moment of inertia - 3

[logistic_guy]

Find the moment of inertia $\displaystyle I$ of a uniform cylinder of radius $\displaystyle R$, and mass $\displaystyle M$ if the rotation axis tangent to its edge and parallel to its central axis.

 

[logistic_guy]

This is a tough one but we have a beautiful theorem called the Parallel-axis Theorem. In short it states that if we know the moment of inertia $\displaystyle (I_{z})$ of a body through its center, then the moment of inertia through an axis parallel to the body's axis is:

$\displaystyle I = I_{z} + Mh^2$

where $\displaystyle h$ is the distance between the two axes.

An axis tangent to the edge of the cylinder has a distance $\displaystyle h = R$ to the center of the cylinder. And from a previous post we know that the moment of inertia of a uniform cylinder through its center is $\displaystyle I_z = \frac{1}{2}MR^2$.

Then, the answer to this problem is:

$\displaystyle I = I_{z} + Mh^2 = \frac{1}{2}MR^2 + MR^2 = \textcolor{blue}{\frac{3}{2}MR^2}$