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Example of rotational motion Example of rotational motion - A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter 42 cm is removed from one edge of the plate as shown in fig.1. Find the position of centre of mass of the remaining portion? Solution :   Area of whole plate = π (56/2) 2 = 784 π   sq. cm. Area of cutout portion = π (42/2) 2 = 441 π   sq. cm. Area of remaining portion, Fig 1 = 784 π – 441 π= 343 π cm 2 As mass is proportional to area Mass of cutout portion/mass of remaining portion =   m 1 /m 2 = 441 π/343 π= 9/7 Let C 2 be centre of mass of remaining portion and C 1 be centre of mass of cutout portion. O is centre of mass of the whole disc. OC 1 = r 1 = 28-21= 7 cm. OC 2   = r 2 = ? Equating moments of masses about O, we get m 2 × r 2 = m 1 × r 1 r 2 = 9