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Question

A circular hole of radius R/2 is cut from a circular disc of radius R. The disc lies in the xy-plane and its centre coincides with the origin. If the remaining mass of the disc is M, then
(a) determine the initial mass of the disc, and
(b) determine its moment of inertia about the z-axis.
981773_bcc2cbaf787c4aeb817e263321610194.png

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Solution

Let σ be the surface mass density of the plate given by σ=MπR2π(R2)2=43MπR2
Now, initial mass of the plate, m1=σ(πR2)=43M
Mass of the plate removed, m2=σ(πR24)=M3
The given arrangement may be considered as a combination of a circular plate of mass (M/3)
The moment of inertia of the big disc of radius R and mass m1 about O and parallel to the z-axis is
Io1=m1R22
The moment of inertia of the small disc of mass m2 and radius R/2 about O and parallel to z-axis
Io1=m2(R/2)22=m2R28
The moment of inertia of the small disc about O (using parallel axis theorem)
Io1=m2R28+m2R24=3m2R28
As the part of the disc of mass m2 is removed from disc m1, hence the net moment of inertia of structure is
Io=Io1Io2=m1R223m2R28
Substituting the values of m1 and m2, we get
I0=1324MR2

1028850_981773_ans_c9b5043371274f32ad510966aa798a4d.png

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