Answer:
Suppose
mass per unit area of circular plate = m.
\[\therefore
\] Total mass of circular plate,
\[M=m\times
\pi {{\left( d/2 \right)}^{2}}=\pi m{{\left( \frac{56}{2} \right)}^{2}}=784\pi
m\]
This
is supposed to be concentrated at the centre O of the disc,
Mass
of cut off portion, \[{{M}_{1}}=m\times \pi {{\left( \frac{{{d}_{1}}}{2}
\right)}^{2}}=\pi m{{\left( 42/2 \right)}^{2}}=441\pi m\]
This
is supposed to be concentrated at \[{{O}_{1}}\], where \[{{O}_{1}}O=AO-A{{O}_{1}}=28-21=7cm\].
Mass
of remaining portion of disc, \[{{M}_{2}}=M-{{M}_{1}}=784\pi m-441\pi
m=343\,\pi m\,\]
Let
it be concentrated at \[{{O}_{2}}\] where \[O{{O}_{2}}=x\].
As \[{{x}_{cm}}=\frac{{{M}_{1}}{{x}_{1}}+{{M}_{2}}{{x}_{2}}}{{{M}_{1}}+{{M}_{2}}}\] and
\[{{x}_{cm}}=0(at\,O)\] \[\therefore
\] \[{{M}_{1}}{{x}_{1}}+{{M}_{2}}{{x}_{2}}=0\]
\[441\pi
m\times 7+343\pi mx=0\] or \[x=\frac{-441\pi m\times
7}{343\,\pi m}=-9cm\]
Hence,
centre of mass of remaining portion of disc is at 9 cm from centre of disc on
the right side of 0.
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