Questions: Find the center of mass of the lamina in the following figure if the circular portion of the lamina has twice the density of the square portion of the lamina.

Solution Steps

We need to find the center of mass of a composite lamina consisting of a square and a semicircle. The semicircle has twice the density of the square.

• Square: Side length = 10 units, centered at (5, 0)
• Semicircle: Radius = 5 units, centered at (15, 0)

• Square Area (A1): \( A_1 = 10 \times 10 = 100 \) square units
• Semicircle Area (A2): \( A_2 = \frac{1}{2} \pi (5)^2 = \frac{25\pi}{2} \) square units

• Square Mass (M1): \( M_1 = \sigma \times A_1 = \sigma \times 100 \)
• Semicircle Mass (M2): \( M_2 = 2\sigma \times A_2 = 2\sigma \times \frac{25\pi}{2} = 25\sigma\pi \)

• Square Center of Mass (x1, y1): \( (x_1, y_1) = (5, 0) \)
• Semicircle Center of Mass (x2, y2): \( (x_2, y_2) = (15, \frac{4r}{3\pi}) = (15, \frac{20}{3\pi}) \)

• Total Mass (M): \( M = M_1 + M_2 = 100\sigma + 25\sigma\pi \)

• x-coordinate: \[ \bar{x} = \frac{M_1 x_1 + M_2 x_2}{M} = \frac{100\sigma \cdot 5 + 25\sigma\pi \cdot 15}{100\sigma + 25\sigma\pi} = \frac{500\sigma + 375\sigma\pi}{100\sigma + 25\sigma\pi} = \frac{500 + 375\pi}{100 + 25\pi} \]
• y-coordinate: \[ \bar{y} = \frac{M_1 y_1 + M_2 y_2}{M} = \frac{100\sigma \cdot 0 + 25\sigma\pi \cdot \frac{20}{3\pi}}{100\sigma + 25\sigma\pi} = \frac{0 + 25\sigma\pi \cdot \frac{20}{3\pi}}{100\sigma + 25\sigma\pi} = \frac{500/3}{100 + 25\pi} = \frac{500}{3(100 + 25\pi)} \]

Final Answer