Answer:
The situation is shown in Fig. 7.104. Let O be the CM of the
original circular portion, that
of the circular e hole cut out and that
of the remaining shaded portion.
Let m be the mass per unit area of the
disc.
Fig. 7.104
Mass of the original disc,
Mass of the circular hole cut,
Mass of the remaining portion,
Masses and may be assumed
to be concentrated at and respectively
and 0 is their CM.
Moment of about O
= Moment of about O
or
or
or
Thus the CM of the resulting portion
lies at R/6 from the centre of the original disc in a direction opposite to the
centre of the cut out portion.
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