Derivation and Formula for Moment of Inertia of a Hollow Sphere
The moment of inertia of a hollow sphere is a key rotational property relevant for competitive exams like JEE Main. It expresses how mass is distributed with respect to its axis, affecting how the spherical shell responds to torque. This concept is widely tested, both for straightforward formula recall and for deeper understanding involving derivation and application. Candidates must distinguish it from the solid sphere, grasp axis-dependence, and master calculation shortcuts. You’ll see this arise in rotational dynamics, especially when comparing different rigid bodies or solving advanced rolling questions.
In JEE contexts, a hollow sphere means a thin spherical shell with all its mass on the surface and negligible wall thickness. Rotational motion problems often pivot on this case, so precise understanding is critical to avoid confusion with other geometries like solid spheres or discs.
Formula and Units for Moment of Inertia of a Hollow Sphere
The standard formula for the moment of inertia of a hollow sphere (mass M, radius R) about any diameter is:
| Axis | Moment of Inertia | SI Unit |
|---|---|---|
| About diameter | I = (2/3)MR2 | kg·m2 |
| About tangent | I = (2/3)MR2 + MR2 | kg·m2 |
Here, I is moment of inertia, M is the mass of the sphere, and R is its radius. For axes not passing through the center, apply the parallel axis theorem. Keeping SI units consistent is mandatory in JEE numericals.
Stepwise Derivation: Moment of Inertia of a Hollow Sphere
For JEE, you should know how the moment of inertia of a hollow sphere is derived. The process starts by considering a thin mass element on the shell, summing inertia over the sphere. This is a typical exam derivation—follow each step for scoring derivations.
- Take a hollow sphere of radius R and total mass M.
- Choose a thin ring at angle θ, thickness Rdθ, ring radius r = R sinθ.
- Surface area of the sphere is 4πR2.
- Elemental area dA = 2πr × Rdθ = 2πR2sinθ dθ.
- Elemental mass dm = (M / 4πR2) × dA = (M/2)sinθ dθ.
- Moment of inertia for the ring: dI = r2dm = R2sin2θ × (M/2)sinθ dθ = (M R2/2)sin3θ dθ.
- Integrate dI from θ = 0 to θ = π to cover whole sphere.
- Result: I = (2/3)MR2.
This finished formula is used in most rotational inertia JEE questions focused on shells or hollow spheres. Try practicing a hand-derivation to solidify each integral step. For axis through the tangent, the parallel axis theorem adds another term MR2.
Hollow Sphere vs Solid Sphere: JEE Comparison Table
| Object | Moment of Inertia (I) | Key Difference |
|---|---|---|
| Hollow sphere | (2/3)MR2 | All mass at surface |
| Solid sphere | (2/5)MR2 | Mass distributed throughout |
Notice that the moment of inertia of a hollow sphere is greater than a solid sphere of same mass and radius. JEE MCQs frequently test this relation, so memorize the contrast and physical meaning.
Worked Example: Calculating Moment of Inertia of a Hollow Sphere
Let’s calculate for a hollow sphere with M = 55.0 kg, R = 0.120 m about its diameter.
- Formula: I = (2/3)MR2
- Plug values: I = (2/3) × 55.0 × (0.120)2
- Compute R2: (0.120)2 = 0.0144 m2
- Multiply: 55.0 × 0.0144 = 0.792 kg·m2
- I = (2/3) × 0.792 = 0.528 kg·m2
For JEE, always present the correct sequence, use SI units, and clearly box your final answer. Vedantu solutions consistently follow this approach for clarity.
- Apply moment of inertia principles to analyze rotational bodies.
- Contrast with results for a disc or cube for broader understanding.
- Connect to the perpendicular axis theorem when asked for axes other than diameter.
- Reference semi-circular bodies as part of mixed-shape assemblies.
- Cite rotational motion to integrate with torque and angular momentum topics.
Applications, Pitfalls, and Revision for JEE
The moment of inertia of a hollow sphere is used in centripetal force, rotational energy, and rolling without slipping problems. You’ll see it applied in physics experiments comparing energy outcomes for shells vs solid objects, especially when interpreting torque or acceleration during rolling motions.
- Don’t confuse a hollow sphere with a solid sphere: their formulas differ.
- Avoid plugging in mass at the center—all mass is at the surface for shells.
- Always check which axis is specified—use parallel axis theorem if tangent or any non-diameter axis is used.
- Units must always be kg·m2, not grams or centimeters for JEE.
