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Question

Three rings each of mass M and radius R are arranged as shown. The moment of inertia of the system about YY' will be


A
3MR2
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B
32MR2
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C
5MR2
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D
72MR2
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Solution

The correct option is D 72MR2

Total moment of inertia (Itotal)=I1+I2+I3

For ring-1:
Moment of inertia about yy, Iyy=MR22

For ring-2:
I2=IAA+Iyy=MR22+MR2=32MR2

For ring 3:
ring 3 is symmectrical as ring 2,
so,
I3=MR22+MR2=32MR2

Hence, Itotal=I1+I2+I3
Itotal=MR22+32MR2+32MR2
Itotal=7MR22

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