Answer:
Suppose
the circular disc of radius a with centre O is made up of
(i)
circular section of radius b with centre \[{{O}_{1}}\] and (ii) remaining
portion of disc with c.m at \[{{O}_{2}}\].
Taking
\[O\] as origin, and \[{{O}_{1}}\], \[{{O}_{2}}\] on X-axis, (y = 0, z = 0),
Fig. 5(HT).4, the position of c.m of disc is given by
\[{{x}_{cm}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}\] ??
(i)
If \[\sigma
\] is surface density of material of the disc,
\[{{m}_{1}}=\pi
{{b}^{2}}\sigma \], \[{{x}_{1}}=c\]
\[{{m}_{2}}=\pi
\left( {{a}^{2}}-{{b}^{2}} \right)\sigma \]
\[{{x}_{2}}=?\]
\[{{m}_{1}}+{{m}_{2}}=\pi
{{a}^{2}}\sigma \]
\[{{x}_{cm}}=0\]
From
(i), \[0=\frac{\pi {{b}^{2}}\sigma \times c+\pi \left(
{{a}^{2}}-{{b}^{2}} \right)\sigma \times {{x}_{2}}}{\pi {{a}^{2}}\sigma }\]
\[\therefore
\] \[{{x}_{2}}=\frac{-c{{b}^{2}}}{\left(
{{a}^{2}}-{{b}^{2}} \right)}\]
Hence c.m
of rest of the portion of the disc lies on the left of O, at a distance \[{{x}_{2}}\]
as shown in the Fig. 5(HT).4.
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