Compound physical pendulum - max velocity

[FEAnalyst]

TL;DR Summary

Which formula will give correct results for the maximum velocity of a pendulum taking into account both rod and bob?

Hi,

I'm working on a simple benchmark problem for FEA. It's a pendulum initially positioned at an angle of $45^{\circ}$ and then subjected to gravity. I'm interested in the maximum velocity (when the pendulum is in the vertical position). So far, I've been using this formula: $$v=\omega \cdot l=\sqrt{\frac{3g}{l} \cdot \left( \cos{\left( \theta_{start} \right)}- cos{\left(\theta_{end} \right)} \right)} \cdot l$$ where: $l$ - pendulum length, $g$ - gravitational acceleration, $\theta_{start}=45^{\circ}$ - starting angle, $\theta_{end}=180^{\circ}$ - final angle. The geometry was just a cuboid (here shown from the side):

I obtained a very good agreement between FEA and hand calculation for this case. But now I'd like to change the geometry to more pendulum-like:

The problem is that the aforementioned formula doesn't provide correct results for this case, even though it only needs the pendulum length, not mass as input. I've been looking for some equations for compound physical pendulums since the geometry above likely represents this kind of problem, but I haven't found any formulas equivalent to the one I've been using so far (for the maximum velocity based on angles). In FEA, I can't make the rod massless so I may have to account for the masses of both components even when the rod is very thin.

Do you know where I can find such a formula ? Should I change the assumptions for the analytical calculations ?

 

[Baluncore]

Do you know where I can find such a formula ?

Conservation of energy is a simple solution for a massless rod pendulum.
Equate the kinetic energy at the bottom with the initial potential energy when released.
Depending on the shape and mass distribution in the bob, you may have to account for rotation of the bob in your model.

Note that the period of oscillation is not dependent on length alone, there is also a circular error, that is a function of the angle. There are more exact period solutions available for significant angles of swing.
https://en.wikipedia.org/wiki/Pendulum#Period_of_oscillation

 

[sophiecentaur]

TL;DR Summary: Which formula will give correct results for the maximum velocity of a pendulum taking into account both rod and bob?

Should I change the assumptions for the analytical calculations ?

Have you read and understood the analytical calculations in the first case? The equations of motion in both cases can be boiled down to a torque and a moment of Inertia. Are you aware of that? The mass cancels out both times.

@Baluncore yes, I try to use Energy when convenient too and both approaches should give the same result.

 

[FEAnalyst]

Conservation of energy is a simple solution for a massless rod pendulum.

Shouldn’t I account for the mass of the rod ? It’s included in the simulation. Then the problem turns into a compound physical pendulum, right? I can find some equations for the period of oscillation but not for the maximum velocity. And the initial angle is not included there.

 

[sophiecentaur]

It’s included in the simulation.

It's up to you if you want to rely on a simulation. the theory is not hard and, no doubt the simulation has some supporting documentation. Can you show me a formula for the oscillation that includes the mass?
BTW what is the mass of a massless rod?
What is the purpose of this exercise for you? Are you trying to learn some mechanics?

 

[FEAnalyst]

Can you show me a formula for the oscillation that includes the mass?

From what I've found (https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/mit8_01scs22_chapter24.pdf), it's: $$T=2 \pi \sqrt{\frac{\left( \frac{1}{3} m_{r}+m_{d} \right) l^{2}+ \frac{1}{2}m_{d}R^{2}}{\left( \frac{1}{2} m_{r}+m_{d} \right)gl}}$$ where: $m_{r}$ - mass of the rod, $m_{d}$ - mass of the disk, $R$ - radius of the disk, $l$ - length of the rod. But as you can see, it assumes that the bob is a disk, not a sphere like in my case. I could adjust my case to use a disk though, not a big issue.

Can you show me a formula for the oscillation that includes the mass? BTW what is the mass of a massless rod?

The point is that I want to include the mass of the rod so it's not massless anymore. That's because the rod has mass in the simulation.

