1. Chapter at a Glance
Real bodies do not just slide — they also spin. System of Particles and Rotational Motion introduces the centre of mass and the rotational versions of Newton's laws: torque instead of force, moment of inertia instead of mass, and angular momentum instead of linear momentum. It ends with rolling motion. This is one of the most important NEET mechanics chapters, usually worth two questions.
- Moment of inertia of standard bodies; parallel and perpendicular axis theorems.
- Radius of gyration.
- Torque as r × F; τ = Iα.
- Conservation of angular momentum (skater, discs joined together).
- Rolling bodies on an incline — who reaches first, speeds and KE ratios.
2. Centre of Mass
- The centre of mass is the point that moves as if the whole mass were concentrated there and all external forces acted on it: M acm = Fext.
- For particles: xcm = Σmixi/Σmi (similarly for y and z).
- For two particles, the centre of mass lies on the line joining them, closer to the heavier one: m₁r₁ = m₂r₂.
- Internal forces cannot change the motion of the centre of mass — the pieces of an exploding shell keep their centre of mass on the original parabola.
- For uniform symmetric bodies (rod, ring, disc, sphere), the centre of mass is at the geometric centre; it may lie outside the material (ring).
3. Torque and Rotational Kinematics
- Angular displacement θ, angular velocity ω = dθ/dt, angular acceleration α = dω/dt. For constant α: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ.
- Links to linear quantities: v = ωr, at = αr.
- Torque τ = r × F, magnitude rF sinθ = force × perpendicular distance. Unit N m.
- Rotational form of Newton's second law: τ = Iα.
- Equilibrium of a rigid body: net force = 0 (no translation) and net torque = 0 (no rotation). A couple (equal, opposite, non-collinear forces) gives torque without net force.
4. Moment of Inertia
Moment of inertia I = Σmr² measures how hard it is to change a body's rotation. It depends on the mass, how the mass is distributed, and the axis. Writing I = MK² defines the radius of gyration K.
| Body | Axis | I | NEET Note |
|---|---|---|---|
| Thin ring | Centre, ⊥ plane | MR² | All mass at distance R |
| Thin ring | Diameter | ½MR² | Perpendicular axis theorem |
| Disc / solid cylinder | Centre, ⊥ plane / own axis | ½MR² | K = R/√2 |
| Disc | Diameter | ¼MR² | Half of ½MR² |
| Solid sphere | Diameter | ⅖MR² | K = √(2/5) R |
| Hollow sphere | Diameter | ⅔MR² | Larger than solid sphere |
| Thin rod (length L) | Centre, ⊥ rod | ML²/12 | — |
| Thin rod | End, ⊥ rod | ML²/3 | Parallel axis: + M(L/2)² |
- Parallel axis theorem: I = Icm + Md², where d is the distance between the parallel axes.
- Perpendicular axis theorem (flat bodies only): Iz = Ix + Iy.
5. Angular Momentum and Rolling
- Angular momentum L = r × p; for a rigid body about a fixed axis L = Iω. τ = dL/dt.
- Conservation of angular momentum: if τext = 0, Iω is constant. A skater or diver pulls in the arms (I decreases) to spin faster.
- Rotational KE = ½Iω²; work done by a torque W = τθ; power P = τω.
- Rolling without slipping: vcm = ωR. The contact point is momentarily at rest; the top point moves at 2vcm.
- Total KE of a rolling body = ½mv²(1 + K²/R²).
| Body rolling down an incline | K²/R² | Acceleration | Order of arrival |
|---|---|---|---|
| Solid sphere | 2/5 | (5/7) g sinθ | First |
| Disc / solid cylinder | 1/2 | (2/3) g sinθ | Second |
| Hollow sphere | 2/3 | (3/5) g sinθ | Third |
| Ring | 1 | (1/2) g sinθ | Last |
6. Formula Master Table
| Linear quantity | Rotational analogue |
|---|---|
| Displacement s | θ |
| Mass m | I = Σmr² |
| Force F = ma | τ = Iα |
| Momentum p = mv | L = Iω |
| KE ½mv² | ½Iω² |
| Work Fs, power Fv | τθ, τω |
| Parallel axis | I = Icm + Md² |
| Rolling on incline | a = g sinθ/(1 + K²/R²); v = √(2gh/(1 + K²/R²)) |
7. Master Summary Table
| Concept | Key point | NEET Note |
|---|---|---|
| Centre of mass | Macm = Fext | Unchanged by internal forces |
| Torque | τ = r × F | Zero if force passes through the axis |
| Moment of inertia | Depends on axis and mass distribution | Ring > disc for same M, R |
| Radius of gyration | I = MK² | Disc: R/√2; solid sphere: √(2/5)R |
| Angular momentum | Conserved if τext = 0 | KE may change even when L is conserved |
| Rolling | v = ωR | Smaller K²/R² reaches bottom first |
8. Mind Map
- Σmx/Σm
- Motion under Fext
- r × F
- τ = Iα
- Equilibrium
- Standard values
- Axis theorems
- Radius of gyration
- L = Iω
- Conservation
- Rolling
9. Tips and Tricks
- Rolling KE split (translational : rotational): ring 1 : 1, disc 2 : 1, solid sphere 5 : 2.
