Question

From a uniform circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed as . Calculate the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the centre of the disc.

 

Answer 1

Parallel Axis Theorem

The moment of inertia of an object about an axis through its centre of mass is the minimum moment of inertia for an axis in that direction in space. The moment of inertia about an axis parallel to that axis through the centre of mass is given by,

I= Icm + (md)^2 

 

The moment of inertia of removed part abut the axis passing through the centre of mass and perpendicular to the plane of the disc = Icm + md^2

I disc =1/2m{r}^2

= [m × (R/3)^2]/2 + m × [4R^2/9] = m{R}^2/2

 

Therefore, the moment of inertia of the remaining portion = moment of inertia of the complete disc – moment of inertia of the removed portion

 

= 9m(R)^2/2 – mR(^2)/2 = 8m(R^2}/2

 

Therefore, the moment of inertia of the remaining portion (I remaining) = 4m(R)^2.

 Formula used 

I= Icm + md^2

Icm - centre of mass at disc centre 

d- distance from disc centre