Calculating moment of inertia of a complicated system

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The discussion focuses on calculating the moment of inertia for a wheel system used in a lab experiment investigating the relationship between moment of inertia and speed. The user initially misunderstood the role of magnets in the wheel's structure and seeks clarification on how to compute the moment of inertia accurately. Key equations involving mass, height, and velocity are provided, along with the need to break down the system into its components and apply the parallel axis theorem. There is also a concern about increasing error as the magnets move toward the center of the wheel, prompting questions about the nature of this error. Understanding these calculations is crucial for the experiment's success.
iancheung228
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Homework Statement


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 and then apply the moment of inertia to the equation mgh=1/2mv^2+I system (omega)^2.

Homework Equations


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.

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.

Screen Shot 2016-06-13 at 9.30.43 PM.png
Screen Shot 2016-06-13 at 9.31.06 PM.png
Screen Shot 2016-06-13 at 9.30.43 PM.png
Screen Shot 2016-06-13 at 9.31.06 PM.png


The Attempt at a Solution

 
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iancheung228 said:
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

Can you explain what you mean by "I thought the magnets were creating the 'hollow' part of the disc" ?

The moment of inertia of a composite is the sum of the moments of inertia of the parts. So break it down into parts and look up the moment of inertia of each. You will need to apply the parallel axis theorem.
 
CWatters said:
Can you explain what you mean by "I thought the magnets were creating the 'hollow' part of the disc" ?

The moment of inertia of a composite is the sum of the moments of inertia of the parts. So break it down into parts and look up the moment of inertia of each. You will need to apply the parallel axis theorem.
yup i figured after having a conservation with another person on physics forum. is it true that there's increasing error as my magnet move towards the center of the wheel?
 
Error between what and what?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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