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Question

A circular plate of uniform density has a diameter of 56 cm. A circular portion of diameter 42 cm is removed from the original circular plate as shown in figure. Find the position of the COM of the remaining portion, from the centre of the original plate.


A
7 cm towards left
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B
7 cm towards right
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C
9 cm towards left
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D
9 cm towards right
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Solution

The correct option is C 9 cm towards left
Let M be the mass of parent/original circular plate with radius 28 cm
M1 be the mass of removed plate

Let mass per unit area be m, then
M=m.π(28×102)2 kg
M1=m×π(21×102)2 kg
MM1=(2821)2=72×4272×32=169 ...(1)

Replacing the removed mass by ve mass and assuming the masses to be at their respective centres of mass, COM of remaining part is given by:
xcom=Mx1M1x2MM1

Here:
Massx-coordinate of COM from origin OParentM0RemovedM1+7 cm


O origin taken at centre of original plate

From Eq. (i) and (ii):
xcom=M(0)(M1)(7)MM1
xcom=(M1)(7)169M1M1
xcom=7M17M19
xcom=9 cm

Due to symmetry along xaxis, COM of remaining plate will lie along xaxis at distance 9 cm towards left from the centre of original plate.

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