How to calculate moment of inertia of this irregular system?

In summary, the conversation revolved around a lab investigating the relationship between moment of inertia and speed. The individual conducting the lab shared details about their experiment, including using a photogate to measure time and calculating the moment of inertia of the magnet. Suggestions were made regarding calculating the correct moment for the two discs and adding the magnets, and using the parallel axis theorem to compensate for the off-center magnets.
  • #1
Homework posted in wrong forum, so no template
hello all, I am currently doing a lab investigating the relationship between the moment of inertia and speed.
I have attached some pictures of my 'wheel' that is going to be rolled down a ramp. FYI i will be using a photogate to measure the time it takes the black strip to pass through. Initially, I thought the magnets were creating the 'hollow' part of the disc, but apparently i am wrong, and I am wondering how to calculate the moment of inertia of the magnet.

I am desperate for help. As my teacher has not yet taught anything on this specific topic.
 

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  • #2
Can you come up with a piece wise density function? So the density is this from r = 2cm to r =4 cm, then the magnets have a density of that and they're centered at 5 cm along the x with radius 1cm or something like that.

If you can, then all you have to do is run the integral and sum them.

Worth noting: You only need to calculate the moment of inertia about the axis that you're planning on having rotation about. So you only need to worry about radial density, and you can assume all the mass lies in a plane.
 
  • #3
BiGyElLoWhAt said:
Can you come up with a piece wise density function? So the density is this from r = 2cm to r =4 cm, then the magnets have a density of that and they're centered at 5 cm along the x with radius 1cm or something like that.

If you can, then all you have to do is run the integral and sum them.

Worth noting: You only need to calculate the moment of inertia about the axis that you're planning on having rotation about. So you only need to worry about radial density, and you can assume all the mass lies in a plane.
Hello, thank you for your suggestion,
i actually went on the internet and tried to derive the moment of inertia of the solid wheels, which is 1/2 m r^2 .
for the moment of inertia for the 8 magnets, I could either assume they are point mass, and use this formula I=mr^2. however, as i move the magnets towards the center of rotation, the error will increase.
thus, i went online again and found this formula mass of magnets( distance from front of magnet to center of wheel^2+2distance from back of magnet to center^2)
After i found the moment of inertia of the system. I will sub it into the equation mgh= 1/2mv^2 (translation)+ 1/2 I(omega)^2
 
  • #4
If your circle is small, I wouldn't assume that they are point masses. But I suppose it is a lab, so experimental error haha
 
  • #5
BiGyElLoWhAt said:
If your circle is small, I wouldn't assume that they are point masses. But I suppose it is a lab, so experimental error haha
But for some reason, I can't get like a reasonable value with these formulas.
MgH=v^2(m/2+Isystem/2r^2)
where M is the mass of the whole wheel system 0.369kg.
g is 9.8
H is the initial height on the ramp which is 0.105m
V is measured with a photogate ,0.10m/0.11746s= 0.85135m/s
radius of the circle is 8cm.
I of the system is (mr^2/2)+(mr^2/2)+8(mass of magnets (0.019kg(0.53m^2+2(0.078m^2))
where the mass of bgi wheel is 0.142kg and small wheel 0.074kg.
 
  • #6
What I think is happening (maybe this is significant, and maybe it isn't), you're neglecting the holes in the board. Can you calculate what the 2 wheels would weigh if they were just the wheels, with no holes or magnets? Also, you have a giant hole in the center. Your 1/2mr^2 is not valid.
Calculate the correct moment for the 2 discs. Sum them. Then add the magnets.
 
  • #7
y
BiGyElLoWhAt said:
What I think is happening (maybe this is significant, and maybe it isn't), you're neglecting the holes in the board. Can you calculate what the 2 wheels would weigh if they were just the wheels, with no holes or magnets? Also, you have a giant hole in the center. Your 1/2mr^2 is not valid.
Calculate the correct moment for the 2 discs. Sum them. Then add the magnets.
um

um maybe the photo isn't too clear, but the wheel is actually a clear plastic, i used black tape so the photogate can sense it. in other words, the wheel is not a hollow cylinder.
 
  • #8
OK, so is it solid all the way through? Even where the magnets are placed? I was thinking that there was a hole in the center.
 
  • #10
BiGyElLoWhAt said:
OK, so is it solid all the way through? Even where the magnets are placed? I was thinking that there was a hole in the center.
yes the plastic disc is solid.
 
