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What is the moment of inertia?

Energy is necessary for the moving of stationary objects. Likewise, when a stationary object rotates, it also needs energy. To obtain that energy, we need a physical quantity that represents the difficulty of rotating the object relative to the axis of rotation. To obtain that energy, we need a physical quantity that represents the difficulty of rotating the object relative to the axis of rotation. This difficulty of rotation is called the moment of inertia. While the physical quantity representing the difficulty of moving the object with respect to the force is “mass”, the rotating version thereof is “moment of inertia”.

This moment of inertia $I_ {j}$ is to be written

$$I_j=m_{j}r_{j}^{2}$$

where the distance from the mass point $m_ {j}$ to the rotation axis stands for $r_ {j}$.

All objects can be regarded as a mass of mass. Therefore, we can obtain the moment of inertia $I$ of any object by integrating the above expression as the integral range of the whole object.

Furthermore, by using this moment of inertia $I$ and the angular velocity $ω$ of the rigid body, it is possible to express the kinetic energy $K$ of rotation of the object.

$$K=\frac{1}{2}I\omega^{2}$$

You can remember this formula easily by comparing the above equation about $K$ with the below equation about the kinetic energy E that you know.

$$E=\frac{1}{2}mv^{2}$$

Practice

In order to obtain the kinetic energy due to the rotational motion of an object, it is necessary to obtain the moment of inertia of the object firstly. Let’s solve the following practice. Through it, I would like you to become accustomed to integrating the whole object as an integral range in order to calculate the moment of inertia.

The Moment of Inertia of a disk

Practice: Find the moment of inertia of the disk with radius $R$ by areal density $σ$. It is assumed that the axis of rotation passes perpendicularly through the center of the disk.

How to Integrate of a disk

First of all, find the mass $M$ of this disc by using areal density $σ$. To find it, we should integrate areal density $σ$.

The area between the circle of the radius \ (r \) around the axis of rotation of the disk and the circle of the radius $r + dr$ with the same center is obtained if you multiply the very small radius $dr$ and the circumference of the circle $2πr$. When the circle of radius $r$ slightly inflates, its inflated part is the area between the circles of radius $r$ and $r + dr$. Therefore the area of the disk can be obtained by integrating this $2πr · dr$ with respect to $r$ from 0 to R. (Areal density) × (area) = (mass), so  specifically the mass of the disc $M$ is obtained as follows.

$$M=\int_0^R σ･2πrdr=2πσ\int_0^R r dr=πσR^2$$

As mentioned earlier, the moment of inertia \ (I \) can be obtained by integrating both sides of \(I_j=m_ {j}r_ {j}^{2}=r^2·2πσrdr)\ .

$$I=\int_0^R r^2･2πσrdr=\frac{πσ}{2}R^4=\frac{M}{2}R^2$$