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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.

shvd
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  • Your explanation is somewhat confusing. For a start, the opening sentence contains errors. The key issue here is that the $M$ in the final answer is the mass of the disc with the hole in it. You need to start with finding an expression for the density of the material and then obtain expressions for the MI of the whole disc and the hole – David Quinn Jan 25 '23 at 19:03
  • Sorry about the error, somebody edited to latex and wrote - instead of + and i didn't check it appropriately . Anyhow thanks for the comment! I'll try to solve it by adjusting my expressions for the mass. – shvd Jan 25 '23 at 19:18

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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?

David Quinn
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  • When i use these values for the mass in my calculations for the moment of inertia i get (M/2)(R^4-r^4-2d^2*R^2)/(R^2-r^2) which already includes the R^2-r^2 but is still wrong. – shvd Jan 26 '23 at 01:13
  • You can simplify your result to get the correct expression by using $R^4-r^4=(R^2-r^2)(R^2+r^2)$ – David Quinn Jan 26 '23 at 10:06