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?
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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?
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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=21MR2
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=29MR2
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=21MR2
Therefore, the moment of inertia of residue disc is-
I=I1−I2
⇒I=29MR2−21MR2
⇒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.