A circular plate $A$ of radius $1.5 \mathrm{r}$ is removed from one edge of a uniform circular plate $B$ of…
A circular plate $A$ of radius $1.5 \mathrm{r}$ is removed from one edge of a uniform circular plate $B$ of radius $2 r$. The distance of centre of mass of the remaining portion from the centre of the plate $B$ is
$\frac{5 \mathrm{r}}{12}$
$\frac{9 \mathrm{r}}{14}$
$\frac{3 r}{4}$
$\frac{7 r}{8}$
Solution
Radius of removed circular plate, $\mathrm{A}=1.5 \mathrm{r}$
Radius of circular plate, $\mathrm{B}=2 \mathrm{r}$
Let mass of circular plate $A, M_1=M$
So surface mass density, $\sigma=\frac{M}{\pi(2 r)^2}=\frac{M}{4 \pi r^2}$
Then mass of removed circular plate, A
$\begin{aligned} & \mathrm{M}_2=\sigma \mathrm{A} \\ & =\frac{\mathrm{M}}{4 \pi \mathrm{r}^2} \times \pi(1.5 \mathrm{r})^2 \\ & =\frac{\mathrm{M}}{4} \times 1.5 \times 1.5=0.5625 \mathrm{M}\end{aligned}$
The distance of centre of mass of the remaining portion from the centre of the plate $B$ is
$\begin{aligned} & X_{c m}=\frac{M_1 x_1-M_2 x_2}{M_1-M_2} \\ & =\frac{M \times 0-0.5625 M \times 0.5 g}{M-0.5625 M}=\frac{9 r}{14}\end{aligned}$