Question

Q14. From a uniform disc of radius R and mass 9 M, a small disc of radius R/3 is removed. What is the moment of inertia of remaining disc about an axis passing through the centre of disc and perpendicular to its plane?​

Answer 1

Question:From a uniform disc of radius R and mass 9 M, a small disc of radius R/3 is removed. What is the moment of inertia of remaining disc about an axis passing through the centre of disc and perpendicular to its plane?​

STEP BY STEP Explanation:  4MR2

As we know that the moment of inertia of complete disc about a perpendicular axis passing through centre O is given as-

I=21​MR2

Whereas M and R are the mass and radius of the disc respectively.

Given that the mass of disc is 9M and radius is R.

Therefore,

I1​=21​×9M×R2=29​MR2

Now, mass of cut out disc of radius 3R​-

m=πR29M​×π(3R​)2

⇒m=M

Now using the theorem of parallel axis, the moment of inertia of cut out disc about the perpendicular axis passing through cenre O is,

I2​=21​×M×(3R​)2M(32R​)2

⇒I2​=21​MR2

Therefore, the moment of inertia of residue disc is-

I=I1​−I2​

⇒I=29​MR2−21​MR2

⇒I=4MR2

Hence the moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc passing through O is 4MR2.