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Question

From a uniform disc of radius R and mass 9M, four small discs of radius R3 are removed as shown in figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane and passing through the centre of the disc is nMR2. Then, n is



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Solution

Given, mass of complete disc =9M
and radius =R
Let σ be the mass per unit area.
9M=σ×πR2=σπR2

The mass of each small disc
=σ×π(R3)2=σπR29=M

Now, let us consider the above system as a complete disc of mass 9M and 4 discs of negative mass M super imposed on it.
I1= Moment of inertia of bigger disc along axis perpendicular to the plane passing through centre.
I2= MOI of each smaller disc along axis perpendicular to the plane passing through centre.
Then, I1=12(9M)R2=9MR22
I2=12(M)(R3)2+M(2R3)2
=MR218+4MR29=9MR218=MR22

MOI of remaining disc about the axis =I14I2
=92MR22MR2=2.5MR2

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