NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
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
Options
- AR( 2 - π ) 2 3 π + 2
- BR( 1 - π ) 2 2 π + 1
- C2 R π 2 3 π + 2
- D3 R π 2 2 π + 1
Correct answer
A. R( 2 - π ) 2 3 π + 2
Step-by-step solution
Side of square = R cos 4 5 ∘ = R 2 Area of square = R 2 2 X COM = π × R 2 × σ × 0 + π × R 2 4 - σ × R 2 + R 2 2 × σ × R 2 π × R 2 × σ + π × R 2 4 - σ + R 2 2 × σ = R( 2 - π ) 2 ( 3 π + 2 ) ∴ The centre of mass of the system is at a distance of R( 2 - π ) 2 ( 3 π + 2 ) from the centre O towards the plate as shown in the figure