A circular hole of 50 mm diameter is cut out from a circular disc of 100 mm diameter as shown in the figure below. Find the centre of gravity of the section from A.

To find the center of gravity (centroid) of the shaded area from point A, we will treat the shaded area as a composite of a large circular disc and a smaller circular hole. We'll use point A as our origin (0,0). First, let's define the properties of the two shapes: Large Circular Disc (LC): Diameter D_LC = 100 mm Radius R_LC = (100)/(2) = 50 mm Area A_LC = R_LC^2 = (50)^2 = 2500 mm^2 Centroid x_LC: Since A is the leftmost point of the large circle and the diameter is 100 mm, the center of the large circle is at x_LC = 50 mm from A. Centroid y_LC: The line A-B is the axis of symmetry, so y_LC = 0 mm. Circular Hole (H): Diameter D_H = 50 mm Radius R_H = (50)/(2) = 25 mm Area A_H = R_H^2 = (25)^2 = 625 mm^2 Centroid y_H: The hole is also symmetric about the line A-B, so y_H = 0 mm. Centroid x_H: The diagram's dimensioning for the hole's position is ambiguous. However, the visual representation clearly shows the hole is offset to the right, not concentric with the large circle. A common interpretation for such offset holes, consistent with the visual, is that the right edge of the hole is tangent to the right edge of the large circle. The right edge of the large circle is at x = 100 mm from A. If the right edge of the hole is tangent to this, then x_H + R_H = 100 mm. x_H + 25 = 100 x_H = 75 mm. Alternatively, if the 50 mm dimension from A to the center of the hole is interpreted as the distance from A to the left edge* of the hole, then x_H - R_H = 50 mm, which gives x_H - 25 = 50 mm, so x_H = 75 mm. This interpretation is consistent with the visual and leads to a centroid shift. Step 1: Calculate the total area of the shaded region. The shaded area A_shaded is the area of the large circle minus the area of the hole. A_shaded = A_LC - A_H A_shaded = 2500 mm^2 - 625 mm^2 = 1875 mm^2 Step 2: Calculate the x-coordinate of the centroid (x) of the shaded area. The formula for the centroid of a composite area with a cutout is: x = A_LC x_LC - A_H x_HA_LC - A_H Substitute the values: x = (2500 mm^2)(50 mm) - (625 mm^2)(75 mm)1875 mm^2 x = 125000 mm^3 - 46875 mm^31875 mm^2 x = (125000 - 46875) mm^31875 mm^2 x = (78125)/(1875) mm Simplify the fraction: x = (125)/(3) mm Step 3: Calculate the y-coordinate of the centroid (y) of the shaded area. Since both the large circle and the hole are symmetric about the line A-B (our x-axis), their individual centroids lie on this axis. Therefore, the centroid of the composite shaded area will also lie on this axis. y = 0 mm The center of gravity of the section from A is at: x = (125)/(3) mm or 41.67 mm y = 0 mm Last free one today — make it count tomorrow, or type /upgrade for unlimited.