A uniform rectangular thin sheet $$ABCD$$ of mass $$M$$ has length $$a$$ and breadth $$b$$, as shown in the figure. If the shaded portion $$HBGO$$ is cut-off, the coordinates of the centre of mass of the remaining portion will be:

To find the center of mass of the remaining portion after removing the shaded rectangle HBGO, we treat the removed section as a negative mass.

Let the total mass of the original sheet ABCD be M.

• Original Area: $$A = a \times b$$
• Removed Area (HBGO): $$A_{rem} = \frac{a}{2} \times \frac{b}{2} = \frac{A}{4}$$
• Removed Mass: $$m = \frac{M}{4}$$

Center of Mass Coordinates

The center of mass (x, y) of a composite system is given by:

$$x_{cm} = \frac{M X - m x_1}{M - m}$$

1. Calculation for x-coordinate:

• Center of original sheet (X): $$a/2$$
• Center of removed portion ($$x_1$$): $$3a/4$$
• Remaining mass: $$M - M/4 = 3M/4$$

$$x = \frac{M \left(\frac{a}{2}\right) - \frac{M}{4} \left(\frac{3a}{4}\right)}{\frac{3M}{4}}$$

$$x = \frac{\frac{a}{2} - \frac{3a}{16}}{\frac{3}{4}} = \frac{\frac{5a}{16}}{\frac{3}{4}}$$

$$x = \frac{5a}{12}$$

2. Calculation for y-coordinate:

Due to symmetry between the x and y dimensions:

• Center of original sheet (Y): $$b/2$$
• Center of removed portion ($$y_1$$): $$3b/4$$

$$y = \frac{M \left(\frac{b}{2}\right) - \frac{M}{4} \left(\frac{3b}{4}\right)}{\frac{3M}{4}}$$

$$y = \frac{5b}{12}$$

Final Result:

The coordinates of the centre of mass of the remaining portion are:

$$\boxed{\left( \frac{5a}{12}, \frac{5b}{12} \right)}$$