From a uniform disc of radius R, a small disc of radius R2 has been cut out from the left having centre at (−R2,0) and is placed on the right with centre at (R2,0) as shown in the figure. Find the centre of mass of the resulting system.
A
xcom=R2
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B
xcom=R4
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C
ycom=0
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D
ycom=−R2
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Solution
The correct options are Bxcom=R4 Cycom=0
Let M be the total mass of disc. Mass of the cut out portion (m), m=MπR2.π(R2)2 ⇒m=M4
After the smaller disc has been cut from the original and is placed on the right of the centre of the disc, the remaining portion is considered to be a system of three masses. The three masses are: M concentrated at (0,0), −m(=M/4) concentrated at (−R2,0) and m(=M/4) concentrated at (R2,0)
COM of the new system: xcom=Mx1−mx2+mx3M−m+m[∵x1=0,x2=−R2,x3=R2] xcom=M(0)−M4(−R2)+M4(R2)M=R4,ycom=0 [by symmetry]