Four-Bar Parallel Linkage Pendulum

[erobz]

The EoM of the lower bob, or of the ground ?
There is no simple way to guess the characteristics of the ground material.
Different materials will necessitate different models.
It is possible to over-extend a model.

So basically...what I'm taking away from this is we should just pick up with

$$ KE = \frac{1}{2} m_l v_x^2 $$

For the lower portion of the bob, just after impact assuming the quantity of energy lost from impact ( through sound/heat) is:

$$ KE_{lost} = \frac{1}{2} m_l v_y^2 $$

and continue on with the analysis treating the impact as a black box so the model doesn't become "over-extended"?

 

[erobz]

I would be more interested in how the mechanism could operate above 90° with the three rods suggested.
I would design it with one link on either side of the bob, with a chain or stepped belt and two equal sized sprockets, so the bob would remain horizontal while it passes over the top without tangling it's rods. That also eliminates the problem of parallel link length and clearance as the bob is passing 90°.

Yeah, to me the actual design characteristics of the machine are not worth examining without having any deliverables. Examining and understanding ( which you already seem to ) the physics in more detail has wider applications (IMO).

 

[grvlfun]

I would be more interested in how the mechanism could operate above 90° with the three rods suggested.
I would design it with one link on either side of the bob, with a chain or stepped belt and two equal sized sprockets, so the bob would remain horizontal while it passes over the top without tangling it's rods. That also eliminates the problem of parallel link length and clearance as the bob is passing 90°.

It just works because of the twisting and rotation of the bob > 90° is possible due to enough play in every axis and because the construction lacks a great amount of precision. However I have done some finetuning to the mechanism, based on your input and now it effectively shows a behaviour that it could get blocked once it goes past 80°-90° as the bob doesn't know if it should stay parallel or rotate.

Do you mean 2 links in total, one on each side or 4 in total with 2 on each side? And would you please elaborate or show a quick drawing of what you mean by "chain or stepped belt and two equal sized sprockets", as I can't imagine how this should look.

 

[Baluncore]

Support stand not shown at ends of fulcrum pin. Bob details not shown.
A loop of chain, belt or cord round two sprockets keeps bob level.
Upper sprocket attached to the frame. Lower to the bob.
A dog's-leg bend in one rod passes through the loop of belt.
The rods are fixed to the two cross pins so the rods remain parallel.

 

[grvlfun]

Hello People, it's me again.

First of all, thank you for your suggestion Baluncore and sorry that I didn't answer anymore. However it isn't possible for me to build it like this.

Second: I've got quite a difference in the calculated linear velocity of the bob vs. the actual measured linear velocity of the bob. The measured velocity is about 35-40% lower than the calculated.
What am I missing here?! Could this be because of the upper/lower plate as both are just connected by linear rails to allow the free movement? So that the 'leverage' of the rods can't act on the lower plate?

 

[Baluncore]

What am I missing here?

It will depend on how well your computation models your physical device.
1. What is the weight of the bobs and guide for the lower bob?
2. Does the bob assembly remain horizontal?
3. What is the weight of the rods?
4. How do you calculate the moment of inertia for rods?
5. Is there significant friction in the joints?
A picture of the fulcrum attachments, and one of the bob assembly would help.

 

[grvlfun]

Well, I think I've found the mistake.
I used video analysis to measure the velocity based on the fps, but I took the average velocity (ω = θ/t) instead of the velocity at the specific point I did the computation for.
Now the question is, if it is a constant acceleration or a changing acceleration?

 

[Baluncore]

Now the question is, if it is a constant acceleration or a changing acceleration?

The acceleration due to gravity is pretty much constant. The acceleration of the bob is a maximum at 90°, but falls to zero as the rods pass vertical with maximum velocity.

 

[grvlfun]

It's the tangential acceleration which causes the change in velocity? Is there any method to find out how much time t is needed to reach the velocity at a certain point when starting from v = 0?

My first approach would have been:
t = θ*2/ω
but this is only for a constant acceleration.

The second approach would be integration. Calculate the angular velocity for every 1° step and then use integration to find area under the curve which then should be the time?!

 

[erobz]

You should be able to solve for $v(\theta)$ ( velocity of the bob $v$ as a function of the angle $\theta$) from the CoE arguments presented earlier in the thread.

Then it follows that:

$$ v(\theta) = r \frac{d \theta}{dt} $$

$$ \Downarrow $$

$$ t = r \int \frac{d \theta}{ v( \theta )} $$

Thats only going to hold until the angle "$\theta_c$ " that the bob contacts the surface. Then you will have another set of EOM until the bob leaves the surface at "$- \theta_c$"

 

[grvlfun]

Thank you, but I'm having quite some trouble calculating it.
Let's take the potential energy of the bob and rods together which would then be $m*g*l*(1-cos(θ_i))$, the PE at contact would be $m*g*l*(1-cos(θ_c))$.

The equation for the angular velocity ω would be $ω = \sqrt {\frac {mgl(1-cos(θ_i)) - mgl(1-cos(θ_c))} {0.5m_{bob}r^2_{bob} + 0.5I_{rods}}}$

So for the equation $t = \int {\frac{d \theta} {ω(\theta)}}$ this would result in $t = \int {\frac{1}{\sqrt {\frac {mgl(cos(θ_c)) - mgl(cos(θ_i))} {0.5*m_{bob}*r^2_{bob} + 0.5*I_{rods} } } } d\theta}$

Is this correct?

 

[Baluncore]

What is the MoI of a rod rotating about an end point?
Where does the mass of the rods enter your equation?