- Mass and radius do not decide which body wins an incline race — only the shape (K²/R²) does.
- When two rotating bodies are joined, use I₁ω₁ + I₂ω₂ = (I₁ + I₂)ω; KE is always lost.
- With L constant, KE = L²/2I: halving I doubles the KE.
- Applying the perpendicular axis theorem to a sphere or cylinder (only for plane bodies).
- Using the parallel axis theorem between two axes when neither passes through the centre of mass.
- Thinking friction does work in pure rolling — the contact point does not slip.
10. Important One-liners
- Moment of inertia plays the role of mass in rotation.
- The centre of mass of a ring lies at its centre, where there is no mass.
- A diver curls up to spin faster — conservation of angular momentum.
- Torque is zero when the force passes through the axis of rotation.
- For the same mass and radius, a ring has a larger moment of inertia than a disc.
- In pure rolling, the lowest point of the wheel is momentarily at rest.
- Kepler's law of areas follows from conservation of angular momentum.
11. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take g = 10 m s⁻². Click the button to reveal each answer and its working.
Average level
Find the moment of inertia of a 2 kg thin ring of radius 0.5 m about (a) an axis through its centre perpendicular to its plane, (b) a diameter.
Show answer & solution
Concept: I = MR²; perpendicular axis theorem gives the diameter value.
- (a) I = 2 × 0.25 = 0.5 kg m².
- (b) Iz = Ix + Iy and Ix = Iy, so Id = 0.5/2 = 0.25 kg m².
Answer: 0.5 kg m²; 0.25 kg m²
A force F = (2î + 3ĵ) N acts at a point whose position vector is r = (î + 2ĵ) m. Find the torque about the origin.
Show answer & solution
Concept: τ = r × F; for vectors in the x–y plane, τz = xFy − yFx.
- τz = (1)(3) − (2)(2) = 3 − 4 = −1.
- τ = −1 k̂ N m (magnitude 1 N m, clockwise).
Answer: −k̂ N m
A 2 kg mass is at x = 0 and a 3 kg mass at x = 5 m. Locate their centre of mass.
Show answer & solution
Concept: xcm = Σmx/Σm.
- xcm = (2 × 0 + 3 × 5)/(2 + 3) = 15/5.
- = 3 m — closer to the heavier 3 kg mass, as expected.
Answer: x = 3 m
A flywheel with I = 0.4 kg m² spins at 20 rad s⁻¹. Find its angular momentum and rotational kinetic energy.
Show answer & solution
Concept: L = Iω; K = ½Iω².
- L = 0.4 × 20 = 8 kg m² s⁻¹.
- K = ½ × 0.4 × 400 = 80 J.
Answer: 8 kg m² s⁻¹; 80 J
A wheel speeds up uniformly from 10 rad s⁻¹ to 30 rad s⁻¹ in 5 s. Find its angular acceleration and the angle turned.
Show answer & solution
Concept: rotational equations of motion.
- α = (30 − 10)/5 = 4 rad s⁻².
- θ = average ω × t = (10 + 30)/2 × 5 = 100 rad.
Answer: 4 rad s⁻²; 100 rad
Advanced level
A solid cylinder rolls without slipping from rest down a slope, descending a vertical height of 3 m. Find its speed at the bottom.
Show answer & solution
Concept: mgh = ½mv²(1 + K²/R²) with K²/R² = ½.
- gh = ½v² × 1.5 = 0.75v².
- v² = gh/0.75 = 4gh/3 = 4 × 10 × 3/3 = 40.
- v ≈ 6.32 m s⁻¹ (slower than √(2gh) = 7.75 m s⁻¹ for sliding).
Answer: about 6.32 m s⁻¹
A disc (M = 2 kg, R = 0.2 m) spins at 50 rad s⁻¹. An identical disc at rest is gently placed on it so that they rotate together. Find the common angular speed and the kinetic energy lost.
Show answer & solution
Concept: no external torque ⇒ angular momentum is conserved; KE is not.
- I of each disc = ½ × 2 × 0.04 = 0.04 kg m².
- 0.04 × 50 = (0.08)ω ⇒ ω = 25 rad s⁻¹.
- KE before = ½ × 0.04 × 2500 = 50 J; after = ½ × 0.08 × 625 = 25 J.
- Loss = 25 J (half, turned into heat by friction between the discs).
