Moment of Inertia – IIT‑JEE Physics Notes
Goal: Understand the moment of inertia (MOI) of rigid bodies, know the main theorems, be able to compute the MOI for standard shapes, and solve JEE‑style problems.
1. Simple Explanation
The moment of inertia (I) measures how difficult it is to change a body’s rotation about a given axis.
Just as mass (m) resists linear acceleration ((F=ma)), the MOI resists angular acceleration ((\tau = I\alpha)).
[ \tau = I,\alpha ]
- Units: (\text{kg},\text{m}^2) (SI)
- Physical picture: Masses farther from the axis contribute more to (I) (because of the (r^2) factor).
2. Key Points
| Item | Formula / Concept | Notes |
|---|---|---|
| Definition | (\displaystyle I = \int r^2,dm) for a continuous body | (r) = distance from the axis |
| Discrete mass | (\displaystyle I = \sum_{i} m_i r_i^2) | Use when body is split into point masses |
| Units | (\text{kg},\text{m}^2) | Always check units |
| Parallel‑axis theorem | (\displaystyle I = I_{\text{cm}} + Md^2) | (I_{\text{cm}}) = MOI about centre‑of‑mass axis, (M) = total mass, (d) = distance between axes |
| Perpendicular‑axis theorem | For a flat lamina in the (xy)-plane: (I_z = I_x + I_y) | Only for planar (thin) bodies |
| Addition theorem | (I_{\text{total}} = I_{\text{body}} + I_{\text{object}}) | For composite bodies (sum of parts) |
| Standard shapes (about a given axis) | ||
| Thin rod, length (L) | (I_{\text{end}} = \frac{1}{3}ML^2), (I_{\text{mid}} = \frac{1}{12}ML^2) | |
| Solid cylinder, radius (R), height (h) | About central axis: (I = \frac12 MR^2) About diameter in the base: (I = \frac34 MR^2) |
|
| Solid sphere, radius (R) | About any diameter: (I = \frac{2}{5}MR^2) | |
| Thin hoop, radius (R) | (I = MR^2) | |
| Flat disk, radius (R) | About central axis perpendicular to plane: (I = \frac12 MR^2) | |
| Point mass (m) at distance (r) | (I = mr^2) |
Tip: For symmetric bodies, the MOI about any diameter is the same.
3. Worked Example
Problem (IIT‑JEE style)
A thin uniform rod of mass (M) and length (L) is hinged at one end. A small mass (m) is attached at a point (a) from the hinge ((0 < a < L)). Find the moment of inertia of the system about the hinge.
Solution
We treat the system as two independent masses:
- Rod – MOI about the hinge (I_{\text{rod}} = \frac{1}{3}ML^2).
- Point mass – MOI (I_{\text{mass}} = m a^2).
Total moment of inertia:
[ \boxed{I_{\text{total}} = \frac{1}{3} M L^2 + m a^2} ]
Checking Units:
(M) in kg, (L) in m → (ML^2) → kg·m².
(m a^2) → kg·m². ✔️
If the hinge is not at the end:
Suppose the hinge is at a distance (d) from the end. Use the parallel‑axis theorem for the rod:
[ I_{\text{rod,hinge}} = I_{\text{cm}} + M d^2, ] where (I_{\text{cm}} = \frac{1}{12} M L^2).
4. Quick Revision Summary
| Concept | Formula |
|---|---|
| MOI (continuous) | (I = \int r^2,dm) |
| MOI (discrete) | (I = \sum m_i r_i^2) |
| Parallel‑axis | (I = I_{\text{cm}} + M d^2) |
| Perpendicular‑axis (planar) | (I_z = I_x + I_y) |
| Addition | (I_{\text{composite}} = I_1 + I_2 + \dots) |
| Thin rod (end) | (\frac13 ML^2) |
| Thin rod (mid) | (\frac1{12}ML^2) |
| Solid cylinder (central) | (\frac12 MR^2) |
| Solid sphere (any diameter) | (\frac25 MR^2) |
| Thin hoop | (MR^2) |
| Disk (central) | (\frac12 MR^2) |
Quick Checks for JEE Problems
- Identify axis – Is it through the center, end, or off‑center?
- Use symmetry – Parallel‑axis or perpendicular