- #1

Duke Le

- 7

- 0

## Homework Statement

Prove the formula for inertia of a ring (2D circle) about its central axis.

## Homework Equations

[tex] I = MR^2 [/tex]

Where:

M: total mass of the ring

R: radius of the ring

## The Attempt at a Solution

- So I need to prove the formula above.

- First, I divide the ring into 4 section so I only need to calculate the rotational inertia of the first quadrant (which has only positive coordinates), then multiply by 4.

- Then, I divide the quadrant into infinite arc length, each is determined by an interval on Oy axis. Each interval is determined by its position y and its height dy. So I integrate along the Oy axis.

+ Since dy is very small, dy ≈ arc length.

+ The differential mass dm then equals: [tex] dm = \frac{dy}{2 \pi R}M [/tex]

* Explanation: [tex] \frac{dy}{2 \pi R} [/tex] is the portion of the ring that belongs to this arc length.

+ This mass has distance R to the central axis.→ The rotational inertia for this mass is: [tex] dI = \frac{dy}{2 \pi R}MR^2 [/tex]

→ The rotational inertia for ring is:

[tex]

I = 4 \int_{0}^{R}\frac{dy}{2 \pi R}MR^2

[/tex]

- Solving for this, I get [tex] I = \frac{2MR^2}{\pi} [/tex]

**4. Conclusion:**

- Could someone tell me where I went wrong with my calculation ? Please show me the mistake and correction but not an entire different solution.

Thank you in advance.

This is the first time I ask a question on physicsforum, so please forgive me if there're any mistakes.

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