# How to calculate the moment of inertia for complex 3D shapes

#### jm090693

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Hi,

I need to calculate the moment of inertia for the component in the attached image so that i can calculate the angular momentum. Is it possible?

Overall i am trying to calculate the forces on this lug as it passes around a 3" radius at 2M a second.

Thanks,

Jack #### Attachments

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#### anorlunda

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Where is the axis of rotation?

Next time, you must post homework in a homework forum and fill out the template.

But even this time, we require that you show some effort before we help. What have you tried so far? What is the axis of rotation?

#### haruspex

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The method is to decompose it into simple shapes, find the MoI of each around the chosen axis, and add them up.
Sometimes it helps to represent a shape as the difference of two shapes, e.g. for a sphere with a hole treat it as a whole sphere then subtract the MoI of the missing piece.

#### jm090693

Hi, I have added some more details here.

I did not post it in the homework section as it isn't homework?

Thanks,

Jack

#### haruspex

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Hi, I have added some more details here.

I did not post it in the homework section as it isn't homework?

Thanks,

Jack
You won't need it to be all that accurate. Just treat the lug as a rectangular plate, LxW say, (length x width). Let the midpoint of the plate be distance R from the centre of rotation.
The moment of inertia would be M(R2+(L2+W2)/12).

Not sure how to figure out the stress, though. On the face of it, the angular motion goes from zero to nonzero, and vice versa, in zero time, which means infinite acceleration. So the elasticity of the material is important.
Do you really need to calculate this to establish the cause? Is there any other feasible cause?
An obvious solution would be to increase the radius of the turn.

In principle, the start and end of the turn would be equally stressful. Do the cracks originate at the leading edge or the trailing edge?

#### jm090693

You won't need it to be all that accurate. Just treat the lug as a rectangular plate, LxW say, (length x width). Let the midpoint of the plate be distance R from the centre of rotation.
The moment of inertia would be M(R2+(L2+W2)/12).

Not sure how to figure out the stress, though. On the face of it, the angular motion goes from zero to nonzero, and vice versa, in zero time, which means infinite acceleration. So the elasticity of the material is important.
Do you really need to calculate this to establish the cause? Is there any other feasible cause?
An obvious solution would be to increase the radius of the turn.

In principle, the start and end of the turn would be equally stressful. Do the cracks originate at the leading edge or the trailing edge?
Hi,

Do you mean treat the leading edge of the lug as a rectangular plate, what about everything behind that front plate or is that excluded?

Basically i want a value which i can compare with the material properties to prove that this is why they are failing. I will then increase the surface area of contact between the lug/chain and reduce the centre of gravity closer to the chain attachment before re-calculating. See attached showing some calculations for the acceleration at the tip of the lugs.

Absolutely, if the radius of that corner could be increased there would be no problem. Unfortunately due to the application it needs to change direction in a very small window. I believe the stress occurs when the chain changes 90 degrees direction and then once horizontal is in a wear-strip/guide meaning it is fixed rotationally. However the inertia of the lug continues as it is accelerating around the 90 degree bend. The only portion stopping the lug rotating any more past horizontal is the 31mm^2 surface area between the lug/chain. This pressure repeatidly at 450/minute 24/7 is causing a fatigue crack on the lug face. (VIDEO) The yellow lug cracks on the leading edge of the chain attachment & the red lug cracks on the leading edge of the chain attachment. Meaning the above theory must be true as there is no other point on the chain path which has forces in these direcitons.

Thanks in advance,

Jack

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#### haruspex

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treat the leading edge of the lug as a rectangular plate
No, the whole thing beyond the narrow stalk. The thickness of the plate does not matter.
the inertia of the lug continues as it is accelerating around the 90 degree bend
Not quite sure what you mean by that. Once it has got going around the bend there is no angular acceleration until the end. The stressful events are starting and finishing the rotation.
The video seems to show that the belt rate changes hugely. That would exert similar stresses during the belt accelerations.
The yellow lug cracks on the leading edge of the chain attachment & the red lug cracks on the trailing edge
I suggest this is because the red lugs act as buffers, sparing the yellow lugs from the deceleration stress.

L

#### llober

No, the whole thing beyond the narrow stalk. The thickness of the plate does not matter.

Not quite sure what you mean by that. Once it has got going around the bend there is no angular acceleration until the end. The stressful events are starting and finishing the rotation.
The video seems to show that the belt rate changes hugely. That would exert similar stresses during the belt accelerations.

I suggest this is because the red lugs act as buffers, sparing the yellow lugs from the deceleration stress.
I think the fatigue tension that produces the crack is not due to the lug leaving the turn, but during the turn itself, because at the end of the turn the material is at compression where the crack appears, but is being teared appart during the turn.

