Centre of gravity of a solid hemisphere: For a solid hemisphere of radius R resting on its flat face (base), the distance of its centre of gravity (C.G.) from the base measured along the vertical axis is equal to which of the following?
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A3R/8
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BR/2
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C5R/8
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D3R/5
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ER/3
Answer
Correct Answer: 3R/8
Explanation
Introduction / Context:Locating the centre of gravity (C.G.) of common solids is essential in structural stability, ship ballast, and mechanical design. A solid hemisphere is frequently encountered in domes, tank ends, and flywheel caps. Knowing its C.G. relative to the base helps when computing moments and support reactions.
Given Data / Assumptions:
- Homogeneous solid hemisphere of radius R.
- Base is the flat circular face; axis is the symmetry axis orthogonal to the base.
- Uniform density; gravitational field is uniform.
Concept / Approach:
The C.G. of a body of revolution can be found by elemental integration of volume and first moments or by standard results. For a solid hemisphere, the C.G. lies on the axis at a known fraction of R from the base. The classic result is 3R/8 measured from the base toward the centre of curvature.
Step-by-Step Solution:
Consider thin circular discs of thickness dy at distance y from the base.Disc radius = sqrt(R^2 − (R − y)^2); elemental volume dV = π r^2 dy.First moment about the base: ∫ y dV; total volume V = (2/3) π R^3.Compute ȳ = (1/V) ∫ y dV → standard result ȳ = 3R/8.Verification / Alternative check:
Known tables of centres of gravity list for a solid hemisphere: distance from the base = 3R/8; from the centre of the original sphere = 5R/8 toward the base—consistent complementary distances.
Why Other Options Are Wrong:
R/2 and R/3 are characteristic of other shapes; 5R/8 is the complementary distance from the sphere centre, not from the base; 3R/5 overestimates the value.
Common Pitfalls:
Confusing solid hemisphere with thin hemispherical shell (whose C.G. lies at R/2 from the centre); measuring from the wrong reference (base vs. sphere centre).
Final Answer:
3R/8