Prerequisites
Before studying this topic, review:
- Centre of Mass - Understanding COM concept
- Kinetic Energy - Energy in motion
The Hook: Why Can’t You Stop a Spinning Figure Skater?
Watch any figure skating championship - when skaters pull their arms in during a spin, they suddenly become a blur of motion! Push them while arms are extended: they slow down easily. Try the same when arms are tucked in: impossible!
In Jawan (2023), when Shah Rukh Khan’s character spins a metal rod to deflect bullets, he holds it at the center. Why? Try spinning a cricket bat by holding it at the end versus the center - the difference is HUGE!
The mystery: Why does the same object become harder or easier to rotate depending on how you hold it?
The answer lies in Moment of Inertia - the rotational equivalent of mass!
The Core Concept
What is Moment of Inertia?
Moment of inertia is the measure of an object’s resistance to rotational acceleration about an axis - just like mass resists linear acceleration.
$$\boxed{I = \sum m_i r_i^2}$$where:
- $I$ = moment of inertia (kg·m²)
- $m_i$ = mass of particle $i$
- $r_i$ = perpendicular distance of particle $i$ from the axis of rotation
For continuous bodies:
$$\boxed{I = \int r^2\,dm}$$Moment of inertia is “rotational mass”
Just as:
- Mass measures resistance to linear acceleration ($F = ma$)
- Moment of inertia measures resistance to angular acceleration ($\tau = I\alpha$)
Key difference: Moment of inertia depends on:
- Mass (more mass → more I)
- Distribution of mass (farther from axis → more I)
- Axis of rotation (different axis → different I)
Interactive Demo: Visualize Moment of Inertia
See how mass distribution affects rotational resistance.
Memory Tricks & Patterns
The “Distance Matters” Rule
Memory Trick: “Inertia Increases with Distance squared”
$$I \propto r^2$$This explains:
- Why ice skaters slow their spin by extending arms (increases $r$)
- Why spinning a rod from the end is harder than from center
- Why hollow objects have larger $I$ than solid ones of same mass
The “Half-Whole-Third” Pattern for Standard Shapes
For standard shapes, remember these fractions:
| Shape | Axis | Fraction | Formula |
|---|---|---|---|
| Disc/Cylinder | Through center, perpendicular to plane | 1/2 | $\frac{1}{2}MR^2$ |
| Solid Sphere | Through center | 2/5 | $\frac{2}{5}MR^2$ |
| Hollow Sphere (shell) | Through center | 2/3 | $\frac{2}{3}MR^2$ |
| Rod | Through center, perpendicular | 1/12 | $\frac{1}{12}ML^2$ |
| Rod | Through end, perpendicular | 1/3 | $\frac{1}{3}ML^2$ |
| Ring/Hollow Cylinder | Through center, perpendicular | 1 | $MR^2$ |
Mnemonic: “DRISH”
Disc = 1/2 Rod (center) = 1/12 I (Rod end) = 1/3 Sphere (solid) = 2/5 Hollow sphere = 2/3
Pattern recognition:
Ring/Hollow cylinder: All mass at distance $R$ → $I = MR^2$ (maximum for given mass and radius)
From hollow to solid: Inertia decreases as mass comes closer to axis
- Hollow sphere ($\frac{2}{3}MR^2$) > Solid sphere ($\frac{2}{5}MR^2$)
- Ring ($MR^2$) > Disc ($\frac{1}{2}MR^2$)
Rod about end = 4 × Rod about center: $\frac{1}{3}ML^2 = 4 \times \frac{1}{12}ML^2$
This saves calculation time in MCQs!
Moment of Inertia for Common Bodies
1. Point Mass
For a point mass $m$ at perpendicular distance $r$ from axis:
$$\boxed{I = mr^2}$$This is the fundamental formula - all others are derived from this!
