The disk of radius 5cm is cut from the uniform disc of radius 10cm in such a way that the edge of the hole touches the edge of the disc. What is the distance of the center of mass of the remaining portion w.r.t the point O ?
A
2.66cm
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B
6cm
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C
1.66cm
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D
2cm
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Solution
The correct option is C1.66cm
Let ρ be the mass density of the disk. Let m1 be the mass of circular disk of radius 10cm and m2 be the mass of circular disk of radius 5cm. Take O to be the origin for reference. So, x1=0 (COM of circular disk of radius 10 cm) x2=5(COM of removed disk).
Now, XCOM=m1x1−m2x2m1−m2 (we treat the removed portion of disc as having negative mass) where m1=ρ(π(10)2)=100πρ m2=ρ(π(5)2)=25πρ XCOM=100πρ×(0)−25πρ×(5)100πρ−25πρ=−25×5100−25 =−1.66 Due to symmetry, YCOM will be zero.
Therefore, (XCOM,YCOM)=(−1.66,0). So, COM of remaining portion will be shift towards left side on the line joining the points 'O' and 'C'.