Centroid shift after cutting a circular hole from a circular plate From a circular plate of diameter 6 cm (radius 3 cm), a smaller circular disc is cut out whose diameter equals the radius of the original plate (i.e., 3 cm). By how much does the centroid (C.G.) of the remaining lamina shift from the original center?

Introduction / Context:
Composite area centroids are frequently needed in plate-cut detailing, counterweights, and lightweighting by cut-outs. When material is removed, the centroid of the remaining shape shifts toward the heavier side opposite the removed area. Here we analyze a classical circular-plate-with-circular-hole configuration.

Given Data / Assumptions:

• Original plate: radius R = 3 cm.
• Removed disc: diameter = R, so r = R/2 = 1.5 cm.
• The small disc is cut so that it is internally tangent to the outer boundary (its center lies along a radius at distance R − r from the big center).
• Uniform thickness and density (area centroid analysis applies).

Concept / Approach:

For removal, use the negative-area method. Let O be the big-plate center and O′ the hole center on the same radius. The centroid shift x along that radius is obtained from first moments of area about O for the remaining lamina.

Step-by-Step Solution:

Verification / Alternative check:

Symmetry check: the shift must be along the line connecting the two centers and less than d (1.5 cm). The value 0.50 cm satisfies both.

Why Other Options Are Wrong:

0.25 cm and 0.75 cm do not satisfy the area-ratio result; 1.00 cm and 1.50 cm exceed or misrepresent the computed centroidal displacement based on actual areas and geometry.

Common Pitfalls:

Placing the small hole center incorrectly (e.g., at R or at R/2 without the tangent condition); forgetting to subtract the removed area in the denominator; mixing diameters with radii.

Final Answer:

0.50 cm