System of Particles and Rotational Motion for NEET UG
Rotation has its own versions of mass, force and momentum. Once you see the parallels with straight-line motion, the formulas follow naturally.
PhysicsClass 1167 MCQs3 lessons12 flashcards3 memory tricks
Practise System of Particles and Rotational Motion free
What you'll learn
- Centre of mass
Find the centre of mass of a system of particles and know how it moves. - Torque and moment of inertia
Calculate torque, and use moment of inertia and the parallel-axis theorem. - Conservation of angular momentum and rolling
Use Iω = constant and the energy of rolling bodies.
Key formulas
| Two particles on a line | x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂) |
|---|---|
| Many particles | r_cm = Σmᵢrᵢ / Σmᵢ |
| Ring about its axis | I = MR² |
| Disc about its axis | I = ½MR² |
| Solid sphere about a diameter | I = ⅖MR² |
| Rod about its centre | I = ML²/12 |
| Parallel-axis theorem | I = I_cm + Md² |
| Conservation | I₁ω₁ = I₂ω₂ |
Memory tricks
Solid sphere (k²/R² = 2/5) beats disc (1/2), which beats hollow sphere (2/3), which beats ring (1). A smaller k²/R² means more energy goes into moving forward rather than spinning.
L = Iω stays constant with no external torque. Pulling mass inwards lowers I, so ω increases. Dancers and divers use the same trick.
Torque τ = rF sin θ. Pushing far from the axis (large r) gives more turning effect for the same force.
Try 3 NEET-style questions
Q1. A solid sphere and a hollow sphere of the same mass and radius rotate with the same angular speed. The ratio of their rotational kinetic energies (solid : hollow) is
- 5 : 3
- 1 : 1
- 2 : 3
- 3 : 5
Show answer
Answer: (4) 3 : 5
I_solid = (2/5)MR², I_hollow = (2/3)MR². K ∝ I, so 2/5 : 2/3 = 3 : 5.
Q2. A skater spinning at 2 rev/s with arms outstretched pulls in her arms, reducing her moment of inertia to half. Her new rate of spin is
- 4 rev/s
- 1 rev/s
- 2 rev/s
- 8 rev/s
Show answer
Answer: (1) 4 rev/s
Iω = constant, so halving I doubles ω: 4 rev/s.
Q3. The moment of inertia of a thin uniform rod of mass M and length L about an axis through its centre and perpendicular to it is
- ML²/3
- ML²/12
- ML²/2
- ML²
Show answer
Answer: (2) ML²/12
For a thin rod about its centre, perpendicular to its length, I = ML²/12. About an end it is ML²/3.
Practise all 67 questions Free · explanations for every option · see where you go wrong