2. Uniform Rod (Length $L$, Mass $M$)
a) Axis through center, perpendicular to rod:
$$\boxed{I = \frac{ML^2}{12}}$$b) Axis through one end, perpendicular to rod:
$$\boxed{I = \frac{ML^2}{3}}$$Derivation (for center):
Consider element $dx$ at distance $x$ from center. Mass of element: $dm = \frac{M}{L}dx$
$$I = \int_{-L/2}^{L/2} x^2 dm = \int_{-L/2}^{L/2} x^2 \frac{M}{L}dx = \frac{M}{L} \cdot \frac{x^3}{3}\bigg|_{-L/2}^{L/2} = \frac{ML^2}{12}$$3. Ring (Radius $R$, Mass $M$)
a) Axis through center, perpendicular to plane:
$$\boxed{I = MR^2}$$All mass is at distance $R$ from axis!
b) Axis as diameter (in the plane):
$$\boxed{I = \frac{MR^2}{2}}$$4. Disc (Radius $R$, Mass $M$)
a) Axis through center, perpendicular to plane:
$$\boxed{I = \frac{MR^2}{2}}$$Derivation:
Consider a ring of radius $r$ and thickness $dr$. Area of ring: $2\pi r \, dr$
Mass of ring: $dm = \frac{M}{\pi R^2} \cdot 2\pi r \, dr = \frac{2Mr}{R^2}dr$
$$I = \int_0^R r^2 dm = \int_0^R r^2 \cdot \frac{2Mr}{R^2}dr = \frac{2M}{R^2}\int_0^R r^3 dr = \frac{2M}{R^2} \cdot \frac{R^4}{4} = \frac{MR^2}{2}$$b) Axis as diameter (in the plane):
$$\boxed{I = \frac{MR^2}{4}}$$5. Solid Cylinder (Radius $R$, Mass $M$, Length $L$)
a) Axis along cylinder axis:
$$\boxed{I = \frac{MR^2}{2}}$$(Same as disc - length doesn’t matter for this axis!)
b) Axis through center, perpendicular to length:
$$\boxed{I = M\left(\frac{R^2}{4} + \frac{L^2}{12}\right)}$$6. Hollow Cylinder (Inner radius $R_1$, Outer radius $R_2$, Mass $M$)
Axis along cylinder axis:
$$\boxed{I = \frac{M(R_1^2 + R_2^2)}{2}}$$Special case (thin shell): If $R_1 \approx R_2 = R$:
$$I = MR^2$$7. Solid Sphere (Radius $R$, Mass $M$)
Axis through center (any diameter):
$$\boxed{I = \frac{2MR^2}{5}}$$Derivation approach: Consider sphere as collection of discs of varying radii, integrate.
8. Hollow Sphere / Spherical Shell (Radius $R$, Mass $M$)
Axis through center (any diameter):
$$\boxed{I = \frac{2MR^2}{3}}$$9. Rectangular Plate (Length $a$, Width $b$, Mass $M$)
a) Axis along length $a$ (through center):
$$\boxed{I = \frac{Mb^2}{12}}$$b) Axis perpendicular to plate through center:
$$\boxed{I = \frac{M(a^2 + b^2)}{12}}$$Radius of Gyration ($k$)
Definition: Radius of gyration is the distance from the axis where the entire mass can be assumed to be concentrated to give the same moment of inertia.
$$\boxed{I = Mk^2}$$ $$\boxed{k = \sqrt{\frac{I}{M}}}$$Physical meaning: If all mass $M$ were at distance $k$ from axis, the moment of inertia would be the same.
Radius of Gyration for Common Bodies
| Body | Axis | $k$ |
|---|---|---|
| Rod (length $L$) | Through center | $\frac{L}{2\sqrt{3}}$ |
| Rod (length $L$) | Through end | $\frac{L}{\sqrt{3}}$ |
| Disc (radius $R$) | Through center, ⊥ to plane | $\frac{R}{\sqrt{2}}$ |
| Ring (radius $R$) | Through center, ⊥ to plane | $R$ |
| Solid sphere (radius $R$) | Through diameter | $\sqrt{\frac{2}{5}}R$ |
| Hollow sphere (radius $R$) | Through diameter | $\sqrt{\frac{2}{3}}R$ |
Rotational Kinetic Energy
Just as $KE = \frac{1}{2}mv^2$ for linear motion:
$$\boxed{KE_{rot} = \frac{1}{2}I\omega^2}$$where $\omega$ is angular velocity (rad/s).
For rolling motion (translation + rotation):
$$KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$$For pure rolling ($v = R\omega$):
$$KE_{total} = \frac{1}{2}Mv^2\left(1 + \frac{I}{MR^2}\right) = \frac{1}{2}Mv^2\left(1 + \frac{k^2}{R^2}\right)$$Comparing rolling objects down an incline:
Higher the $I/MR^2$ ratio, slower the object (more energy goes to rotation).
