# Moment of Inertia of the Shaded Area About the x-Axis Let's break the problem into two steps: --- ## **Step 1 Divide the Area Into Simple Shapes** The shaded area is the difference between: - A large **triangle** (base = 100 mm, height = 90 mm) - A **semicircle** (diameter = 30 mm, radius = 15 mm) removed from the base ### **Diagram and Dimensions** - Triangle base = \(100\, \text{mm}\) - Triangle height = \(90\, \text{mm}\) - Semicircle radius = \(15\, \text{mm}\) (centered at the base) --- ## **Step 2: Calculate the Moment of Inertia** ### **A. Triangle (about x-axis at base)** The moment of inertia about the base (x-axis) for a triangle: \[ I_{x,\text{triangle}} = \frac{b h^3}{12} \] - \(b = 100\, \text{mm}\) - \(h = 90\, \text{mm}\) \[ I_{x,\text{triangle}} = \frac{100 \times (90)^3}{12} = \frac{100 \times 729000}{12} = \frac{72900000}{12} = 607500\, \text{mm}^4 \] --- ### **B. Semicircle (about x-axis at base)** For a semicircle of radius \(r\) with its flat edge on the x-axis: \[ I_{x,\text{semicircle}} = \frac{1}{2} \left( \frac{\pi r^4}{4} \right) \] - \(r = 15\, \text{mm}\) \[ I_{x,\text{semicircle}} = \frac{1}{2} \times \frac{\pi (15)^4}{4} = \frac{1}{2} \times \frac{\pi \times 50625}{4} = \frac{1}{2} \times 39822.97 = 19911.48 \, \text{mm}^4 \] --- ### **C. Subtract the Semicircle from the Triangle** \[ I_{x,\text{shaded}} = I_{x,\text{triangle}} - I_{x,\text{semicircle}} = 607500 - 19911.48 = 6055088.52\, \text{mm}^4 \] --- ## **Final Answer** \[ \boxed{I_{x,\text{shaded}} = 6.06 \times 10^6 \, \text{mm}^4} \] --- ## **Graphical Representation** ```plaintext /\ / \ /____\ | | <-- semicircle removed at base ``` - Triangle: base = 100 mm, height = 90 mm - Semicircle: diameter = 30 mm, at the base --- ## **Summary Table** | Shape | Equation Used | Value (\(\text{mm}^4\)) | |------------|----------------------------------|-------------------------| | Triangle | \(\frac{b h^3}{12}\) | 6,075,000 | | Semicircle | \(\frac{1}{2}\frac{\pi r^4}{4}\) | 19,911.5 | | **Total** | | 6,055,088.5 | --- ### **Example Calculation** If the semicircle radius was 20 mm, recalculate: \[ I_{x,\text{semicircle}} = \frac{1}{2} \times \frac{\pi (20)^4}{4} = \frac{1}{2} \times \frac{\pi \times 160000}{4} = \frac{1}{2} \times 125663.7 = 62831.85 \text{ mm}^4 \] \[ I_{x,\text{shaded}} = 607500 - 62831.85 = 6012168.15 \text{ mm}^4 \] --- **Final Answer:** \[ \boxed{I_{x,\text{shaded}} = 6.06 \times 10^6 \, \text{mm}^4} \]