The disc of mass M with uniform surface mass density σ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position x/3a/π,x/3a/π where x is _____________. (Round off to the Nearest Integer).
[a is an area as shown in the figure]

Solution

Physics sir · 4 steps

Slice the quarter disc into thin triangular wedges; each wedge's centre of mass sits at 2a/3 along its radius.

1. Read the figure. Here a is the radius of the disc. The shaded quarter runs from θ = 0 to θ = π/2, and its mass is σ × (π a²)/4.
2. Pick a wedge. A thin wedge of angle dθ is almost a triangle with its tip at the centre. Its mass is dm = σ × 1/2a² dθ. Its centre of mass lies on the wedge's radius at 2a/3 from the centre, so its x-coordinate is 2a/3 cos θ.

   x_(cm) = (∫₀^(π/2) 2a/3 cos θ · (σ a²)/2 dθ)/((σ π a²)/4) = 2a/3 × 2/π∫₀^(π/2)cos θ dθ = 4a/3π

3. Use symmetry. The quarter disc is symmetric about the line y = x, so y_(cm) = x_(cm) = 4a/3π. Matching with x/3a/π gives x = 4.
4. So. The centre of mass is at (4a/3π, 4a/3π), so x = 4.

The common slip: 6. Using the quarter-ring result 2a/π instead of the quarter-disc result gives 6a/3π, that is 6, not 4.

The idea to keep. For any disc sector, use thin triangular wedges with their centre of mass at 2R/3. A quarter disc gives 4R/(3π) from each straight edge; a quarter ring gives 2R/π.

Answer from the NTA final answer key. Solution by PrepWiser.