Ranking (fastest to slowest):
- Solid sphere: $I/MR^2 = 2/5$ → fastest
- Disc/Cylinder: $I/MR^2 = 1/2$ → medium
- Ring/Hollow cylinder: $I/MR^2 = 1$ → slowest
Remember: “Solid Speeds, Ring Retards”
When to Use Moment of Inertia
Use moment of inertia when:
- Rotational motion about a fixed or moving axis
- Calculating rotational kinetic energy ($KE = \frac{1}{2}I\omega^2$)
- Finding torque-acceleration relation ($\tau = I\alpha$)
- Rolling motion problems (disc, sphere on incline)
- Problems involving angular momentum ($L = I\omega$)
Key question: “Is the object rotating?” → Yes → Use $I$
Common Mistakes to Avoid
Mistake: Treating $I$ as a constant like mass.
Truth: Unlike mass, $I$ depends on the axis of rotation. Same object, different axis = different $I$!
Example: Rod of length $L$, mass $M$:
- About center: $I = \frac{ML^2}{12}$
- About end: $I = \frac{ML^2}{3}$ (4 times larger!)
Mistake: Using $I = \frac{MR^2}{2}$ for a disc about its diameter.
Correct:
- Perpendicular to plane (through center): $I = \frac{MR^2}{2}$
- Along diameter (in plane): $I = \frac{MR^2}{4}$
Always check: Which axis is the rotation about?
Mistake: Writing $I$ without proper units.
Correct units: kg·m² (SI), g·cm² (CGS)
Check: $[I] = [m][r^2] = \text{kg} \cdot \text{m}^2$
Mistake: Thinking hollow and solid objects have same $I$.
Truth: For same mass and radius, hollow objects have larger $I$ (mass farther from axis).
Example:
- Hollow sphere: $I = \frac{2}{3}MR^2$
- Solid sphere: $I = \frac{2}{5}MR^2$
- Ratio: $\frac{2/3}{2/5} = \frac{5}{3}$ (hollow has 1.67 times more!)
Worked Examples
Level 1: Foundation (NCERT)
Four point masses each of 1 kg are placed at the corners of a square of side 2 m. Find the moment of inertia about an axis passing through the center and perpendicular to the plane.
Solution:
Distance of each mass from center: $r = \frac{\text{diagonal}}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2}$ m
$$I = \sum m_i r_i^2 = 4 \times 1 \times (\sqrt{2})^2 = 4 \times 2 = \boxed{8 \text{ kg·m}^2}$$A uniform rod of mass 2 kg and length 1 m is rotating about an axis through its center and perpendicular to its length with angular velocity 10 rad/s. Find: a) Moment of inertia b) Rotational kinetic energy
Solution:
a) $I = \frac{ML^2}{12} = \frac{2 \times 1^2}{12} = \boxed{\frac{1}{6} \text{ kg·m}^2}$
b) $KE = \frac{1}{2}I\omega^2 = \frac{1}{2} \times \frac{1}{6} \times 10^2 = \frac{100}{12} = \boxed{8.33 \text{ J}}$
A disc of radius 0.5 m and mass 4 kg is rotating about an axis through its center and perpendicular to its plane at 120 rpm. Find the rotational kinetic energy.
Solution:
Convert rpm to rad/s: $\omega = \frac{120 \times 2\pi}{60} = 4\pi$ rad/s
$I = \frac{MR^2}{2} = \frac{4 \times (0.5)^2}{2} = 0.5$ kg·m²
$KE = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.5 \times (4\pi)^2 = 0.25 \times 16\pi^2 = \boxed{39.5 \text{ J}}$
Level 2: JEE Main
A ring and a disc of same mass $M$ and radius $R$ are rotating with same angular velocity $\omega$. Find the ratio of their rotational kinetic energies.
Solution:
Ring: $I_1 = MR^2$, $KE_1 = \frac{1}{2}MR^2\omega^2$
Disc: $I_2 = \frac{MR^2}{2}$, $KE_2 = \frac{1}{2} \times \frac{MR^2}{2} \times \omega^2 = \frac{MR^2\omega^2}{4}$
$$\frac{KE_1}{KE_2} = \frac{\frac{1}{2}MR^2\omega^2}{\frac{1}{4}MR^2\omega^2} = \boxed{2:1}$$Ring has twice the kinetic energy (higher moment of inertia)!
A uniform rod of mass $M$ and length $L$ is rotating about an axis perpendicular to it at distance $L/4$ from one end. Find the moment of inertia.