Anyway, or it's a poltergeist, or is due to the forces produced during the turn... obviously, it's the second reason.

You'll calculate the moi ... and what? If you have to turn the piece that way, you'll have to reinforce the piece just where the crack is appearing, because "experimentally" you have determined that the cross section of the piece there is not enough to handle the stress.

#### jm090693

The stressful events are starting and finishing the rotation.
Yes, precisely. I am concentrating on the stressful event at the end of the rotation. Here the inertia of the lug means it wants to continue rotating or travelling at the higher corner speed. The leading edge of the chain attachment slows the lug down to the linear speed of the horizontal chain. I believe the lug is rotating on the bolt which connects it to the chain attachment at the end of rotation, where the lug want to continue.
It's the stress on the lug i want to calculate at this point.

The video seems to show that the belt rate changes hugely. That would exert similar stresses during the belt accelerations.
My bad i should have explained. This is a high speed camera video, the sections at the start & end are the running speed, the section in the middle has been slowed down so you can see whats going on.

I suggest this is because the red lugs act as buffers, sparing the yellow lugs from the deceleration stress.
I believe the same theory above of the lugs rotating at the end of rotation is happening on the red lugs too, as the cracks are both on the leading edge of the lug.

#### jm090693

I think the fatigue tension that produces the crack is not due to the lug leaving the turn, but during the turn itself, because at the end of the turn the material is at compression where the crack appears, but is being teared appart during the turn.

Anyway, or it's a poltergeist, or is due to the forces produced during the turn... obviously, it's the second reason.

You'll calculate the moi ... and what? If you have to turn the piece that way, you'll have to reinforce the piece just where the crack is appearing, because "experimentally" you have determined that the cross section of the piece there is not enough to handle the stress.
Hi,

the chain is driven, if the lug was cracking during the turn it would be on the trailing edge of the lug. At the end of the rotation the bottom of the lug is in tension as the lug is wanting to continue rotating on the chain attachment.

I need to prove my theory in calculations so i can confirm my new design with calls prior to paying for new moulds.

Thanks,

Jack

#### haruspex

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I think the fatigue tension that produces the crack is not due to the lug leaving the turn, but during the turn itself
No, the stress during the turn is just a bit of tension from centrifugal force. This would be quite modest compared with the very high angular accelerations at start and end of the turn.

#### haruspex

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It's the stress on the lug i want to calculate at this point.
As I indicated in post #5, to do that you need to find the angular acceleration at transition. This is problematic because it depends on the elasticity of the lug material, and even of the chain.
Even without the lugs, the chain itself cannot instantly transition between linear and rotational. If you could inspect the chain shape in great detail you would see some inertia at each end of the turn. The extent of that depends on the chain tension. Maybe we can assume the chain is quite taut and we can ignore this.
That leaves us with the lug elasticity. Any data on that?

L

#### llober

If a CAD programd doesn't give you the MoI of this complex 3D shape, then perhaps you can use an experimental method to find an aproximate value. We can use Python to solve numerically the differential equation, first with an estimated value for your MoI, and iterating until the period (T) found experimentally is close enough to the result T solving the differential equation:

https://repl.it/@llober/Solid-body-pendulum

After, you calculate the MoI at the real axis position, using the parallel axes theorem.

Now, what I think about your problem.

Before and after the turn, you have a Kinetic Energy for the moving object, with mass m and a linear speed v ...

$$K = \frac{1}{2}m \ v^2$$

In the turn, you have a Rotational Energy ...
$$K_r = \frac{1}{2}\ I_s \ ω^2$$

So which is the problem in my opinion? If both energies don't match, then you'll have an inelastic "impact" both when entering the turn, and after when leaving the turn. Greater the mismatch, greater the "hit" to the moving piece.

Doing very, very rough numbers, the tension wheel should rotate, for example, at 30 rad/s (now is at about 45 rad/s). If you need to mantain the same linear speed, the radius of the tension wheel should be 1.5 times greater.

Take it with a grain (or two) of salt.

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#### haruspex

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Before and after the turn, you have a Kinetic Energy for the moving object, with mass m and a linear speed v ...

$$K = \frac{1}{2}m \ v^2$$

In the turn, you have a Rotational Energy ...
$$K_r = \frac{1}{2}\ I_s \ ω^2$$

So which is the problem in my opinion?
The energy is not in itself a problem. Forces are the problem.
If both energies don't match,
Match? As in, same magnitude? What has that to do with anything?!

L

#### llober

The energy is not in itself a problem. Forces are the problem.

Match? As in, same magnitude? What has that to do with anything?!
You mean that Forces and Energies are not related in anyway?

