I was trying to find the moment of inertia of the disk object the following way
$$I_o = I_d-I_h$$ where $$I_0$$ is the moment of inertia of the disk with a hole and $$I_d$$ is the moment of inertia of the disk without a hole and $$I_h$$ is the moment of inertia of the hole
then to find $$I_h$$ I use the parallel axis theorem such that
$$I_h = I_{cm} + md^2$$
is there any other way to calculate the moment of inertia of the hole? because the problem with finding the moment of inertia of the hole with the parallel axis theorem is that the hole does not intersect with the center of the disk
Best Answer
I for the solid disk is found with the
$$ I_{d} = \frac{1}{2}MR^{2} $$
equation. The same equation is used for the $I_{cm}$ of the hole:
$$ I_{cm} = \frac{1}{2}mr^{2} $$
The offset used is $md^{2}$, where I assume $d$ is given.
The $M$ and $m$ will come from the two areas $A$ and $a$, and the (given?) density. Also, mass equations must be multiplied by a thickness, $t$ (maybe also given?) to get a volume.
The final answer is then:
$$ I_{0} = \frac{1}{2}MR^{2} - \frac{1}{2}mr^{2} - md^{2} $$
To get your book's final answer as requested in the comments below, note that as I was discussing above, the masses come from the volume and the density:
$$ M = \rho\pi R^{2}t $$
$$ m = \rho\pi r^{2}t $$
So that it should be clear that
$$ m = \frac{r^{2}}{R^{2}}M $$
Therefore, substituting out $m$ for this:
$$ I_{0} = \frac{1}{2}MR^{2} - \frac{r^{2}}{R^{2}}M(\frac{1}{2}r^{2}+d^{2}) $$