Questions
From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about perpendicular axis, passing through centre?
detailed solution
Correct option is C
ITotal disc = MR22MRemoved = M4(Mass∝area )Iremoved (about the same perpendicular axis)= M4(R2)22+ M4(R2)2 = 3MR232 Removed disc = Itotal- Iremoved = MR22-332MR2 = 1332MR2Talk to our academic expert!
Similar Questions
One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing
through the centre of the original disc. Its moment of inertia about the axis of rotation is
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