- Review fundamental definitions at moment of inertia of continuous bodies for generalizations.
| Revision Aspect | Key Point |
|---|---|
| Primary formula | I = (2/3)MR2 |
| Solid sphere | I = (2/5)MR2 |
| Tangent axis | Apply parallel axis theorem |
| Units | kg·m2 |
| Common error | Formula mix-up with solid sphere or disc |
Before exams, rehearse the derivation, anchor the formula I = (2/3)MR2 in memory, and vigilantly read question wording about axes. Many tricky JEE problems blend rigid body topics and require comparing shell, disc, and solid sphere inertia. That’s why Vedantu JEE Physics revision sheets highlight these patterns for fast last-minute brushing.
- Practice direct calculation using the main result for shells.
- Extend logic to other hollow bodies to spot exam analogies.
- Use mock tests such as rotational motion mock test for full-concept mastery.
- Contrast with annular disc and circle for rapid recall.
In summary, the moment of inertia of a hollow sphere is given by I = (2/3)MR2 about its diameter. Correct application and clear differentiation from related bodies are essential for JEE success. Thorough practice and using quality resources like Vedantu’s worked solutions ensure concept clarity and speed on exam day.
FAQs on Moment of Inertia of a Hollow Sphere – Concepts, Formula & Examples
1. What is the moment of inertia of a hollow sphere?
The moment of inertia of a hollow sphere about its diameter is given by I = (2/3)MR², where M is the mass and R is the radius of the sphere.
Key points:
- This formula applies when the axis is through the centre (diameter).
- It is important in rotational mechanics for calculating rotational energy and dynamics.
- Used in problems for JEE, NEET, and CBSE exams.
2. How do you derive the formula for the moment of inertia of a hollow sphere?
The derivation of the moment of inertia for a hollow sphere involves integrating mass elements at distance R from the axis:
Steps:
- Consider a thin spherical shell of mass M and radius R.
- Since all mass is at distance R from the centre, I = ΣmᵢR² = MR².
- However, integration over the sphere considering rotation gives I = (2/3)MR² about its diameter.
3. What is the formula for the hollow moment of inertia?
The formula for the moment of inertia of a hollow sphere about its diameter is:
- I = (2/3)MR² — about the diameter (central axis)
- I = (2/3)MR² + Md² — about a parallel axis at distance d (using parallel axis theorem)
4. What is the difference between the moment of inertia of solid and hollow spheres?
The main difference is in their mass distribution and resulting inertia formulas:
- Hollow sphere (thin shell): I = (2/3)MR²
- Solid sphere: I = (2/5)MR²
This distinction is frequently tested in MCQs and theory questions.
5. What are the real-world applications of the hollow sphere’s moment of inertia?
The moment of inertia of a hollow sphere is used in:
- Design of sports balls (e.g., table-tennis, cricket balls)
- Planetary and astronomical modeling
- Engineering of rotating machinery and storage tanks
- Physics experiments involving rolling motion
6. How does the axis of rotation affect the moment of inertia of a hollow sphere?
Changing the axis changes the moment of inertia value:
- Through diameter/centre: I = (2/3)MR²
- About a tangent: Use parallel axis theorem ⇒ I = (2/3)MR² + MR² = (5/3)MR²
7. Does wall thickness affect the moment of inertia of a hollow sphere?
Yes, wall thickness affects the moment of inertia.
If the hollow sphere is thin (a spherical shell), use I = (2/3)MR².
If it has significant thickness (thick shell), the moment of inertia needs integration between inner and outer radii:
- Thicker shells have more complex formulas based on their geometry
8. Can the moment of inertia of a hollow sphere be less than that of a solid sphere of same mass and radius?
No, the moment of inertia of a hollow sphere is always greater than that of a solid sphere with same mass and radius.
This is because mass is distributed farther from the axis in the hollow sphere:
- Hollow sphere: I = (2/3)MR²
- Solid sphere: I = (2/5)MR²
9. What mistakes are common when using parallel axis theorem with hollow spheres?
Students often make these errors:
- Forgetting to add Md² term (where d is the distance between axes)
- Using the wrong axis — always check if the axis passes through centre or tangent
- Confusing hollow and solid sphere formulas
10. How is the moment of inertia of a hollow sphere important for JEE and CBSE exams?
The hollow sphere’s moment of inertia is frequently asked:
- In derivation, MCQ, and numerical questions
- To test understanding of rotational dynamics and comparison with solid sphere
- As part of SYLLABUS for JEE Main, NEET, CBSE boards
