COMEDK · Physics · 7. Center of Mass Momentum and Collision
From a circular disc of radius \(R\), a square is cut out with a radius as its diagonal. The centre of mass of remaining portion is at a distance (from the centre)
- A \(\dfrac{R}{(4 \pi-2)}\)
- B \(\dfrac{R}{2 \pi}\)
- C \(\dfrac{R}{\pi-2}\)
- D \(\dfrac{R}{2 \pi-2}\)
Answer & Solution
Correct Answer
(A) \(\dfrac{R}{(4 \pi-2)}\)
Step-by-step Solution
Detailed explanation
Let the radius of the circular disc be \(R\). The area of the disc is \(A_1 = \pi R^2\). The centre of mass of the disc is at the origin \((0, 0)\). A square is cut out such that its diagonal is equal to the radius \(R\). Let the side of the square be \(a\). Then,…
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