Answer: 25 rad s⁻¹; 25 J
A uniform rod of mass 3 kg and length 1 m carries a 0.5 kg point mass at one end. Find the moment of inertia of the system about an axis through the other end, perpendicular to the rod.
Show answer & solution
Concept: add the moments of inertia of the parts about the same axis.
- Rod about its end: ML²/3 = 3 × 1/3 = 1 kg m².
- Point mass at 1 m: mr² = 0.5 × 1 = 0.5 kg m².
- Total I = 1.5 kg m².
Answer: 1.5 kg m²
A 2 kg solid sphere rolls without slipping at 5 m s⁻¹. Find its total, translational and rotational kinetic energies.
Show answer & solution
Concept: Ktrans = ½mv²; Krot = ½Iω² = ½(⅖mR²)(v/R)² = ⅕mv².
- Ktrans = ½ × 2 × 25 = 25 J.
- Krot = ⅕ × 2 × 25 = 10 J.
- Total = 35 J (ratio 5 : 2).
Answer: 35 J total; 25 J and 10 J
A constant torque of 20 N m acts for 10 s on a wheel (I = 4 kg m²) that starts from rest. Find the final angular speed, angle turned and work done, and check the work–energy theorem.
Show answer & solution
Concept: τ = Iα; W = τθ = ΔK.
- α = 20/4 = 5 rad s⁻²; ω = 5 × 10 = 50 rad s⁻¹.
- θ = ½ × 5 × 100 = 250 rad.
- W = τθ = 20 × 250 = 5000 J.
- ΔK = ½ × 4 × 50² = 5000 J ✓.
Answer: 50 rad s⁻¹; 250 rad; 5000 J
12. NEET-pattern Questions with Detailed Answers
These questions are original and modelled on concepts repeatedly asked in NEET UG.
The radius of gyration of a disc about an axis through its centre and perpendicular to its plane is:
- R
- R/2
- R/√2
- √(2/5) R
Show solution
MK² = ½MR² ⇒ K = R/√2. √(2/5) R is for a solid sphere; R is for a ring.
Answer: (C)
A ballet dancer spinning with outstretched arms pulls them in. Her angular speed:
- Decreases
- Increases
- Remains the same
- Becomes zero
Show solution
No external torque, so Iω is constant. Pulling in the arms reduces I, so ω increases.
Answer: (B)
A ring, a disc, a solid sphere and a hollow sphere of equal mass and radius roll down the same incline from rest. The one reaching the bottom first is the:
- Ring
- Disc
- Solid sphere
- Hollow sphere
Show solution
a = g sinθ/(1 + K²/R²); the solid sphere has the smallest K²/R² (2/5), so it has the largest acceleration. The ring (K²/R² = 1) is last.
Answer: (C)
The moment of inertia of a uniform rod (mass M, length L) about an axis through one end perpendicular to it is:
- ML²/12
- ML²/3
- ML²/2
- ML²
Show solution
Parallel axis theorem: ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3.
Answer: (B)
For a disc rolling without slipping, the ratio of rotational to total kinetic energy is:
- 1 : 2
- 1 : 3
- 2 : 3
- 2 : 7
Show solution
Translational ½mv², rotational ¼mv², total ¾mv². Rotational : total = ¼ : ¾ = 1 : 3. (2 : 7 is for a solid sphere.)
Answer: (B)
A shell flying along a parabola explodes in mid-air into several pieces. The centre of mass of the pieces:
- Moves vertically downward
- Continues along the original parabola
- Stops at the point of explosion
- Moves horizontally
Show solution
The explosion is due to internal forces; the only external force is still gravity. So the centre of mass keeps following the same parabola (until a piece hits the ground).
Answer: (B)
If the Earth suddenly shrank to half its radius with no change in mass, the length of a day would become:
- 48 h
- 12 h
- 6 h
- 24 h
Show solution
L = Iω = (⅖MR²)(2π/T) is constant, so T ∝ R². Halving R makes T one-fourth: 24/4 = 6 h.
Answer: (C)
The moment of inertia of a thin ring about a tangent in its own plane is:
- MR²
- (3/2)MR²
- 2MR²
- ½MR²
Show solution
About a diameter I = ½MR². The tangent in the plane is parallel to a diameter at distance R: ½MR² + MR² = (3/2)MR². (2MR² is for a tangent perpendicular to the plane.)
Answer: (B)
13. Quick Revision Summary
- xcm = Σmx/Σm; internal forces do not move the centre of mass.
- τ = r × F = Iα; L = Iω, conserved when τext = 0.
- Ring MR², disc ½MR², solid sphere ⅖MR², hollow sphere ⅔MR², rod ML²/12 and ML²/3.
- I = Icm + Md²; Iz = Ix + Iy for flat bodies.
- Rolling: v = ωR; a = g sinθ/(1 + K²/R²); solid sphere wins the race.