Rotational MotionHard
Question
From a circular disc of radius R and mass 9 M, a small disc of mass M and radius 
is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is
is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is Options
A.
B.MR2
C.4MR2
D.
Solution
Mass of the disc = 9 M
Mass of removed portion of disc = M
The moment of inertia of the complete disc about an
axis passing through its centre O and perpendicular
to its plane is I1 =
MR2Now, the moment of inertia of the disc with removed portion
Therefore, moment of inertia of remaing portion of the disc about O is
I = I1 - I2
= 9
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