What is the moment of inertia of a fidget spinner assuming mass of the spinner as M and radius of the disc end as \(R\)? (Assuming that the Fidget spinner is made of three discs )

Do the disks have holes in them? If so, what are the inner and outer radii. And how are the disks connected to each other? If you look at an actual fidget spinner, there is a complicated connection between them. We'll probably need to simplify the connection.

Model the fidget spinner as three disks with their centers as the vertices of an equilateral triangle. The disk radius is \(R\) and the distance (\(d\)) of each disk center to the origin is \(\frac{2}{\sqrt{3}} R\).

The moment of a disk of mass \(M\) about it's center is \(\frac{M R^2}{2}\). Since all three disks are the same distance from the origin, consider the total mass to be concentrated in one of them. We must use the Parallel Axis Theorem to calculate the moment with respect to the origin.

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## Comments

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TopNewestHow exactly do you want it to be modeled?

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it should consits of only three discs of each radius R and mass of each disc as M/3

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Do the disks have holes in them? If so, what are the inner and outer radii. And how are the disks connected to each other? If you look at an actual fidget spinner, there is a complicated connection between them. We'll probably need to simplify the connection.

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there are no holes and the discs are stuck together using an adhesive.

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I have an answer. This is a pretty nice problem. If you have the answer, perhaps you could post it as a problem.

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Mr.Steven Chase please post your solution.

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Model the fidget spinner as three disks with their centers as the vertices of an equilateral triangle. The disk radius is \(R\) and the distance (\(d\)) of each disk center to the origin is \(\frac{2}{\sqrt{3}} R\).

The moment of a disk of mass \(M\) about it's center is \(\frac{M R^2}{2}\). Since all three disks are the same distance from the origin, consider the total mass to be concentrated in one of them. We must use the Parallel Axis Theorem to calculate the moment with respect to the origin.

\[I = \frac{M R^2}{2} + Md^2 = \frac{M R^2}{2} + \frac{4}{3} MR^2 = \boxed{\frac{11}{6} MR^2} \]

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Actually, there is also one disc with centre at the origin.

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In that case it's easier. \(\frac{25}{2} M R^2\)

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so added up all moment of inertia of all the discs along the axis at origin right?

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Yes, that's right. If you agree with my answer, you can post it. Thanks

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In that case, how about posting this as a question ?

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