I = m * L² / 3 ; which is not the centre of mass.

 

[grvlfun]

I'll try to write my steps in detail.
- mass of the bob = 2.332 kg
- mass of one rod = 1.72 kg
- nbr rods = 4
- length of rod = 0.9m, width = 0.025m
- radius turning axis to bob = 0.85m --> 0.85/2 = 0.425 radius from CoG of rod to axis.
- angle before impact $\theta_c$ = 20°

Potential energy:
- PE rods = $1.72*4*9.81*0.425*(1-cos\theta) = 28.68(1-cos\theta)$
- PE bob = $2.332*9.81*0.85*(1-cos\theta) = 19.45(1-cos\theta)$
- total PE = $PE_{bob} + PE_{rods} = 48.13(1-cos\theta)$

MoI rods:
- I_rod = $1/12*1.72*(0.9^2 + 0.025^2) = 0.1161$
- MoI + parallel axis = $I_{rod} + 1.72*0.425^2 = I_rod + 0.3106 = 0.4267$
- total MoI I^rods= $4*0.4267 = 1.7068$

Kinetic energy:
- KE bob $0.5*m*v^2 = 0.5*m*r^2*ω^2 = 0.5*2.332*0.85^2*ω^2 = 0.842*ω^2$
- KE rods $0.5*I_{rods}*ω^2 = 0.5*1.7068*ω^2 = 0.853*ω^2$

Equation for angular velocity:
$PE_i = PE_c + KE_{bob} + KE_{rods}$
$=> 48.13(1-cos\theta_i) - 48.13(1-cos\theta_c) = 0.8424*ω^2+0.8534*ω^2$
$=> 48.13*cos\theta_c - 48.13*cos\theta_i = 1.695*ω^2$
$=> ω = \sqrt { \frac {48.13*cos\theta_c - 48.13*cos\theta_i} {1.695}}$

For an angle $\theta_i = 75°$ and $\theta_c = 20°$ this would result in $ω = 4.39 \frac {rad}{s}$
... if the calculations are correct...

And $=> ω = \sqrt { \frac {48.13*cos\theta_c - 48.13*cos\theta_i} {1.695}}$ would then be used for the integration to find the time needed to reach the angular velocity??

 

[erobz]

I'll try to write my steps in detail.
- mass of the bob = 2.332 kg
- mass of one rod = 1.72 kg
- nbr rods = 4
- length of rod = 0.9m, width = 0.025m
- radius turning axis to bob = 0.85m --> 0.85/2 = 0.425 radius from CoG of rod to axis.
- angle before impact $\theta_c$ = 20°

Potential energy:
- PE rods = $1.72*4*9.81*0.425*(1-cos\theta) = 28.68(1-cos\theta)$
- PE bob = $2.332*9.81*0.85*(1-cos\theta) = 19.45(1-cos\theta)$
- total PE = $PE_{bob} + PE_{rods} = 48.13(1-cos\theta)$

MoI rods:
- I_rod = $1/12*1.72*(0.9^2 + 0.025^2) = 0.1161$
- MoI + parallel axis = $I_{rod} + 1.72*0.425^2 = I_rod + 0.3106 = 0.4267$
- total MoI I^rods= $4*0.4267 = 1.7068$

Kinetic energy:
- KE bob $0.5*m*v^2 = 0.5*m*r^2*ω^2 = 0.5*2.332*0.85^2*ω^2 = 0.842*ω^2$
- KE rods $0.5*I_{rods}*ω^2 = 0.5*1.7068*ω^2 = 0.853*ω^2$

Equation for angular velocity:
$PE_i = PE_c + KE_{bob} + KE_{rods}$
$=> 48.13(1-cos\theta_i) - 48.13(1-cos\theta_c) = 0.8424*ω^2+0.8534*ω^2$
$=> 48.13*cos\theta_c - 48.13*cos\theta_i = 1.695*ω^2$
$=> ω = \sqrt { \frac {48.13*cos\theta_c - 48.13*cos\theta_i} {1.695}}$

For an angle $\theta_i = 75°$ and $\theta_c = 20°$ this would result in $ω = 4.39 \frac {rad}{s}$
... if the calculations are correct...

And $=> ω = \sqrt { \frac {48.13*cos\theta_c - 48.13*cos\theta_i} {1.695}}$ would then be used for the integration to find the time needed to reach the angular velocity??

Why is $\theta_c$ in your expression for $\omega( \theta )$ ? It should be a limit of the integration, not in the integration itself.

An accompanying diagram would be helpful.

Another thing: Is the virtual rod pinned to the bob's CoM, or is it pinned above it?

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

EDIT:
I see. you are using $\theta_c$ 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.

 

[erobz]

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

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

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

$$ KE_r ( \omega ) = \frac{1}{2} \left( { I_r }_G + m_r \frac{r^2}{4} \right) \omega^2 $$

$$ KE_b ( \omega ) = \frac{1}{2} M_b r^2 \omega^2 $$

Let:

$\frac{1}{2} \left( { I_r }_G + m_r \frac{r^2}{4} \right) = \beta$

$\frac{1}{2} M_b r^2 = \varphi$

Then

$$ \omega( \theta) = \sqrt{ \kappa - rg \left[ \left( M_b + m_r \right) - \left( M_b + \frac{1}{2} m_r \right) \cos \theta \right] } \sqrt{ \frac{1}{ \beta + \varphi} } $$

Thats what I'm getting.

Then:

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

 

[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:

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