Moment of Inertia of a Disc About Different Axes with Solved Problems
The moment of inertia of a disc is a key concept in rotational mechanics and is crucial for JEE Main Physics. It measures a disc’s resistance to changes in rotational motion when a torque is applied. Understanding this property helps you solve problems related to rolling, angular momentum, and rotational kinetic energy, which frequently appear in exams.
In real-life, discs are found in flywheels, CDs, and pulleys. For JEE, you need to differentiate the moment of inertia for a solid disc, ring, or hoop, as well as how axis location affects the value. Let’s explore its formulae, derivation, and applications using logical stepwise explanations.
Formulae and Unit Table: Moment of Inertia of a Disc and Related Shapes
| Object/Axis | Formula | Unit (SI) |
|---|---|---|
| Disc, through centre and perpendicular axis | (1/2)MR2 | kg·m2 |
| Disc, through any diameter | (1/4)MR2 | kg·m2 |
| Ring, central axis | MR2 | kg·m2 |
| Hoop, central axis | MR2 | kg·m2 |
| Rod, perpendicular bisector | (1/12)ML2 | kg·m2 |
Derivation of the Moment of Inertia of a Disc About Its Centre
To derive the moment of inertia of a disc about its central axis (perpendicular to its plane), consider a uniform solid disc of mass M and radius R.
Divide the disc into thin concentric rings of radius r and thickness dr. Each ring’s mass is dm = σ·2πr·dr, where σ = M/πR² is the mass per unit area.
Moment of inertia for each ring about the central axis is dI = dm·r².
Integrate from r = 0 to r = R:
- I = ∫0R r²·(σ·2πr·dr)
- = 2πσ ∫0R r³ dr
- = 2πσ [r4/4]0R
- = 2πσ·R4/4
- Since σ = M/πR², substitute and simplify:
- I = (1/2)MR²
The final result: the moment of inertia of a disc about its central (perpendicular) axis is (1/2)MR².
Moment of Inertia of a Disc: Different Axes Explained
- Through centre, perpendicular to plane: (1/2)MR² (standard axis for rotation).
- Through any diameter (in disc’s plane): (1/4)MR². This uses the perpendicular axis theorem.
- Through edge, perpendicular to plane: Use the parallel axis theorem: I = (1/2)MR² + MR² = (3/2)MR².
- Through a tangent (in plane): I = (1/4)MR² + MR² = (5/4)MR².
- Disc with a central hole: Subtract the moment of inertia of the missing (hole) area from the full disc.
For JEE, always check if the given axis passes through the mass centre or is offset; this changes your calculation method.
Moment of Inertia of a Disc with a Central Hole
A disc with a hole of radius r1 (hole) and r2 (outer disc) and mass M has:
- Full disc (r2): Ifull = (1/2)M2r2²
- Hole (r1): Ihole = (1/2)M1r1²
Subtract the hole’s inertia from the full disc. Let M1 = M·(r1²/r2²) by area-mass proportionality.
- I = (1/2)MR2² − (1/2)M(r1²/r2²)·r1²
Example: A disc of mass 2 kg, outer radius 0.3 m, and hole radius 0.1 m.
- Full disc inertia: Ifull = (1/2) × 2 × (0.3)2 = 0.09 kg·m²
- Hole’s mass: Mhole = 2 × (0.1)2/(0.3)2 ≈ 0.222 kg
- Hole inertia: Ihole = (1/2) × 0.222 × (0.1)2 ≈ 0.0011 kg·m²
- Disc with hole: 0.09 − 0.0011 = 0.0889 kg·m²
If the axis is not at the centre, always use the proper parallel axis correction. Check condition and units before substituting values.
Applications and Example Problems on Moment of Inertia of a Disc
- Calculating rotational kinetic energy: Use rotational kinetic energy of disc is (1/2)Iω2.
- Finding acceleration of a rolling disc: Apply Newton’s second law with both linear and rotational forms.
- Predicting motion in rotational motion.
- Example: What is the moment of inertia about a tangent parallel to the disc’s diameter? Use parallel axis: I = (1/4)MR² + MR² = (5/4)MR².
- Physics Olympiad and JEE questions often ask you to compare the disc’s inertia to that of a ring or a rod.
For practice, you can find additional problems in the rotational motion practice paper and moment of inertia of annular disc sections.
| Shape | Key Axis | Formula |
|---|---|---|
| Disc | Perpendicular through centre | (1/2)MR2 |
| Ring | Perpendicular through centre | MR2 |
| Hoop | Perpendicular through centre | MR2 |
| Rod | Midpoint perpendicular | (1/12)ML2 |
Common pitfalls when solving moment of inertia of a disc problems include mixing up the axis (centre vs. diameter vs. edge) and confusing disc with ring/hoop results. Always check which shape and axis the question refers to.
