Four holes of radius R are cut from a thin square plate of the side 4R and mass M in XY plane as shown. Then moment of inertia of the remaining portion about z-axis is=? Plz explain clearly!!

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Answer 1 · 2018-04-12T13:24:18

#(64 - 15 pi)/24 MR^2 #

For the #4R times 4R# square plate, I get #I_(p: zz) = 8/3 MR^2# using a standard formula for a rectangular plate:

That's for rotation around an axis perpendicular to it's centre.

The plate has density #sigma = M/(16R^2)# per unit area.

A single disc, therefore, has mass:

#m_d = sigma pi R^2 = pi/16 M#

A disc's inertia about it's centre is therefore:

#I_(d: cm) = 1/2 m_d R^2 = pi/32 M R^2#

Using the parallel axis theorem (see drawing):

#I_(d:zz)=I_(d:cm)+m_d d^2#

#=pi/32 M R^2+ pi/16 M (sqrt 2 R)^2 = (5 pi)/32 MR^2#

So the inertia of the hollowed out plate is:

#I_(n et : zz) = I_(p: zz) - 4I_(d:zz) = (64 - 15 pi)/24 MR^2 #