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System of Particles and Rotational Motion – Class 11 Physics notes, important questions and MCQs

NCERT Class 11 Physics, chapter 6. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Extended bodies are studied using the centre of mass, torque, angular momentum and moment of inertia. Rotational motion has equations similar to linear motion.

Centre of mass and torque

Centre of mass: r_cm = Σmᵢrᵢ/M; it moves as if all external forces act on it.
Torque τ = r × F; angular momentum L = r × p = Iω; τ = dL/dt.
With no external torque, L is conserved (a skater pulling in arms spins faster).

Moment of inertia

I = Σmr²; ring about axis MR², disc ½MR², solid sphere (2/5)MR², rod about centre ML²/12.
Parallel axis: I = I_cm + Md²; perpendicular axis (plane lamina): I_z = I_x + I_y.
Rotational KE = ½Iω²; rolling KE = ½mv²(1 + k²/R²).

Key terms

Torque: turning effect of a force, τ = rF sin θ
Moment of inertia: rotational analogue of mass, I = Σmr²
Radius of gyration: distance k from the axis at which the whole mass may be assumed concentrated, I = Mk²

Formulas

τ = Iα
L = Iω
I = I_cm + Md²

Important questions with answers

Where is the centre of mass of two particles of 2 kg at x = 0 and 3 kg at x = 5 m?2 marks

x_cm = (2 × 0 + 3 × 5)/(2 + 3) = 15/5 = 3 m.

State the theorems of parallel and perpendicular axes and use one to find the MI of a disc about a diameter.3 marks

Parallel axes: I = I_cm + Md². Perpendicular axes (plane lamina): I_z = I_x + I_y. For a disc, I_z (axis through centre ⟂ plane) = ½MR², and by symmetry I_x = I_y, so 2I_d = ½MR² and I_d = ¼MR².

Derive the acceleration of a body rolling down an inclined plane without slipping and compare a ring, disc and solid sphere on an incline of 30° (g = 10).5 marks

Energy conservation: mgh = ½mv²(1 + k²/R²), so v² = 2gh/(1 + k²/R²). Along the slope (s = h/sin θ), a = g sin θ/(1 + k²/R²). With sin 30° = ½: ring (k²/R² = 1): a = 2.5 m/s²; disc (½): a = 5/1.5 ≈ 3.33 m/s²; solid sphere (2/5): a = 5/1.4 ≈ 3.57 m/s². The sphere reaches the bottom first, then the disc, then the ring.

Quick revision

No external torque ⇒ L conserved.
I depends on axis and mass distribution.
Rotational KE = ½Iω².

Exam tips

Rolling down an incline: solid sphere reaches first, then disc, then ring.
Use the analogy: x ↔ θ, v ↔ ω, a ↔ α, m ↔ I, F ↔ τ, p ↔ L — every linear formula has a rotational partner.

Practice MCQs

Rotational analogue of mass is
1. moment of inertia
2. torque
3. angular momentum
4. angular velocity
Show answer
Answer: (a) moment of inertia. Moment of inertia.
SI unit of torque is
1. N
2. N m
3. J/s
4. kg m²
Show answer
Answer: (b) N m. Force × distance.
Moment of inertia of a ring about its axis is
1. ½MR²
2. MR²
3. (2/5)MR²
4. MR²/12
Show answer
Answer: (b) MR². All mass at R.
A skater pulls in her arms; her angular speed
1. increases
2. decreases
3. stays same
4. becomes zero
Show answer
Answer: (a) increases. I decreases, L constant.
A torque of 20 N m acts on a wheel of moment of inertia 4 kg m². Angular acceleration is
1. 5 rad/s²
2. 80 rad/s²
3. 16 rad/s²
4. 0.2 rad/s²
Show answer
Answer: (a) 5 rad/s². τ/I.
MI of a uniform rod about one end is
1. ML²/12
2. ML²/2
3. ML²/3
4. ML²
Show answer
Answer: (c) ML²/3. ML²/12 + M(L/2)².
Which reaches the bottom of an incline first when rolled together
1. solid sphere
2. disc
3. ring
4. all together
Show answer
Answer: (a) solid sphere. Smallest k²/R².
A solid sphere rolls without slipping. The fraction of its total KE that is rotational is
1. 2/5
2. 2/7
3. 1/2
4. 5/7
Show answer
Answer: (b) 2/7. (1/5)/(7/10).

All chapters of Class 11 Physics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.