# Solve the following :

Question:

A square plate of edge $d$ and a circular disc of diameter $d$ are placed touching each other at the midpoint of an edge of the plate as shown in figure. Locate the center of mass of the combination, assuming same mass per unit area for the two plates.

Solution:

Area density of square $=\rho_{s}=\frac{m_{S}}{d^{2}}$

(Mass/unit area)

Mass/unit area of circle $=\rho_{C}=\frac{4 m_{C}}{\pi d^{2}}$

$\rho_{s}=\rho_{C} \Rightarrow \frac{m_{s}}{d^{2}}=\frac{4 m_{C}}{\pi d^{2}}$

$\Rightarrow m_{C}=\frac{\pi}{4} m_{s}$

C.O.M of square $=\left(\frac{d}{2}, 0\right.$

C.O.M of circle $=\left(\frac{3 d}{2}, 0\right)$

$=\frac{x_{c} m_{C}+x_{S} m_{S}}{m_{C}+m_{S}}$

$\frac{\frac{d \pi}{2} m_{S}+\frac{3 d}{m} m_{S}}{\left(\frac{\pi}{4}+1\right) m_{S}}$

$=\frac{d}{2} \cdot \frac{\left(\frac{\pi}{4}+3\right)}{\frac{\pi}{4}+1}=\frac{d}{2}\left(\frac{\pi+12}{\pi+4}\right)$

From C.O.M of disc it is

$\frac{d}{2}\left(\frac{\pi+12}{\pi+4}\right)-\frac{d}{2}=\frac{d}{2}\left(\frac{\pi+12-\pi-4}{\pi+4}\right)$

$=\frac{\mathrm{d}}{2} \cdot \frac{8}{\pi+4}=\frac{\mathrm{d} .4}{\pi+4}$ distance away.