What is the purpose of this exercise for you? Are you trying to learn some mechanics?

In short words, I just want to use hand calcs to verify the correctness of the solution obtained from the simulation. Also for educational purposes but it's also supposed to be sort of a benchmark problem (test for the simulation software with a known analytical solution) as well.

 

[sophiecentaur]

where: mr - mass of the rod, md - mass of the disk

Ah yes. What counts here is the ratio of masses because you need to know the MI of each component and also the Torque due to gravity. Top and bottom include the masses so the mass dimension cancels out.
The MI in your formula includes the MI of the elements on the rod (mass of rod and mass of diisc on the end) and the MI of the disc. You need to replace the formula for MI of a disc (${m_d R^2}/2$) with the formula for the MI of a sphere (${2m_d R^2 /5}$). (Check these formula separately by a google search) You will notice that the MI of a sphere is less than for a disc because more of the mass of a sphere is near the centre.

The version with a disc is probably more common because that's the layout of a pendulum clock, probably.

 

[FEAnalyst]

@sophiecentaur Thanks for the explanation, I may even stay with the disk (as you said - it resembles a clock pendulum), I’ll see. But the question is - how can I calculate the maximum velocity of the pendulum when I have only this formula for the oscillation period ?

 

[sophiecentaur]

@sophiecentaur Thanks for the explanation, I may even stay with the disk (as you said - it resembles a clock pendulum), I’ll see. But the question is - how can I calculate the maximum velocity of the pendulum when I have only this formula for the oscillation period ?

@Baluncore suggested how to do that. You would need to find the Potential energy at the release height (height of CM of the whole thing above lowest point (h)) and equate that to the kinetic energy at the bottom which is

Mgh = E = ${(2πf)^2 I} /2$
solve it for f (the rotational frequency passing through the lowest point.).

The velocity of the bottom of the pendulum (you choose which bit) will be 2πfR.
You should check this against other web pages about rotating bodies and pendulums. Don't just use any formula blindly.

This method is much easier than solving the equation of motion for a pendulum which is not true shm.

 

[FEAnalyst]

@sophiecentaur So the formula for the velocity would just be: $$v=d \cdot \sqrt{2 \cdot \frac{mgh}{I}}$$ where $m=m_{r}+m_{d}$ and $I=\frac{1}{3} m_{r} l^{2}+m_{d} l^{2}+ \frac{1}{2}m_{d}R^{2}$ ? I assume here that $d$ means the distance from the pivot to the bottom of the bob (total length of the pendulum) while $R$ stands for the radius of the disk.

I would also have to determine $h$ somehow.

 

[sophiecentaur]

You take the rotational speed in 'revs per second' and multiply by the distance from the pivot to whichever point you choose along the length. I suspect you didn't understand my general equation for rotational energy. The I in the formula is the total moment of inertia about the pivot point. Read around more and don't try to get away with the limited stuff in this thread. You want to learn about this topic? Get reading.

 

[FEAnalyst]

You take the rotational speed in 'revs per second' and multiply by the distance from the pivot to whichever point you choose along the length

That's what I meant: $$mgh=\frac{1}{2}I \omega^{2}$$ $$\omega=\sqrt{\frac{2mgh}{I}}$$ $$v=d \cdot \omega=d \cdot \sqrt{\frac{2mgh}{I}}$$ I choose $d$ to be the distance from the pivot to the lowest point of the pendulum (bottom of the disk).

The I in the formula is the total moment of inertia about the pivot point.

Isn't it what I wrote: $I=\frac{1}{3} m_{r} l^{2}+m_{d} l^{2}+ \frac{1}{2}m_{d}R^{2}$ ? It's from the MIT pdf but it also makes sense to me since we sum the moment of inertia of the rod about the pivot with the moment of inertia of the disk (cylinder) plus Steiner's theorem term to account for the offset between the axis of the disk and the pivot axis.

