There is a thin uniform disc of radius R and mass per unit area σ , in which a hole of radius R / 2 has been…

There is a thin uniform disc of radius R and mass per unit area σ, in which a hole of radius R/2 has been cut out as shown in the figure. Inside the hole, a square plate of same mass per unit area σ is inserted so that its corners touch the periphery of the hole. The distance of the centre of mass of the system from the origin is

  1. R(2-π)23π+2
  2. R(1-π)22π+1
  3. 2 R π 2 3 π + 2
  4. 3 R π 2 2 π + 1

Solution

Side of square = R cos 4 5 ∘ = R 2

Area of square = R 2 2

XCOM=π×R2×σ×0+π×R24-σ×R2+R22×σ×R2π×R2×σ+π×R24-σ+R22×σ

=R(2-π)2(3π+2)

∴  The centre of mass of the system is at a distance of R(2-π)2(3π+2) from the centre O towards the plate as shown in the figure

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