Question

A uniform square sheet has a side length of $2R$. A circular sheet of maximum possible area is removed from one of the quadrants of the square sheet. The centre of mass of the remaining portion from the centre of the original sheet is at a distance of :

• A. $\frac{\pi R}{\sqrt{2}[16-\pi ]}$
• B. $\frac{\pi }{[16-\pi ]}$
• C. $\frac{R}{\pi [16-\pi ]}$
• D. $\frac{R\pi }{16-\pi }$

Answer 1

The correct option is A $\frac{\pi R}{\sqrt{2}[16-\pi ]}$

Let $m_1$ be the mass of circular sheet of radius $R$ and $m_2$ be the mass of the remaining portion of square sheet after removing the circular sheet.

Side of square $=2R$

Radius of circle= $\frac{R}{2}$

Let $x_1$ and $x_2$ be the x components of centre of masses of $m_1$ and $m_2$ respectively.

Let $y_1$ and $y_2$ be the y components of centre of masses of $m_1$ and $m_2$ respectively.

$x_1$ = $\frac{R}{2}$

$y_1$ = $\frac{R}{2}$

$x_2=x$

$y_2=y$

Let the centre of square sheet be at origin.

Let $m$ be the mass per unit area.

$m_1=m\pi \frac{R^2}{2^2}$

$m_2=m$($4R^2-\pi \frac{R^2}{2^2}$)

Centre of mass of $m_1$ and $m_2$ is at origin.

0=$\pi \frac{R^2}{2^2}\frac{R}{2}+(4R^2-\pi \frac{R^2}{2^2})x$

$x$ =-$\pi \frac{R}{2(16-\pi )}$

Similarly $y=-\pi \frac{R}{2(16-\pi )}$

Centre of mass of $m_2$ is at ($\pi \frac{R}{2(16-\pi )}$,$\pi \frac{R}{2(16-\pi )}$)

Distance of it from origin is $\pi \frac{R}{\sqrt{2}\times (16-\pi )}$