b) Determine the coordinates of the centroid for the shape given below

The overall shape can be considered as:

• Shape 1: A rectangle with width $250 \text{ mm}$ and height $100 \text{ mm}$.
• Shape 3: A right-angled triangle on top of the rectangle, with base $250 \text{ mm}$ and height $150 \text{ mm} - 100 \text{ mm} = 50 \text{ mm}$. Its vertices are $(0, 100)$, $(250, 100)$, and $(250, 150)$.
• Shape 2: A semi-circle removed from the bottom right, with a diameter of $100 \text{ mm}$ (radius $r = 50 \text{ mm}$). Its diameter lies on the x-axis, centered at $x = 150 \text{ mm} + 50 \text{ mm} = 200 \text{ mm}$.

For Shape 1 (Rectangle):

$$A_1 = 250 \text{ mm} \times 100 \text{ mm} = 25000 \text{ mm}^2$$ $$x_1 = \frac{250 \text{ mm}}{2} = 125 \text{ mm}$$ $$y_1 = \frac{100 \text{ mm}}{2} = 50 \text{ mm}$$

For Shape 3 (Triangle):

$$A_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 250 \text{ mm} \times 50 \text{ mm} = 6250 \text{ mm}^2$$

The centroid of a triangle is the average of its vertices' coordinates.

$$x_3 = \frac{0 + 250 + 250}{3} = \frac{500}{3} \text{ mm} \approx 166.67 \text{ mm}$$ $$y_3 = \frac{100 + 100 + 150}{3} = \frac{350}{3} \text{ mm} \approx 116.67 \text{ mm}$$

For Shape 2 (Semi-circle, removed):

$$A_2 = -\frac{1}{2} \pi r^2 = -\frac{1}{2} \pi (50 \text{ mm})^2 = -1250\pi \text{ mm}^2$$

The centroid of a semi-circle is located at its axis of symmetry. The x-coordinate is the center of its diameter. The y-coordinate is $\frac{4r}{3\pi}$ from the diameter.

$$x_2 = 200 \text{ mm}$$ $$y_2 = \frac{4 \times 50 \text{ mm}}{3\pi} = \frac{200}{3\pi} \text{ mm} \approx 21.22 \text{ mm}$$

For Shape 1:

$$A_1 x_1 = 25000 \text{ mm}^2 \times 125 \text{ mm} = 3125000 \text{ mm}^3$$ $$A_1 y_1 = 25000 \text{ mm}^2 \times 50 \text{ mm} = 1250000 \text{ mm}^3$$

For Shape 3:

$$A_3 x_3 = 6250 \text{ mm}^2 \times \frac{500}{3} \text{ mm} = \frac{3125000}{3} \text{ mm}^3$$ $$A_3 y_3 = 6250 \text{ mm}^2 \times \frac{350}{3} \text{ mm} = \frac{2187500}{3} \text{ mm}^3$$

For Shape 2:

$$A_2 x_2 = -1250\pi \text{ mm}^2 \times 200 \text{ mm} = -250000\pi \text{ mm}^3$$ $$A_2 y_2 = -1250\pi \text{ mm}^2 \times \frac{200}{3\pi} \text{ mm} = -\frac{250000}{3} \text{ mm}^3$$

Total Area:

$$\sum A_i = A_1 + A_3 + A_2 = 25000 + 6250 - 1250\pi = (31250 - 1250\pi) \text{ mm}^2$$ $$\sum A_i \approx 31250 - 1250(3.14159) \approx 31250 - 3926.99 = 27323.01 \text{ mm}^2$$

Sum of moments about y-axis:

$$\sum A_i x_i = 3125000 + \frac{3125000}{3} - 250000\pi = \left(\frac{12500000}{3} - 250000\pi\right) \text{ mm}^3$$ $$\sum A_i x_i \approx 4166666.67 - 785398.16 = 3381268.51 \text{ mm}^3$$

Sum of moments about x-axis:

$$\sum A_i y_i = 1250000 + \frac{2187500}{3} - \frac{250000}{3} = \left(1250000 + \frac{1937500}{3}\right) \text{ mm}^3$$ $$\sum A_i y_i = \frac{3750000 + 1937500}{3} = \frac{5687500}{3} \text{ mm}^3$$ $$\sum A_i y_i \approx 1895833.33 \text{ mm}^3$$

$$\bar{x} = \frac{\sum A_i x_i}{\sum A_i} = \frac{3381268.51 \text{ mm}^3}{27323.01 \text{ mm}^2} \approx 123.75 \text{ mm}$$ $$\bar{y} = \frac{\sum A_i y_i}{\sum A_i} = \frac{1895833.33 \text{ mm}^3}{27323.01 \text{ mm}^2} \approx 69.39 \text{ mm}$$

The coordinates of the centroid are:

$$\boxed{\mathbf{(\bar{x}, \bar{y}) = (123.75 \text{ mm}, 69.39 \text{ mm})}}$$