I wonder what part of what I said is not true about the kinetic and rotational energies ... or do you mean that during the turn, the object mantains both energies, so is the sum of the initial Kinetic plus the Rotational? #### Attachments

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#### haruspex

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I would like to know what part of what I said is not true about the kinetic and rotational energies
You wrote that the problem is if the energies do not match. I asked what you meant by that.
You mean Forces an Energies don't have anything to do, one concept with the other, and are not related in anyway?
Of course they are related, but there is no limit to the energy the object may have without there being any significant stress.

L

#### llober

You wrote that the problem is if the energies do not match. I asked what you meant by that.

Of course they are related, but there is no limit to the energy the object may have without there being any significant stress.
Ok, let's say that the object enters the turn with a kinetic energy of 1 unit ... this object is not in free movement, because is being pulled by the chain, but when it enters the turn, its energy (kinetic) is 1 unit.

While in the turn, its energy, due to its moment of inertia and angular speed forced by the chain, has to be 3 units.

Can you tell me where it gets this "extra 2-energy units"? You tell it, of course because there're acting forces. The forces act suddenly, because in a very short time, the energy of the object has to increase from 1 unit to 3 units.

At the exit of the turn, the energy of the object has to decrease suddenly from 3 units to 1 unit.

(The 3:1 ratio is a rough estimate I've made, based on the drawings of the machine and its elements).

Did you understand it?

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#### haruspex

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The forces act suddenly, because in a very short time, the energy of the object has to increase from 1 unit to 3 units.
Sure, but it's not the increase that matters, it's the rapid change in direction. By your argument, there would be no stress if the energies were of equal magnitude.
But it is good to see that you are now acknowledging the importance of the rate of change rather than just the magnitude.

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L

#### llober

Sure, but it's not the increase that matters, it's the rapid change in direction. By your argument, there would be no stress if the energies were of equal magnitude.
But it is good to see that you are now acknowledging the importance of the rate of change rather than just the magnitude.
Wow, I didn't negate the importance of the rate of change, because what I said is that the energy of the body has to change from 1 to 3 (for example) at the very start of the turn, and that implicitly means "instantly", which, of course, is not zero time but a short time.

And the magnitudes do matter, because the rate is about how those magnitudes change in time ... and we're talking about the, more or less, same "short" time (undetermined), but obviously is not the same to change from 1.0 to 1.01, than from 1.0 to 3.0 in the "same" short time ... I don't get how you can't not understand this.

You talk about forces, I talk about energies ... we're talking about the same physics, but we're watching the problem from different points of view, perhaps because I'm an engineer. In my vision, I can give an order of magnitude of the root cause of the problem ... in your vision, it's absolutely obscure, and due to complex forces that act in very short undetermined times ... you can't give any number, not even approximate numbers with that approach.

I don't care the forces or its distribution to see what is happening in this case... I know that are there, in a very complex interation, which presumably can only be solved doing a numerical finite element simulation of the case.

By the way, yes, I'm saying that if you design properly the system, and the Kinetic energy of the piece when enters the turn is more or less equal to the Rotational energy that will have during the turn (which is a desing choice), then the unique forces that will "smoothly" play are the centripetal forces, that are unavoidable because, of course, you have to turn. At least, I hope that you realize this.

Sir, you see the little times and uncertain forces, but don't see the overall picture.

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#### haruspex

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if ... the Kinetic energy of the piece when enters the turn is more or less equal to the Rotational energy that will have during the turn, then the unique forces that will "smoothly" play are the centripetal forces
No. The forces that execute the transition create large torques. During the turn, the torques have disappeared and only the (much weaker) centripetal force remains.

what I said is that the energy of the body has to change from 1 to 3 (for example) at the very start of the turn,
Not until post #17. Until then you implied it was all down to the magnitude of the change. Indeed, you originally stated (post #8) that the main stress was during the turn, not at the transition between linear and rotational.

L

#### llober

No. The forces that execute the transition create large torques. During the turn, the torques have disappeared and only the (much weaker) centripetal force remains.

Not until post #17. Until then you implied it was all down to the magnitude of the change. Indeed, you originally stated (post #8) that the main stress was during the turn, not at the transition between linear and rotational.
Yes, I said in post #8 that the strees was in the turn, and I was wrong because I didn't realize the huge mismatch between the entering kinetic energy and the rotational energy that the piece has to achieve of a sudden, and I assumed only the centripetal forces, that is how the system should work if properly designed. Errare humanum est!

"The forces that execute the transition..." ... yes, there's a transition, big and problematic transition. For example, at the exit of the turn, where the energy has to go from 3 (example) to 1 ...that energy is totally lost ... the effect is a "hit", where that extra energy will be dissipated by unelastic deformation forces, vibration, heat, sound waves, etc ...

And, by the way, you implied in post #14 that forces and energies are not related!

