta_c} \frac{d \theta}{ \sqrt{ \kappa - rg \left( \left( M_b + m_r \right) - \left( M_b + \frac{1}{2} m_r \right) \cos \theta \right) } }$

This results in the same equation I had, just written a little different? However, with $\kappa$ there is still the fact that you provide the starting angle $\theta_o$ and the contact angle $\theta$ ?

 

[erobz]

Shouldn't it be $PE_{rod} = m_r r g ( \frac {1} {2} - \frac {1}{2} \cos \theta )$ ?

No. The CoM of the rod is at $r - \frac{r}{2} \cos \theta$ ( see diagram )

This results in the same equation I had, just written a little different? However, with $\kappa$ there is still the fact that you provide the starting angle $\theta_o$ and the contact angle $\theta$ ?

$\kappa$ is just a constant in the integrand, and it is to be treated as such. Yes it is a function of $\theta_o$ ( the lower limit of the integration - the initial condition) , but that has no other significance when integrating with respect to $\theta$. $\theta_c$ is just another constant ( like $\theta_o$ ) but it is the upper limit of integration - the final condition. $\theta_o \geq \theta \geq \theta_c$ in other words $\theta$ changes from $\theta_o$ to $\theta_c$.

 

[grvlfun]

No. The CoM of the rod is at $r - \frac{r}{2} \cos \theta$( see diagram )

That would be the case if the rod is falling to the ground? The CoM of the rod can just 'fall' r/2 and therefore for me it should be $PE_{rod} = m_rg\frac {r} {2} (1-cos\theta) => m_rgr(\frac {1} {2} - \frac {1} {2} cos\theta)$

And I will try the integration and see what I'm getting as a result, thank you.

 

[erobz]

That would be the case if the rod is falling to the ground? The CoM of the rod can just 'fall' r/2 and therefore for me it should be $PE_{rod} = m_rg\frac {r} {2} (1-cos\theta) => m_rgr(\frac {1} {2} - \frac {1} {2} cos\theta)$

No.

$PE = 0$ is an arbitrary datum that is fixed. We are defining the potential energy of some mass(es) relative to that datum. The $PE(\theta)$ of either the rod or the bob is defined by where the CoM of the rod or the bob are with respect to our datum for some arbitrary angel $\theta$.

Do you agree that the potential energy of the bob vs $\theta$ is given by?

$$ PE_{bob} = M_b g \left( r - r \cos \theta \right) $$

This is with respect to the chosen datum, what do you think is different about defining the potential energy of the rod with respect to our datum?

 

[grvlfun]

I'm sorry, but this confuses me.

The change in height is $\frac {r} {2} - \frac {r} {2} cos\theta = \frac {r} {2} (1 - cos \theta) = r (\frac {1}{2} - \frac {1} {2} cos \theta)$ as I've seen it in different pendulum and physical pendulum problems.

Change in height from $\theta_o$ to $\theta$ is $r (\frac {1}{2} - \frac {1} {2} cos \theta_o) - r (\frac {1}{2} - \frac {1} {2} cos \theta)$.

r is measured from the fulcrum on, and the distance of the fulcrum axis to the CoM of the rod is r/2. With your equation, when the pendulum is at angle $\theta = 0$ the rod would still have a PE of $m_rgr(1-\frac {1}{2} cos0) = m_rgr(0.5)$, but at the bottom a pendulum has a PE = 0.

 

[erobz]

I'm sorry, but this confuses me.

The change in height is $\frac {r} {2} - \frac {r} {2} cos\theta = \frac {r} {2} (1 - cos \theta) = r (\frac {1}{2} - \frac {1} {2} cos \theta)$ as I've seen it in different pendulum and physical pendulum problems.

Change in height from $\theta_o$ to $\theta$ is $r (\frac {1}{2} - \frac {1} {2} cos \theta_o) - r (\frac {1}{2} - \frac {1} {2} cos \theta)$.

r is measured from the fulcrum on, and the distance of the fulcrum axis to the CoM of the rod is r/2. With your equation, when the pendulum is at angle $\theta = 0$ the rod would still have a PE of $m_rgr(1-\frac {1}{2} cos0) = m_rgr(0.5)$, but at the bottom a pendulum has a PE = 0.

We aren't taking about the changes between initial and final so let's let them alone for now. They might be confusing the issues here.

If you put the PE = 0 at the fulcrum what do you get for the equations?

 

[erobz]

I'm going to be honest, choosing the PE= 0 at the fulcrum results in an algebraically much cleaner perhaps more understandable picture, with minimal algebraic effort. However, it's none the less equivalent to the result shown above (my bad for not being efficient). Once you re-write the equations w.r.t the fulcrum, either try to prove that identity, or just plug in some values to verify for yourself...I did.

 

[grvlfun]

I must say that I completely lost track now with the PE = 0 at the fulcrum.
I just put your integral into wolfram and I'm getting the same values as with $PE_{rod} = m_rgr(0.5-0.5cos\theta)$ but also there I'm not sure if I did it correctly...

 

[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:

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