{%hackmd 5xqeIJ7VRCGBfLtfMi0_IQ %}

# Rotational inertia

## Question

Given two objects, say a solid cylinder and a hollow cylinder, rolling on a ramp, which one goes faster?

## Experiments

You need: [handout](https://www.math.nsysu.edu.tw/~chlin/math-runway/rotational-inertia.pdf)

1. Roll the object to see the real-world answer.
2. Approximate rotational inertia.
3. Use calculus to find the rotational inertia.

## Intuition

Galileo told us that two objects, regardless of their weights, freely falls at the same height will hit the ground at the same time. When the same experiment is observed on a ramp, two kinds of motion may happen—[sliding](https://en.wikipedia.org/wiki/Sliding_(motion)) and [rolling](https://en.wikipedia.org/wiki/Rolling). Based on what we learned, the force dragging the object is $mg\sin\theta$.


(Source: [Wikipedia](https://en.wikipedia.org/wiki/Sliding_(motion)))

With Newton's second law of motion $F = ma$, the acceleration is independent of the weight.

However, in the language of rolling, the "weight" is replaced by the rotational inertia $I = mr^2$, where $r$ is the distance between the particle and the pivot point. With again $F = mg\sin\theta$, the acceleration depends on the angle $\theta$ of the ramp and the ratio $\frac{I}{m}$. This means objects with larger rotational inertia $I$ roll slower, while objects with smaller $I$ roll faster.

When particles on an object are of different distances to the pivot point, we use calculus

$$
I = \int\rho r^2\, dV,
$$

where $\rho$ is the density, to find the rotational inertia.

For each of the object, we may approximate a rough value for $I$ and calculate the precise value for $I$. Take the solid cylinder as an example, you may consider the cylinder as $7$ parts, where the weight of each part is $\frac{1}{7}M$. We consider all these parts as a particle at their centers, so the distances from the centers to the pivot point are $0$ for the central part and $\frac{2}{3}R$ for the remaining parts. As a result, an approximation could be

$$
\frac{1}{7}M\cdot 0^2 + \frac{1}{7}M\cdot \left(\frac{2}{3}R\right)^2\cdot 6 = \frac{24}{63}MR^2.
$$

For the precise value, the density is $\frac{M}{\pi R^2}$ since the area of the disc is $\pi R^2$. Thus,

$$
\begin{aligned}
I &= \int \frac{M}{\pi R^2} \cdot r^2 \, dA \\
&= \frac{M}{\pi R^2}\int r^2 \cdot r\, drd\theta \\
&= \frac{M}{\pi R^2} \cdot \frac{r^4}{4} \big|_{r=0}^R \big| _{\theta=0}^{2\pi} \\
&= \frac{1}{2} MR^2.
\end{aligned}
$$

Note that here $M$ is the total weight and $R$ is the radius of the cylinder, while $r$ is a variable indicating the distance from the particle to the pivot point.

The answers for the handout are as follows.

| object | approximation | precise value |
| -------- | -------- | -------- |
| solid cylinder | 24/63 | 1/2 |
| hol cylinder | 1 | 1 |
| solid cylinder | 4/(9 + 6sqrt(2)) | 2/5 |
| solid cylinder | 2/3 | 2/3 |

The number listed above are the ratio only, while the rotational inertia is the ratio times $MR^2$.

## More questions to think about

1. For other objects, approximate and calculate $I$.

## Resources

1. [Wikipedia: List of moments of inertia](https://en.wikipedia.org/wiki/List_of_moments_of_inertia)