  • #11
sorry
BiGyElLoWhAt said:
If that's the case, then this is all you should need.
https://en.wikipedia.org/wiki/Parallel_axis_theorem
I'm sorry for dragging on, I was thinking this was a lot more complicated than it was.
can you explain this theorem to me as I have not yet learn the moment of inertia at school and i don't really understand this part
 
  • #12
So then ##I_{sys} = I_{disc\ 1} + I_{disc\ 2} + I_{mags}##
And I_mags is not at zero, so you need to shift it.
 
  • #13
BiGyElLoWhAt said:
So then ##I_{sys} = I_{disc\ 1} + I_{disc\ 2} + I_{mags}##
And I_mags is not at zero, so you need to shift it.
I think i will read the parallet axis theorem again. tahksn
 
  • #14
So basically, it tells you how to move the axis. You have 2 places where there are magnets, and they're off center. So you need to compensate for that. You know the moment of inertia of a disc, so shift it, and add it to the original moment. It's a rather simple formula.
You have ##I_{d}= I_{cm} + md^2##
 
  • #15
BiGyElLoWhAt said:
So basically, it tells you how to move the axis. You have 2 places where there are magnets, and they're off center. So you need to compensate for that. You know the moment of inertia of a disc, so shift it, and add it to the original moment. It's a rather simple formula.
You have ##I_{d}= I_{cm} + md^2##
but i thought the magnets and the wheel share the rotation axis
 
  • #16
They do rotate about the same axis, but the axis the magnets rotate about is not it's center of mass. You know the moment of inertia for a disc about it's center of mass ##I_{cm}##, and you have a formula that allows you to shift that axis by a distance d that uses the formula that you already know.
 
  • #17
iancheung228 said:
thus, i went online again and found this formula mass of magnets( distance from front of magnet to center of wheel^2+2distance from back of magnet to center^2)
You should replace this part of the formula with the parallel axis theorem. which should not include the distance to the edges, only to the center of the magnets.
 
  • #18
BiGyElLoWhAt said:
You should replace this part of the formula with the parallel axis theorem. which should not include the distance to the edges, only to the center of the magnets.[/QUOTE
so in other words, you are assuming the magnets are point mass? but wouldn't it because invalid as the magnets move closer to the center of the wheel
 
  • #19
I am not assuming that the magnets are a point mass.
It is not mr^2. It is ##I_{cm} + mr^2## for the magnets. The I_cm takes care of the fact that they are not point masses.
 
  • #20
i think i understand the parallel axis theorem now. Thank you
i have another question thought, how do i find the initial moment of inertia of the magnets?
 
  • #21
They're solid discs, aren't they? ;-)
 
  • #22
BiGyElLoWhAt said:
They're solid discs, aren't they? ;-)
Haha, I am not sure what you meant by that but let me try to explain to you what i understand
So I parallel = Icm +Md^2
so mass is the mass of magnet
and d is the distance from the center of magnet to the center of wheel
but how do you calculate the Icm, just i=mr^2?, but in that case we are assuming its pooint mass,
and if you really want an accurate moment of inertia of magnet, you need to have a experession for the arc on the magnets in order to integrate, which is impossible for me to come up the expression lol
 
  • #23
You have some solid discs. Call the axis of rotation z. You only care about the moment about z, right? So as far as you're concerned, all of the mass lies in the x-y plane, and the dimension of the magnets in z doesn't matter. So you have 2 flat, solid discs, exactly like the wheels, except they have different mass and r. You can calculate the moment of inertia if the magnets were rotating about their own center, right? It'd be the same as the wheel. Then apply the parallel axis theorem.
 
  • #24
Maybe this will help. Z is the axis that get's calculated (cm), and d is the shift from the P.A. theorem.
 

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1. What is moment of inertia?

Moment of inertia is the measure of an object's resistance to changes in its rotational motion. It depends on the mass and distribution of mass of the object.

2. How do I calculate the moment of inertia of a system?

To calculate the moment of inertia of an irregular system, you will need to use the formula I = ∑mr², where I is the moment of inertia, m is the mass of each component, and r is the distance of each component from the axis of rotation. Add up the individual moments of inertia for each component to get the total moment of inertia for the system.

3. Can I use the same formula for any shape or system?

No, the formula for moment of inertia may vary depending on the shape and distribution of mass of the system. For an irregular system, you may need to use the parallel axis theorem, which takes into account the distance between the axis of rotation and the center of mass of the system.

4. What units are used for moment of inertia?

Moment of inertia is typically measured in units of kg·m² or g·cm². The units will depend on the units used for mass and distance in the formula.

5. Why is moment of inertia important in physics?

Moment of inertia is an important concept in physics because it helps us understand and predict the rotational behavior of objects. It is used in many areas of physics, such as in calculating the torque required to rotate an object or in predicting the stability of a rotating object.

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