# Where is wrong in this proof for rotational inertia ?

• Duke Le
In summary, the formula for inertia of a ring (2D circle) about its central axis can be found by integrating along the Oy axis.
Duke Le

## Homework Statement

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

## Homework Equations

$$I = MR^2$$
Where:
M: total mass 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: $$dm = \frac{dy}{2 \pi R}M$$
* Explanation: $$\frac{dy}{2 \pi R}$$ 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: $$dI = \frac{dy}{2 \pi R}MR^2$$

→ The rotational inertia for ring is:
$$I = 4 \int_{0}^{R}\frac{dy}{2 \pi R}MR^2$$

- Solving for this, I get $$I = \frac{2MR^2}{\pi}$$

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

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Ho Duke! Welcome to PF!
Duke Le said:
dm=dy2πRM
This is unnecessary.

You have assumed the ring to be made up of infinite small masses 'dm'. All these masses are equidistant from the center. What is the moment of inertia of individual dm about the central axis? This would be much simpler.
Duke Le said:
Please show me the mistake and correction but not an entire different solution.
Ok. If you want to use your method, your integration limits are wrong. You are moving "along" the ring, not from 0 to R. So what should be the upper limit of y? Is this hint enough?

cnh1995 said:
Ho Duke! Welcome to PF!
Ok. If you want to use your method, your integration limits are wrong. You are moving "along" the ring, not from 0 to R. So what should be the upper limit of y? Is this hint enough?

I think I have found the mistake. My intention for integrating from 0 to R was that dy ~= arc length from y to y+dy (moving vertically in y+ direction), I guess that approximation is wrong here :D

The upper limit should be R*pi/2

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Duke Le said:
I think I have found the mistake. My intention for integrating from 0 to R was that dy ~= arc length from y to y+dy (moving in y+ direction), I guess that approximation is wrong here :D

The upper limit should be R*pi/2

This is why polar coordinates were invented!

PeroK said:
This is why polar coordinates were invented!

I haven't really learned polar coordinate yet, only its definition and simple calculation. I did try integration using dθ and Rcos(θ) though, it was:
$$I = 4\int_{0}^{\frac{\pi }{2}} \frac{d\theta }{2 \pi}MR^2$$
and it gave the correct result. I just had no idea why the other formula didn't work.

Is the approximation dy ≅ arc length from y to y + dy ever used ? Or is it just wrong ? Thank you.

Duke Le said:
Is the approximation dy ≅ arc length from y to y + dy ever used ? Or is it just wrong ? Thank you.

It's not an approximation. When ##y## is close to 1, then ##\Delta y \approx 0##.

PeroK said:
It's not an approximation. When ##y## is close to 1, then ##\Delta y \approx 0##.

I'll add a picture because I don't have enough vocabulary yet.

For the dy ≅ arc length approximation, I meant: dy ≅ a when dy is very small.
Is this approximation valid ?

Duke Le said:
I'll add a picture because I don't have enough vocabulary yet.
View attachment 112532

For the dy ≅ arc length approximation, I meant: dy ≅ a when dy is very small.
Is this approximation valid ?
No.

PeroK said:
No.

Okay that's where things went wrong. Thanks a lot !

## 1. What is rotational inertia?

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to change in its rotational motion. It depends on the mass and distribution of mass of the object.

## 2. How is rotational inertia calculated?

Rotational inertia is calculated by multiplying the mass of the object by the square of its distance from the axis of rotation. This distance is also known as the object's radius of gyration.

## 3. What is wrong in the given proof for rotational inertia?

The given proof may be incorrect due to incorrect calculations or assumptions, improper application of formulas, or neglecting external forces such as friction or air resistance.

## 4. How does rotational inertia affect rotational motion?

Higher rotational inertia makes it more difficult to change an object's rotational motion. This means that more force is required to accelerate or decelerate the object, and it will also take longer to stop or change directions.

## 5. Can rotational inertia be changed?

Yes, rotational inertia can be changed by altering the object's mass or its distribution of mass. For example, changing the shape of an object or moving its mass closer to or further from the axis of rotation will affect its rotational inertia.

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