From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
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See the figure ,
OO1 = R/2
Let mass/area of disc = m
So, mass of disc(M) = area of disc × mass per unit area
= πR²m
Mass of the portion removed form the disc ( m') = area of removing part × mass per unit area
= π(R/2)²m = mr²/4 = M/4
If the center of mass of remaining part from the centre O is at a distance D ,
D ={ M×0 - m'×(R/2)}/(M-m')
= {-M/4×R/2}/(M-M/4)
= -R/6
Hence, centre of mass of the remaining part at R/6 to the left of centre O.
OO1 = R/2
Let mass/area of disc = m
So, mass of disc(M) = area of disc × mass per unit area
= πR²m
Mass of the portion removed form the disc ( m') = area of removing part × mass per unit area
= π(R/2)²m = mr²/4 = M/4
If the center of mass of remaining part from the centre O is at a distance D ,
D ={ M×0 - m'×(R/2)}/(M-m')
= {-M/4×R/2}/(M-M/4)
= -R/6
Hence, centre of mass of the remaining part at R/6 to the left of centre O.
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