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

Given that:

Mass of dics =  9M 

Mass of removed part m=​///πR9M​2×π(R/3)3=M

moment of inertia of the disc without hole about axis perpendicular to plane through center will be:-

I1​=29MR2​

Moment if inertia of hole  about axis passing through center of disc will be:-

I2​=2M(3R​)2​+M(32R​)2

The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc will thus be:-

I=I1​−I2​

I=29MR2​−⎡​2M(3R​)2​+M(32R​)​2=4MR2

Answer =4MR2

 

The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc will thus be:-

answer= 4MR^2