Let M and R be the mass and radius of a disc. A small disc of radius R/3 is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis passing through the centre O and perpendicular to the plane of disc is 4xMR2. The value of x is . [figure]
Answer
9
Solution
Mass per unit area = M R2
Mass of the removed disc m = M R2 ( R3 )2 = M9
Its centre O′ is at distance d = R - R3 = 2R3 from O.
M.I. of the removed disc about the axis through O (parallel axes theorem):
I2 = 12m ( R3 )2 + m ( 2R3 )2
I2 = M9 ( R218 + 4R29 ) = M9 R22 = MR218
M.I. of the complete disc: I1 = 12MR2
I = I1 - I2 = MR22 - MR218
I = 818MR2 = 49MR2
x = 9
Answer: 9