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From  a uniform disc of  radius R a small disc of radius R/3 is cut . The mass of the remaining portion is M . find the moment of inrtia of such a disc about the axis passing through the geometrical centre of original disc and perpendicular to the the plane of the disc .
10 months ago

Area of the remaining disc = [R2 – (R2/9)]
A = (8R2/9)
Mass of the remaing disc = M
Mass per unit area of the material =  = 9M/(8R2)
Mass of the part remaining = (R2/9) = M/8

So let us assume a M/8 and -M/8 plate on the part from where we removed the small disc.
Now we have a disc of mass 9M/8 with radius R and another disc of mass -M/8 with radius R/3.

So Moment of Inertia of larger disc about its axis =9MR2/16
Now let us assume the distance between the centers of the two disc be d.
Now, moment of inertia of smaller disc about its axis = -MR2/144
Moment of inertia of smaller disc about center of original disc = (-MR2/144) + (-Md2/8)

So, total moment of inertia about the centre of original disc = (9MR2/16) – (MR2/144) – (Md2/8)
I = (5MR2/9) – (Md2/8)
10 months ago
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