In post #17 I included this drawing... You put words in my mouth that I did not say! According to may view, the centripetal forces (done by the walls) do turn the ball. What you don't seem to realize is that the kinetic energy at the entry of the turn, is the rotational energy during the turn (if the body moves freely).

If you substitute the value of ω by v/r, and assume that the ball is a "point mass", you'll see that's the same thing.

And for me, given that you have to make a turn and you go to a certain fixed speed, then the best choice is to balance the kinetic entering energy with the rotational energy ... or this is not true?

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#### haruspex

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you implied in post #14 that forces and energies are not related!
I did not. I wrote that the energy is not in itself a problem. Specifically, it is the rate at which the energy is transferred from one mode to another, i.e. the forces.
where the energy has to go from 3 (example) to 1 ...that energy is totally lost .
No, most of the KE lost by each flange would be transferred back into the belt.
In post #17 I included this drawing
I assume in this diagram that the ball is frictionless, so is not rotating on its own centre. Or that might as well be the case. That means there is not essentially any rotational KE involved. Yes, you can choose to view it as rotational KE by choosing a an axis off to the side, as you have done, but you can do exactly the same with constant velocity motion.
the best choice is to balance the kinetic entering energy with the rotational energy
Try this: a rod is moving laterally at a constant speed. One end encounters a rubber stop and bounces elastically. This converts much of the linear KE to rotational, but there is no net change in KE.
By your reasoning, there is no stress on the rod, right?

L

#### llober

The definition of Rotational energy is
$$E_{rot}=\frac{1}{2}Iω^2$$
where $I$ is the moment of inertia around the axis of rotation considered.

https://en.wikipedia.org/wiki/Rotational_energy

In post #5, you correctly stated that if we consider the object to be like a plate, then the moment of inertia would be:
$$M(R^2+\frac{L^2+W^2}{12})$$
So, the rotational energy, as is officially defined, is:
$$E_{rot}=\frac{1}{2}Mω^2(R^2+\frac{L^2+W^2}{12})=\frac{1}{2}MR^2ω^2 + \frac{1}{2}Mω^2\frac{L^2+W^2}{12}$$
As we know that $ω=\frac{v_{cm}}{R}$
$$E_{rot}=\frac{1}{2}Mv_{cm}^2+\frac{1}{2}Mω^2\frac{L^2+W^2}{12}$$
And you can say that this is the Kinetic energy of the center of mass moving around the center of the rotation, plus an additional energy coming from the rotation of the object around its own center of mass.

In our case, we consider a plate of about 6" x 2.5". I'll work in SI units from now on, so the plate is 0.15m x 0.06m. The rotational speed for a 2m/s linear speed and 0.044m tension wheel radius is 45.5 rad/s. The radius to the center of mass is 0.127m. The energy per Kg is
$$E_{rot}=\frac{1}{2}(0.127)^2(45.5)^2 + \frac{1}{2}(45.5)^2·\frac{(0.15)^2+(0.06)^2}{12}=19 \ m^2s^{-2}$$
Before entering the turn, the linear Kinetic energy per Kg is
$$\frac{1}{2}(2)^2 = 2 \ m^2s^{-2}$$
So we end having 9 times more energy just "one instant" after entering the turn, that the energy we had during the linear movement just "one instant" before entering the turn (oops .. I said 3 times in a previous post... sorry).

And, to make it worst, that energy is taken away from the object as soon as it leaves the turn, which at 45.5 rad/s happens 0.04 s later.

The "minimum stress" to the system, imho, would be to reduce the rotational speed of the drive wheel (if that is feasible), and have the same energy before and during the turn, which doesn't mean that no stress would exist, but means that the own kinetic energy of the entering object would suffice for the turn.

But, there's one caveat: if we want to mantain the speed of the drive chain at 2 m/s, if we reduce the rotational speed, we have to increase the radius and so the moment of inertia is affected!

At the end, in this particular case, it's not possible to mantain the 2 m/s drive chain speed and balance exactly the initial kinetic energy with the rotational energy, but we can get quite close to the objective

For example, with a drive wheel radius of 8.2" (from 1.73" initial) and a rotational speed of 9.6 rad/s (91.7 rpm), we have a rotational energy that is only 2 times greater than the linear kinetic energy for 2 m/s speed (originally, was 9 times greater).

#### haruspex

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which doesn't mean that no stress would exist, but means that the own kinetic energy of the entering object would suffice for the turn
Your insistence that the magnitude of the change in energy is relevant is not based on any physical law that I am aware of.
Consider an inertial frame of reference moving with the linear portion of the belt approaching the turn. In this frame, we go from no KE to the same rotational KE (but a different linear KE). If the energies matched, somehow, in the lab frame they do not in this frame, yet the accelerations, forces and stresses are exactly the same.

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