 

[sophiecentaur]

That's what I meant:

Sorry: misunderstanding of symbols in there. You seem to know more than your original responses imply. So where's the problem? Are you getting some disagreements in your sums?

I remember meeting Steiner's therorem first in O Level Statistics at school but no one pointed out the mechanical connection and the theorem wasn't given a name for me, at the time.

 

[FEAnalyst]

So where's the problem?

I've tried using this formula in my case and the maximum velocity doesn't agree with the simulation. Here are my inputs:
- total mass $m=31.1983 \ kg$
- distance from the center of mass to the pivot point in the initial configuration: $h=0.6792898 \ m$
- moment of inertia about the pivot point: $I=29.6747 \ kg \cdot m^{2}$
All those values were calculated analytically and confirmed by taking measurements in 3D modeling software so they are correct. The total length of the pendulum is $1.09577 \ m$ and I take it as $d$ since I'm looking for the maximum velocity at the lowest point of the pendulum (bottom of the bob).
Thus, the maximum velocity is: $$v_{max}=d \cdot \sqrt{\frac{2mgh}{I}}=1.09577 \cdot \sqrt{\frac{2 \cdot 31.1983 \cdot 9.81 \cdot 0.6792898}{29.6747}}=4.10175 \ \frac{m}{s}$$
While from the simulation I get $v_{max} \approx 6.219 \ \frac{m}{s}$
So I wonder if the formula for the maximum velocity is correct.

 

[sophiecentaur]

If you know the for gpe before and after then that’s the max KE.
KE in your formula looks right. Are all your dimensions correct? Tip of disc needs radius added.
How far is the disagreement between calculation and sim ?

 

[FEAnalyst]

Are all your dimensions correct?

Yes, here are the dimensions:

Tip of disc needs radius added.

Actually, it's still a sphere instead of a disk but I accounted for this in calculations (and they were confirmed by measurements in 3D modeling software so there's no mistake here).

How far is the disagreement between calculation and sim ?

It's $4.10175 \ \frac{m}{s}$ from the analytical calculations vs $6.219 \ \frac{m}{s}$ from the simulation so really big difference (around $41 \%$).

There is a chance that the simulation is wrong but it gave me the correct result for the initial geometry (cuboid) so I'm still more suspicious of the analytical approach.

 

[sophiecentaur]

(and they were confirmed by measurements in 3D modeling software so there's no mistake here).

The modelling software may not 'do' centres of mass. Are you sure that's in the right place at the start and at the end? Is h the right value to be using? Are your omegas and your fs appropriate? There are so many places where there can be an error - and, let's face it, the problem just has to be simple. Sorry to tell my grandma how to suck eggs but have you drawn out the two situations with labelled heights and stuff? Your software will have done the drawing out for you but print it out and use pencil to put the exactly appropriate symbols where you know they should be.
Rely on your care and don't assume the software is getting the 'thinking' right for you. It strikes me that you are nearly there.

 

[sophiecentaur]

so I'm still more suspicious of the analytical approach.

I just read this. How could the analytical approach be wrong? If you give it the wrong numbers then the answer will be wrong but you have total control of your inputs and if you tell it the right thing it will not be wrong. Which is why I was suggesting long winded drawing of all the positions of pivots and CMs. Your print out doesn't show them and it only shows one position for the pendulum.

 

[FEAnalyst]

I know what caused this large discrepancy - $h$ was incorrectly defined. It should go from the center of mass in the initial position (at an angle of $45^{\circ}$) to the center of mass in the lowest (vertical) position:

Not just to the pivot point.

 

[sophiecentaur]

I know what caused this large discrepancy - $h$ was incorrectly defined. It should go from the center of mass in the initial position (at an angle of $45^{\circ}$) to the center of mass in the lowest (vertical) position:

Not just to the pivot point.

Well done. You just made me feel really good then - like the midwife at someone's first baby!!

At one point you expressed doubt about the simulation; well justified. But you need never have doubt about the tried and tested analysis. The faults are always your own and you can correct them.