# Thread: [SOLVED] Moment of inertia?

1. ## [SOLVED] Moment of inertia?

A uniform disc has mass M and radius R. A circular hole of radius r is made in the disc with its centre at a distance x from the centre of the disc. Find the moment of inertia of the remaining portion of the disc about an axis passing through the centre of the hole and perpendicular to the plane of the disc.

2. Originally Posted by fardeen_gen
A uniform disc has mass M and radius R. A circular hole of radius r is made in the disc with its centre at a distance x from the centre of the disc. Find the moment of inertia of the remaining portion of the disc about an axis passing through the centre of the hole and perpendicular to the plane of the disc.
Parallel axis theorem plus thinking of hole as negative mass was the trick I used a loong time ago

3. ## Just dotting the i's and crossing the t's

Isomorphism is correct but just thought I'd add the explicit solution for absolute clarity and so this can get a SOLVED on it:

Spoiler:

Let the moment of inertia of about the centre of the large disc:
the large (unpunctured) disc be $I_A$ ;
the small cut out disc be $I_B$;
the punctured large disc be $I_{AB}$.

Then we must have:

$I_{AB} + I_B = I_A$

where

$I_A = \frac{1}{2} M R^2$

and

$I_B = \frac{1}{2}m r^2 + m x^2$ (Parallel axis theorem)

so

$I_{AB} = I_A - I_B = \frac{1}{2}M R^2 - \frac{1}{2} m r^2 - m x^2$

but

$m = \frac{r^2}{R^2}M$ (same density and similarity of shape)

so

$\color[rgb]{0,0,1}\boxed{I_{AB} = \frac{1}{2}\left(R^2 - \frac{r^4}{R^2} - \frac{2 r^2 x^2}{R^2} \right)M}$