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 $(\frac{x}{3}\frac{R}{\pi },\frac{x}{3}\frac{R}{\pi })$, then, $x$ is _______.

$($Round off to the nearest integer$)$
$(R$ is the radius of disc$)$

Answer 1

We know that the centre of mass of the quarter disc is at the position $(\frac{4}{3}\frac{R}{\pi },\frac{4}{3}\frac{R}{\pi })$.

So, $x=4$