Make sure to use SI units for mass and radius, double check the final units (kg·m2), and avoid plugging values too early—keep everything symbolic until the last step.
For a deeper comparison of shapes and their practical significance, see the moment of inertia of ring page and the moment of inertia summary.
In engineering and everyday technology, the moment of inertia of a disc is widely used (for example, in vehicle wheels and machinery flywheels). JEE Main often frames conceptual and numerical questions that test your understanding of how different axes or modifications like holes affect inertia.
Be sure to review related topics like the perpendicular axis theorem, rotational motion, and kinetic theory for well-rounded exam readiness using Vedantu’s resources.
Summary: Remember, the core formula for the moment of inertia of a disc through its centre is (1/2)MR². Adjust it for other axes using the right theorems. Build confidence by solving a variety of practice problems, paying close attention to axis position, mass, and radius each time.
FAQs on Moment of Inertia of a Disc: Concepts, Formula & Derivation
1. What is the moment of inertia of a disc about its central axis?
The moment of inertia (I) of a uniform solid disc about its central axis is given by I = (1/2)MR². Here, M is the mass of the disc and R is its radius. This formula shows how the rotational inertia depends on both the mass and the square of the radius of the disc.
2. What is the formula for the moment of inertia of a disc?
The standard formula for the moment of inertia of a solid disc about an axis perpendicular to its plane and passing through its center is I = (1/2)MR². For other axes, different formulas apply, such as:
- About a diameter: I = (1/4)MR²
- About an axis tangent to the disc in the plane: I = (3/4)MR²
3. What is the unit of moment of inertia for a disk?
The unit of moment of inertia in the SI system is kilogram meter squared (kg·m²). It quantifies the mass distribution about the axis of rotation and is crucial for rotational motion calculations.
4. How is the moment of inertia of a disc derived?
To derive the moment of inertia of a disc about its center, calculus is used to sum the contributions of all infinitesimal mass elements at varying distances. The process involves:
- Dividing the disc into thin concentric rings of radius r and thickness dr
- Calculating the mass element: dm = (2πr dr) × (M/πR²)
- Integrating r² dm from r = 0 to r = R
- The result gives I = (1/2)MR²
5. What is the difference between the moment of inertia of a disc and a ring?
A solid disc and a thin ring of the same mass and radius have different moments of inertia because their mass is distributed differently.
- Disc (about central axis): I = (1/2)MR²
- Ring (about central axis): I = MR²
- The ring has greater moment of inertia since all its mass is at maximum distance from the axis.
6. How does the axis affect the moment of inertia of a disc?
The position and direction of the rotation axis directly change the moment of inertia of a disc. Key examples are:
- Central/perpendicular axis: I = (1/2)MR²
- About diameter (in-plane): I = (1/4)MR²
- About tangent to disc: I = (3/4)MR²
- Use parallel axis theorem to shift axes.
7. What is the moment of inertia of a disc with a hole at its center?
The moment of inertia of a disc with a central hole is calculated by subtracting the inertia of the missing inner disc from the solid disc.
- Formula: I = (1/2)MouterRouter² – (1/2)MinnerRinner²
- Where Mouter, Router are mass and radius of the whole disc, Minner, Rinner are that of the hole.
8. What is the moment of inertia of a disc about its diameter?
The moment of inertia of a disc about one of its diameters (in the plane of the disc) is I = (1/4)MR². This value is half of the moment of inertia about the central perpendicular axis.
9. Can the moment of inertia of a disc be calculated about any axis using the parallel and perpendicular axis theorems?
Yes, the parallel and perpendicular axis theorems allow calculation of the moment of inertia of a disc about axes not passing through its center or lying in its plane.
- Parallel axis theorem: I = Icenter + Md², where d is distance between axes
- Perpendicular axis theorem: Iz = Ix + Iy, for planar objects like discs
10. What is the importance of the moment of inertia of a disc in real-world applications?
Understanding the moment of inertia of a disc is critically important for designing rotors, flywheels, gears, and various rotating machinery. It determines:
- Rotational stability and acceleration
- Energy storage in flywheels
- Performance in sports equipment like frisbees
- Design safety and efficiency in rolling systems
11. What errors should be avoided when solving disc moment of inertia problems in competitive exams?
Key errors to avoid in competitive exams include:
- Mixing up the formulas for disc, ring, and hoop
- Not checking the axis of rotation specified
- Ignoring the presence of holes or non-uniform mass
- Forgetting to apply the parallel axis theorem when needed
- Unit conversion mistakes – always use kg·m²
12. Can the moment of inertia be negative for a disc?
No, the moment of inertia is always a positive quantity for a disc or any rotating body. It represents the sum of positive mass elements times the square of their distance from the axis (r²), which cannot be negative.
