Question

A disc (of radius $rcm$ ) of uniform thickness and uniform density $\sigma$ has a square hole with sides of length $l=\frac{r}{\sqrt{2}}cm.$ One corner of the hole is located at the centre of the disc and centre of the hole lies on $y$ -axis as shown. Then the $y$ -coordinate of position of centre of mass of disc with hole (in $cm$ ) is

• A. $-\frac{r}{2(\pi -1/4)}$
• B. $-\frac{r}{4(\pi -1/4)}$
• C. $-\frac{r}{4(\pi -1/2)}$
• D. $-\frac{3r}{4(\pi -1/4)}$

Answer 1

Answer: (C)

This disc can be assumed to be made of a complete uniform disc and a square plate with same negative mass density
$Y_{cm}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}=\frac{\left(\pi r^2\right)\sigma (0)+{\ell }^2(-\sigma )(r/2)}{\pi r^2\sigma +{\ell }^2(-\sigma )}$
$=\frac{-{\ell }^{2}r}{2\left(\pi r^2-{\ell }^2\right)}=\frac{-\frac{r^3}{2}}{2\left(\pi r^2-\frac{r^2}{2}\right)}=\frac{-r}{4\left(\pi -\frac{1}{2}\right)}$