Calculate the moment of inertia of a disk of radius $R$ with a hole of radius $r$ and center a distance $d$ from the original disk.

Question

I calculated the inertia using $I_{\rm tot}= I_d+I_h$ where $I_{tot}$ is the inertia of the disk without a hole $\left(\frac12 MR^2\right)$ in reference to its center of mass (origin), $I_d$ is the inertia of the disk with a hole (what we solve for), $I_o$ is the inertia of the hole in reference to the axis used in $I_{tot}$. I first set up an equation for $I_o$ around its own center of mass using $\frac12 mr^2$ where i expressed $m$ in terms of the area of the disk and the hole. Then i used Steiner to get $I_o$ in terms of the original axis of rotation. After which I subtracted $I_o$ from $I_{tot}$ to get $I_d= \frac M2\left(R^2-\frac{r^4}{R^2} - \frac{2r^2d^2}{R^2}\right)$. According to my resource the answer should be $I_d= \frac M2\left(R^2+r^2-\frac{2r^2d^2}{R^2-r^2}\right)$ but I have no clue as to where I went wrong.

Answer 1

Hint…If the mass of the disc with the hole in it is $M$, then the density of the material is $$\rho=\frac{M}{\pi(R^2-r^2)}$$

So the mass of the whole disc before the hole is removed is $$\rho\pi R^2=\frac{MR^2}{(R^2-r^2)}$$ and the mass of the hole itself is $$\frac{Mr^2}{(R^2-r^2)}$$

Can you finish?