moment of inertia - 4

[logistic_guy]

Find the moment of inertia $\displaystyle I$ of a thin uniform cylinder (coin) of radius $\displaystyle R$, and mass $\displaystyle M$ if the rotation axis is through its center in the plane of the cylinder.

 

[logistic_guy]

The rotation through the center in the plane of the cylinder just means that the rotation is through the $\displaystyle x$-axis or the $\displaystyle y$-axis, NOT through the $\displaystyle z$-axis.

There is a beautiful theorem which is called the Perpendicular-axis Theorem. It states that the sum of the moments of inertia of a body through the plane axes equal to the moment of inertia through its center.

In other words, this theorem says this:

$\displaystyle I_z = I_x + I_y$

We notice that the moment of inertia through the $\displaystyle x$-axis is symmetric to that through the $\displaystyle y$-axis.

Then,

$\displaystyle I_z = I_x + I_y = I_x + I_x = I_y + I_y = 2I_x = 2I_y$

Here is the crucial part. The moment of inertia through the center of a uniform cylinder, a thin uniform cylinder, a uniform disc, a uniform coin, or whatever similar uniform shape is the same as long as the mass is distributed evenly through the circle area.

Therefore, the moment of inertia of a thin uniform cylinder through its center is: $\displaystyle I_z = \frac{1}{2}MR^2$.

If you are confused why this result is true by saying the uniform cylinder has a height $\displaystyle h$ while the thin uniform cylinder doesn't have a height, or it might have a very tiny one (like the coin), you can always prove this kind of results by using the definition of moment of inertia, that is: $\displaystyle I = \int R^2 \ dm$.

Finally, we can answer the op problem.

The moment of inertia through the center in the plane of the cylinder is:

$\displaystyle I_x = I_y = \frac{1}{2}I_z = \frac{1}{2}\frac{1}{2}MR^2 = \textcolor{blue}{\frac{1}{4}MR^2}$