A uniform disc of mass M and radius R has a circular hole of radius R/2 as shown | TestHub

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A uniform disc of mass M and radius R has a circular hole of radius R/2 as shown. About the shown axis, the remaining moment of inertia is $M R^{2} / C$ . Find C to the nearest integer.

num1 diagram

Answer:

2

Solution:

Original disc: I₀=MR²/2.

Removed disc has mass M/4 and radius R/2.

About the shown axis its inertia is 3MR²/32.

Remaining inertia=MR²/2-3MR²/32=13MR²/32.

Thus C=32/13≈2.46, which rounds to 2.

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

Physics - Rotation Question with Solution | TestHub

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