Solution:
Method: Use parallel axis theorem (we’ll learn this in next topic!)
Axis is at distance $d = L/4 - L/2 = -L/4$ from center (magnitude: $L/4$)
$$I = I_{cm} + Md^2 = \frac{ML^2}{12} + M\left(\frac{L}{4}\right)^2$$ $$I = \frac{ML^2}{12} + \frac{ML^2}{16} = \frac{4ML^2 + 3ML^2}{48} = \boxed{\frac{7ML^2}{48}}$$A ring of mass $m$ and radius $r$ is placed concentrically on a disc of mass $M$ and radius $R$ (where $R > r$). Find the moment of inertia of the system about the common axis.
Solution:
System moment of inertia = Sum of individual moments of inertia
Ring: $I_1 = mr^2$
Disc: $I_2 = \frac{MR^2}{2}$
$$I_{total} = I_1 + I_2 = \boxed{mr^2 + \frac{MR^2}{2}}$$Level 3: JEE Advanced
A ring, disc, and solid sphere, all of same mass and radius, are released from rest at the top of an inclined plane. They roll down without slipping. Find: a) Which reaches the bottom first? b) Ratio of their speeds at the bottom
Solution:
Using energy conservation:
For pure rolling: $v = R\omega$
$$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2\left(1 + \frac{I}{MR^2}\right)$$ $$v = \sqrt{\frac{2gh}{1 + I/MR^2}}$$Ring: $I/MR^2 = 1$ → $v_{\text{ring}} = \sqrt{\frac{2gh}{2}} = \sqrt{gh}$
Disc: $I/MR^2 = 1/2$ → $v_{\text{disc}} = \sqrt{\frac{2gh}{3/2}} = \sqrt{\frac{4gh}{3}}$
Sphere: $I/MR^2 = 2/5$ → $v_{\text{sphere}} = \sqrt{\frac{2gh}{7/5}} = \sqrt{\frac{10gh}{7}}$
a) Order (fastest first): Sphere > Disc > Ring
b) Ratio of speeds:
$$v_{\text{sphere}} : v_{\text{disc}} : v_{\text{ring}} = \sqrt{\frac{10}{7}} : \sqrt{\frac{4}{3}} : \sqrt{1}$$ $$= \sqrt{30} : \sqrt{28} : \sqrt{21}$$ $$\boxed{\approx 5.48 : 5.29 : 4.58}$$Find the moment of inertia of a uniform disc of mass $M$ and radius $R$ with a circular hole of radius $R/2$ cut from it. The center of the hole is at distance $R/2$ from the center of the disc. Axis passes through the center of the original disc, perpendicular to its plane.
Solution:
Method: Original disc - Removed portion
Original disc: $I_1 = \frac{MR^2}{2}$
Removed portion:
- Radius: $r = R/2$
- Mass: $m = M \times \frac{\pi(R/2)^2}{\pi R^2} = \frac{M}{4}$
- Distance from axis: $d = R/2$
Using parallel axis theorem for removed part:
$$I_2 = I_{cm} + md^2 = \frac{mr^2}{2} + md^2$$ $$I_2 = \frac{M/4 \times (R/2)^2}{2} + \frac{M}{4} \times \left(\frac{R}{2}\right)^2$$ $$I_2 = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{MR^2 + 2MR^2}{32} = \frac{3MR^2}{32}$$Remaining portion:
$$I = I_1 - I_2 = \frac{MR^2}{2} - \frac{3MR^2}{32} = \frac{16MR^2 - 3MR^2}{32} = \boxed{\frac{13MR^2}{32}}$$A solid cylinder of mass $M$, radius $R$, and height $h$ rests on a horizontal surface. What minimum horizontal force $F$ applied at the top will cause it to topple? (Assume sufficient friction to prevent slipping)
Solution:
For toppling, cylinder rotates about the edge point of contact.
For toppling to begin: Torque about edge must overcome the restoring torque due to weight.
Torque by $F$: $\tau_F = F \times h$ (clockwise)
Torque by weight: $\tau_w = Mg \times R$ (anticlockwise)
At the verge of toppling:
$$F \times h = Mg \times R$$ $$\boxed{F = \frac{MgR}{h}}$$Note: If $h > R$, toppling is easier (smaller $F$ needed). This is why tall objects topple easily!
