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Why is the formula for calculating the moment of inertia this integral

$$ \int r^2 dm~? $$ I understand the way we derived this formula from looking at the distribution of kinetic energy of a rotating object. I believe classical mechanics should have intuitive sense, and that everything in every formula has a reason it is the way it is and not something else(I repeat, in classical mechanics). Inertia (rotational) as I understand it, tells us how "hard" is it to rotate an object around a perpendicular axis. I i just don't see why $\int r^2 dm$ tell's us that.

For example the center of mass, similar formula but it makes much more sense (to me) $$ \sum_{i=1}^n m_i r_i, $$ because the more mass the $i$-th component of mass has, the closer the center of mass will be to that particular point, and also the further away a given mass is from our origin the further the center of mass, this is all perfectly logical and every part of the formula makes complete sense but in the case of the moment of inertia I simply cannot see why $ \int r^2 dm$ tells me how hard is it to rotate.

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The way I prefer to think about it is from the point of view that an extended object is a distribution of mass, sort of like a probability distribution in statistics.

A moment is a number (usually a sum or integral) that describes the shape of a distribution. If you have all of the moments, you can completely reconstruct the distribution. The $n^\textrm{th}$ moment of a distribution $\rho(x)$ is calculated by taking

$$\int\rho(x)x^n dx$$

If we make the substitution $\rho(x)dx=dm$, where we take $\rho(x)$ to be the density of an object, we get that the zeroth moment is $\int dm=m$, the total mass of the object. The first moment of a distribution is its mean, so the first moment tells you, in weird units, the "average position" of an object (i.e. the position of its center of mass, scaled by the total mass). The second moment of a distribution, $\int \rho(x)x^2 dx$, tells you essentially how "spread out" the distribution is about its mean. You'll notice that the second moment $\int \rho(x)x^2 dx=\int x^2 dm$ is the moment of inertia, which makes intuitive sense; the more "spread out" the object is, the harder it will be to rotate.

So the moment of inertia got its name because it's the second moment of the mass distribution.

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Intuitively you can think of it this way: Moment of intertia measures how hard is to change the angular momentum of the object.

Now, the angular momentum of a point single point of mass $i$ is:

$$ \vec{L_i} = r_i \times m_i \vec{v_i} $$

To make this point rotate faster, you need to apply some torque. The mass part is easy, and one of the $r_i$ is just there also.

But note the effect of $r_i$ in how the torque affects the velocity: the farther the point is from the center of rotation (greater $r_i$), then the least effect the torque is to the velocity $v_i$. This is because this point needs more linear velocity to get the same angular velocity.

The net effect is that the moment of intertia is proportional to the square of $r_i$: one $r_i$ because that's just how the angular momentum is measured and another one because the linear velocity is affected by this amount.

A more mathematical, but not very strict way to look at it is with the analogy between linear and angular momentum. Note that these equations are for a single particle:

$$ \vec{P} = m \vec{v} $$ $$ \vec{L} = I \vec{\omega} $$

But also, from definition:

$$ \vec{L} = \vec{r} \times m \vec{v} $$ $$ \vec{\omega} = {\vec{r} \times \vec{v} \over |\vec{r}|^2} $$

So

$$ I {\vec{r} \times \vec{v} \over |\vec{r}|^2} = \vec{r} \times m \vec{v} $$

Simplify both $\vec{r} \times \vec{v}$ and move the squared distance:

$$ I = |\vec{r}|^2 m $$

Now that you have the momentum of inertia for a single mass particle, integrate over your whole body and you get your original equation.

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