Centre of Mass and Collision Part 1

Given, radius of hole, r=a∕2 and radius of disc, R=a
Let xCM be the centre of mass of system,
m1,x1 be the mass and centre of mass of disc
m2,x2 be the mass and centre of mass of circular hole.
∴‌m2=‌

m1

πR2

πr2
⇒‌m2=‌

m1r2

R2

=‌

m1(a∕2)2

a2

=‌

m1

4

and xCM=‌

m1x1+m2x2

m1+m2

∴‌‌xCM=‌

m1a−(m1∕4)(3a∕2)

m1−(m1∕4)

⇒‌‌xCM=‌

m1a(1−3∕8)

3m1∕4

⇒xCM=‌

5a

6