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 is M R2, so the mass of the removed disc is
m = M R2 ( R3 )2 = M9
Its centre O′ is at distance d = R - R3 = 2R3 from O, so by the parallel axes theorem its M.I. about the axis through O is
I2 = 12m ( R3 )2 + m ( 2R3 )2
I2 = M9 ( R218 + 4R29 ) = M9 R22 = MR218
The M.I. of the complete disc is I1 = 12MR2, so for the remaining part
I = I1 - I2 = MR22 - MR218
I = 818MR2 = 49MR2
x = 9
Answer: 9