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 cm2
As mass is proportional to area
Mass of cutout portion/mass of remaining portion = m1/m2 = 441 π/343 π= 9/7
Let C2 be centre of mass of remaining portion and C1 be centre of mass of cutout portion.
O is centre of mass of the whole disc.
OC1 = r1 = 28-21= 7 cm.
OC2 = r2 = ?
Equating moments of masses about O, we get
m2× r2 = m1× r1
r2 = 9