Question

The disc of mass M with uniform surface mass density $\sigma$ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position $\frac{x a}{3\pi},\frac{x a}{3\pi}$, where x is _______.(Round off to Nearest Integer) [a is an area as shown in the figure]

Answer 1

4

Answer 2

The center of mass for a uniform quarter disc of radius 'a' located in the first quadrant is known to be at $(\frac{4a}{3\pi}, \frac{4a}{3\pi})$. By comparing this standard formula with the given coordinates of the center of mass $(\frac{x a}{3\pi}, \frac{x a}{3\pi})$, we directly find that $x=4$.