Centre of Mass of Hollow and Solid Hemisphere Explained

The Centre of Mass of Hollow and Solid Hemisphere is a classic JEE topic, blending symmetry, calculus, and physical reasoning. Hemispheres appear often in both numerical questions and real-world objects, so mastering their centre of mass formula and derivation is essential for high scores. Many students confuse the results for hollow (shell) and solid hemispheres, as the internal mass distribution is crucial. In this article, you’ll find step-by-step derivations, key formulas, quick contrasts, and solved examples to simplify advanced questions on semicircular and hemispherical objects, directly aligned to the JEE syllabus.

Defining the Centre of Mass of Hollow and Solid Hemisphere

The centre of mass is the point where the entire mass of a body or system can be thought of as concentrated for translational motion. For symmetric bodies like spheres and hemispheres, this point lies on the axis of symmetry. Calculating the Centre of Mass of Hollow and Solid Hemisphere involves integrating over their unique mass distributions, not just using simple averaging.

In JEE Physics, the centre of mass formulas vary for a solid hemisphere (mass distributed throughout) and a hollow hemisphere (mass only on the curved surface). Identifying which formula to use, and when, helps avoid frequent mistakes in exam problems.

• The centre of mass for a solid hemisphere is not at its geometric centre.
• For a hollow hemisphere, mass exists solely on the curved shell, shifting the centre of mass outward.
• Both types are referenced from the flat base, along the symmetry (vertical) axis.
• Symbols: R is the radius; axis convention is from base upward.
• Make sure you apply the correct limits in integration for semi-spherical objects.

Centre of Mass of Solid Hemisphere: Derivation and Formula

Let’s derive the centre of mass for a solid hemisphere of uniform density and radius R, where the base plane is at y = 0. We orient the y-axis vertically upward, passing through the base centre.

1. Consider a differential disc parallel to the base, at height y.
2. The disc's radius at this height is r = √(R² − y²).
3. Disc area = πr²; thickness = dy; so volume element = πr²dy.
4. Mass element: dM = (M / (2/3 π R³)) πr² dy, since total hemisphere volume is (2/3)πR³.
5. The y-coordinate of centre of mass is y_CM = (1/M) ∫ y dM, limit y = 0 to y = R.

Substitute and compute (JEE-level steps):

y_CM = (1/M) ∫₀^R y × (3M/2R³) (R² − y²) dy

y_CM = (3/2R³) ∫₀^R (R²y − y³) dy = (3/2R³) [R²y²/2 − y⁴/4]₀^R

y_CM = (3/2R³) [R² × R²/2 − R⁴/4] = (3/2R³) [R⁴/2 − R⁴/4] = (3/2)(1/4)R = 3R/8

Thus, for a solid hemisphere, the centre of mass is at a distance 3R/8 from the base, along the axis of symmetry. This formula is critical for all JEE numerical and conceptual questions involving hemispherical solids.

• Always set the origin at the base for consistency in calculation.
• The result is independent of mass, depends only on geometry (R).
• Confusing it with the centroid (R/2) is a common pitfall!

Centre of Mass of Hollow Hemisphere: Key Derivation and Insights

For the hollow hemisphere (spherical shell), mass is equally distributed along the curved surface only, not the interior. We choose an axis similarly (origin at base centre, y-axis vertical).

1. Take a thin ring at angle θ from vertical; ring mass dM ∝ sinθ dθ.
2. Vertical position of each element: y = R cosθ.
3. Centre of mass: y_CM = (1/M) ∫ y dM (from θ = 0 to π/2).
4. Integrate: y_CM = (1/M) ∫₀^π/2 R cosθ × M sinθ dθ.
5. y_CM = R ∫₀^π/2 cosθ sinθ dθ = R/2.

Centre of mass of a hollow hemisphere is at a distance R/2 from the base along the axis of symmetry. This is closer to the rim than for a solid hemisphere, as all the mass lies on the outer shell.

• The formula R/2 is a quick shortcut for all shell-type hemisphere problems.
• Ensure you know the difference between “hollow” (surface only) and “solid.”
• Avoid assuming the result for full spheres applies; only half is present.

• Use centre of mass theory page for basics.
• See centre of mass of semicircular ring for similar planar shapes.
• Compare to moment of inertia of a cone for related mass distributions.
• Review moment of inertia basics for rotational analogues.
• Practice with JEE units and power mock tests for reinforcement.

Practical Examples and Common Mistakes in Centre of Mass of Hollow and Solid Hemisphere

Here’s a classic JEE-style question: Find the centre of mass location for a solid hemisphere with radius 8 cm. Using y_CM = 3R/8 = 3×8/8 = 3 cm above the base.

Now, for a hollow hemisphere of same radius: y_CM = R/2 = 8/2 = 4 cm.

• Never average the results between sphere and hemisphere!
• When combining hemispheres and cones, use the formula for each component and weighted mean.
• Beware: wrong axis choice can shift all answers by R (absolute error).
• Gravitation concepts sometimes ask for “height from base”; always clarify your reference.
• Check units and measurement consistency in every calculation.

Try linking your understanding to system of particles and rotational motion for more composite body questions. For advanced problems, the semicircular ring and disc pages are also helpful.

Summary Table: Formulae and Tips for JEE

• Remember: Solid = 3R/8; Hollow = R/2, both above base on axis.
• Use calculus derivations as shown for partial or composite shapes.
• Apply the same origin when combining with other shapes (like cones or discs).
• Build speed with unit radius first; scale up by R in all answers.
• Find more solved examples in work energy practice sets.

Mastering the Centre of Mass of Hollow and Solid Hemisphere not only unlocks a popular JEE concept but strengthens your problem-solving for any body with spherical symmetry. For deep dives and updates, Vedantu's expert-reviewed content keeps you aligned with current exam demands.