From a uniform disk of radius R, a circular hole of radiusR/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.

From a uniform disk of radius R, a circular hole of radiusR/2 is cut o

1.	From a uniform disk of radius R, a circular hole of radiusR/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.
Answer» EXPLANATION: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.

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From a uniform disk of radius R, a circular hole of radiusR/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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From a uniform disk of radius R, a circular hole of radiusR/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.