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 )

This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

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.

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestHow exactly do you want it to be modeled?

Log in to reply

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

Log in to reply

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.

Log in to reply

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

Log in to reply

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

Log in to reply

Mr.Steven Chase please post your solution.

Log in to reply

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.

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply