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Systems of Particles and Rotational Motion

Physics · Class 11NCERT pp. 92–1232 sets × 25 questions

Key facts and formulas for Systems of Particles and Rotational Motion, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.

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Formula sheet for this chapter

IdeaRule or fact
Centre of massX = Σ m x / Σ m; internal forces cannot move it
Torqueτ = r × F, magnitude r F sin θ; zero if line of action passes through the point
Angular momentumL = r × p; particle on a line: m v d; circle: m r² ω; conserved if no external torque
Rotational dynamicsτ = I α; K = ½ I ω²; P = τ ω; θ = ½ α t² from rest
Moments of inertiaRing M R²; disc ½ M R²; rod end M L²/3; rod middle M L²/12; solid sphere 2/5 M R²
TheoremsParallel: I = I_cm + M d²; perpendicular (plane lamina only): I_z = I_x + I_y
Radius of gyrationI = M k²; disc R/√2, ring R, rod middle L/√12
RollingK = ½ m v² (1 + k²/R²); a = g sin θ / (1 + k²/R²); top point 2v
EquilibriumΣF = 0 and Στ = 0; coplanar case has 3 independent conditions
CoupleMoment = F × perpendicular separation; net force zero

Three sample questions

Q1. Masses of 4 kg and 6 kg are 10 m apart. Their centre of mass is at what distance from the 4 kg mass?
  1. 4 m
  2. 6 m
  3. 5 m
  4. 2.5 m
Hint and solution

Hint: Put the 4 kg mass at the origin and use the weighted average of positions. The centre of mass lies closer to the heavier body.

  1. X = (m₁x₁ + m₂x₂) / (m₁ + m₂) — Place the 4 kg mass at x = 0 and the 6 kg mass at x = 10 m.
  2. X = (4 × 0 + 6 × 10) / (4 + 6) — Numerator is 60 kg m and total mass is 10 kg.
  3. X = 60 / 10 = 6 m — The centre of mass is 6 m from the 4 kg mass, closer to the heavier 6 kg body. Option B.
Answer: B. X = (4 x 0 + 6 x 10)/10 = 6 m, nearer the heavier mass.
Q2. Three masses 100 g, 200 g and 300 g are at 0, 30 cm and 60 cm along a rod. Their centre of mass is at
  1. 40 cm
  2. 45 cm
  3. 30 cm
  4. 50 cm
Hint and solution

Hint: Convert nothing; masses can stay in grams. Add each mass times its position, then divide by the total mass.

  1. X = Σ mᵢxᵢ / Σ mᵢ — Centre of mass is the mass weighted average of positions.
  2. Σ mx = 0 + 200 × 30 + 300 × 60 = 24000 — The first mass sits at the origin, so it adds zero. The others add 6000 and 18000.
  3. X = 24000 / 600 = 40 cm — Total mass is 600 g, so divide 24000 by 600. The centre of mass is at 40 cm. Option A.
Answer: A. X = (0 + 200 x 30 + 300 x 60)/600 = 24000/600 = 40 cm.
Q3. Masses of 1 kg, 2 kg and 3 kg lie at (0,0), (3,0) and (0,4) m. The centre of mass is at
  1. (2, 3)
  2. (1, 1)
  3. (1, 2)
  4. (1.5, 2)
Hint and solution

Hint: Find X and Y separately. Only the 2 kg mass has a non zero x, and only the 3 kg mass has a non zero y.

  1. X = Σ mx / Σ m, Y = Σ my / Σ m — Treat the x and y coordinates separately. Total mass is 1 + 2 + 3 = 6 kg.
  2. X = (1×0 + 2×3 + 3×0) / 6 = 6 / 6 = 1 — Only the 2 kg mass at x = 3 contributes.
  3. Y = (1×0 + 2×0 + 3×4) / 6 = 12 / 6 = 2 — Only the 3 kg mass at y = 4 contributes. The centre of mass is (1, 2) m. Option C.
Answer: C. X = (0 + 6 + 0)/6 = 1, Y = (0 + 0 + 12)/6 = 2.

How to use this chapter

Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.

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