Real-World Applications
Figure Skating Spins: Skater reduces $I$ by pulling arms in → conservation of angular momentum → spins faster
Tightrope Walkers: Use long poles to increase $I$ → harder to tip over (more stable)
Flywheels in Engines: Large $I$ → stores rotational energy smoothly
Helicopter Blades: Mass at tips → large $I$ → maintains rotation even if engine fails momentarily
Athletic Equipment:
- Longer bats/golf clubs → larger $I$ → more power but harder to swing
- Gymnasts tuck during somersaults → reduce $I$ → spin faster
Quick Revision Box
| Body | Axis | Moment of Inertia | $k$ (radius of gyration) |
|---|---|---|---|
| Rod (length $L$) | Through center, ⊥ | $\frac{ML^2}{12}$ | $\frac{L}{2\sqrt{3}}$ |
| Rod (length $L$) | Through end, ⊥ | $\frac{ML^2}{3}$ | $\frac{L}{\sqrt{3}}$ |
| Ring (radius $R$) | Through center, ⊥ plane | $MR^2$ | $R$ |
| Disc (radius $R$) | Through center, ⊥ plane | $\frac{MR^2}{2}$ | $\frac{R}{\sqrt{2}}$ |
| Solid Sphere (radius $R$) | Any diameter | $\frac{2MR^2}{5}$ | $\sqrt{\frac{2}{5}}R$ |
| Hollow Sphere (radius $R$) | Any diameter | $\frac{2MR^2}{3}$ | $\sqrt{\frac{2}{3}}R$ |
| Solid Cylinder (radius $R$) | Along axis | $\frac{MR^2}{2}$ | $\frac{R}{\sqrt{2}}$ |
Memory aid: Hollow > Solid, Always!
Practice Problems
Foundation Level
Calculate the moment of inertia of a thin rod of mass 500 g and length 50 cm about an axis through its center.
Three point masses 2 kg, 3 kg, and 4 kg are placed at positions (0,0), (1,1), and (2,1) meters. Find $I$ about z-axis.
A disc of mass 2 kg and radius 20 cm rotates at 300 rpm. Find its rotational KE.
JEE Main Level
Two uniform rods each of mass $M$ and length $L$ are joined to form a cross (+). Find the moment of inertia about an axis passing through the intersection and perpendicular to the plane.
A solid sphere and hollow sphere of same mass and radius roll down an incline of height $h$. Find the ratio of their speeds at the bottom.
A thin uniform circular ring of mass $M$ and radius $R$ is rotating about an axis through its center and perpendicular to its plane with angular velocity $\omega$. Find work required to stop it.
JEE Advanced Level
A uniform disc of mass $M$ and radius $R$ is mounted on a fixed horizontal axle. A block of mass $m$ hangs from a massless string wound around the rim. Find: a) Acceleration of the block b) Tension in the string
Four point masses each of mass $m$ are connected by massless rods to form a square of side $a$. The system rotates about an axis perpendicular to the plane and passing through one corner. Find the moment of inertia.
A hollow cylinder and solid cylinder of same mass and radius are given the same kinetic energy and released on an incline. Which travels farther up the incline?
Connection to Other Topics
Prerequisites you should know:
- Centre of Mass - Understanding mass distribution
- Kinetic Energy - Energy concepts
What’s next:
- Parallel & Perpendicular Axis Theorems - Calculate $I$ for different axes
- Torque - Rotational analog of force
- Angular Momentum - Uses $L = I\omega$
Cross-chapter links:
- Rolling Motion - Applies moment of inertia in energy calculations
- Work-Energy Theorem - Extended to rotation
Teacher’s Summary
Moment of inertia is “rotational mass” - it measures resistance to angular acceleration, just like mass resists linear acceleration
Distance is key: $I \propto r^2$ - mass farther from axis contributes quadratically more to $I$
Memory pattern for standard shapes:
- DRISH: Disc (1/2), Rod-center (1/12), I-Rod-end (1/3), Sphere (2/5), Hollow (2/3)
- Hollow always > Solid for same mass and radius
Different axis → Different $I$ - Unlike mass, moment of inertia depends on the axis of rotation!
JEE pattern: Rolling races, composite bodies, and energy conservation problems are favorites
Quick check: Units must be kg·m² - if not, you’ve made a mistake!
“Mass says ‘how much matter,’ moment of inertia says ‘how matter is distributed around the axis.’”
Master these formulas with DRISH, and you’ll crack 2-3 questions worth 8-12 marks in JEE. More importantly, you’ll understand why figure skaters become a blur when they tuck in - physics in action!