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Chapter 9 · Class 11 Physics

System of Particles and Rotational Motion — Important Questions

34 questions With answers CBSE format

SUMMARY: This chapter explores the dynamics of systems of particles and the principles of rotational motion, including the concepts of center of mass and torque.
KEY TOPICS: center of mass, torque, angular momentum, moment of inertia, rotational kinetic energy, equilibrium of rigid bodies, theorems of perpendicular and parallel axes, rolling motion, conservation of angular momentum, rigid body dynamics

Previous chapter Oscillations Next chapter Thermal Properties of Matter

Multiple Choice Questions (MCQs) 5 questions

Q1 1 Mark

The moment of inertia of a uniform rod of mass M and length L about an axis through its centre and perpendicular to its length is:

AML²/12

BML²/3

CML²/6

DML²

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Correct answer: Option 1 — ML²/12

Q2 1 Mark

Torque is the rotational analogue of:

AVelocity

BForce

CMass

DEnergy

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Correct answer: Option 2 — Force

Q3 1 Mark

The angular momentum of a particle about a point is:

AL = m × v

BL = r × p

CL = m × r

DL = F × r

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Correct answer: Option 2 — L = r × p

Q4 1 Mark

For a body in pure rolling without slipping the linear velocity v of the centre of mass and angular velocity ω are related by:

Av = ω

Bv = rω

Cv = r/ω

Dv = ω/r

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Correct answer: Option 2 — v = rω

Q5 1 Mark

The centre of mass of a uniform circular ring lies:

AOn the ring

BAt its centre

COutside the ring

DCannot be determined

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Correct answer: Option 2 — At its centre

Short Answer Questions 5 questions

Q6 3 Marks

Define centre of mass and write the formula for a system of particles.

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Centre of mass (COM) is the unique point where the entire mass of the system can be considered to be concentrated for translational motion. For a system of particles: x_COM = Σ(m_i x_i)/Σm_i; similarly for y and z. For continuous distributions: x_COM = (1/M) ∫ x dm. The COM of an isolated system moves with constant velocity (Newton's first law for the system).

Q7 3 Marks

Define moment of inertia and write its formula for a discrete distribution.

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Moment of inertia (I) is the rotational equivalent of mass — measures the resistance to angular acceleration about an axis. For point particles: I = Σ m_i r_i² where r_i is the perpendicular distance of mass m_i from the axis. SI unit: kg·m². I depends on the mass distribution AND on the chosen axis.

Q8 3 Marks

State the parallel axes theorem.

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Parallel axis theorem: I about an axis parallel to one through the centre of mass equals the moment of inertia about the COM axis plus M·d² where M is total mass and d is the distance between the two parallel axes. I = I_COM + Md². Useful for finding I about non-COM axes from a known COM moment.

Q9 3 Marks

A wheel of moment of inertia 2 kg·m² is rotating at 10 rad/s. Find its rotational kinetic energy.

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KE_rotational = (1/2) I ω² = (1/2)(2)(10²) = 100 J. The formula is the rotational analogue of (1/2)mv² with I replacing m and ω replacing v.

Q10 3 Marks

State and explain the principle of conservation of angular momentum.

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In the absence of external torque the total angular momentum of a system is conserved: L = Iω = constant. So if I changes (e.g. by a skater pulling arms in) ω must change inversely. This is why skaters spin faster when arms come in. Mathematically follows from τ = dL/dt — with τ_ext = 0 dL/dt = 0.

Long Answer Questions 6 questions

Q11 6 Marks

A solid sphere of radius R rolls down an inclined plane of height h without slipping. Find its velocity at the bottom.

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Energy conservation: mgh = (1/2)mv² + (1/2)Iω². For solid sphere I = (2/5)mR² and v = Rω so ω = v/R. (1/2)Iω² = (1/2)(2/5)mR²(v²/R²) = (1/5)mv². Total KE = (1/2)mv² + (1/5)mv² = (7/10)mv². So mgh = (7/10)mv² ⇒ v = √(10gh/7). Note v < √(2gh) — rolling is slower than sliding because some KE is in rotation.

Q12 6 Marks

State and prove the perpendicular axes theorem for a planar body.

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Perpendicular axes theorem: for a planar body in the xy-plane the moment of inertia about an axis perpendicular to the plane (z-axis) at any point equals the sum of the moments of inertia about two perpendicular axes (x and y) lying in the plane and intersecting at that point: I_z = I_x + I_y. Proof: for an element at (x, y) the distance from x-axis is y from y-axis is x and from z-axis is √(x²+y²). So I_x = ∫y² dm I_y = ∫x² dm and I_z = ∫(x² + y²) dm = I_x + I_y.

Q13 6 Marks

A force F = 5 N is applied perpendicularly to the end of a 0.4 m long rod. Calculate the torque about the other end. Then if the rod has moment of inertia 0.2 kg·m² about that end, find its angular acceleration.

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Torque τ = F × r = 5 × 0.4 = 2 N·m. Angular acceleration α = τ/I = 2/0.2 = 10 rad/s² (rotational analogue of a = F/m).

Q14 6 Marks

Discuss the kinetic energy of a rigid body in rolling motion.

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Total KE of a rolling body = translational KE + rotational KE = (1/2) M v_COM² + (1/2) I ω². For pure rolling v_COM = R ω so ω = v/R. Substituting: KE_total = (1/2) M v² + (1/2)(I/R²) v² = (1/2) M v² (1 + I/(MR²)). The factor (1 + I/(MR²)) depends on shape: ring → 2; disc → 3/2; sphere → 7/5. So a sphere rolls fastest down a slope a ring slowest (more KE goes into rotation for higher I/MR² shapes).

Q15 6 Marks

A skater with arms outstretched has moment of inertia 4 kg·m² and rotates at 2 rad/s. When she pulls her arms in her moment of inertia becomes 1 kg·m². Find her new angular velocity and the change in rotational kinetic energy.

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By conservation of angular momentum: I₁ ω₁ = I₂ ω₂ ⇒ 4(2) = 1 ω₂ ⇒ ω₂ = 8 rad/s. Initial KE = (1/2)(4)(2²) = 8 J. Final KE = (1/2)(1)(8²) = 32 J. Change = 32 − 8 = 24 J (increase). The extra energy comes from the work the skater's muscles did pulling her arms in against the centripetal pseudo-force.

Q16 6 Marks

Compare centre of mass and centre of gravity with the help of a table.

Assertion–Reason Questions 5 questions

Q17 1 Mark

Assertion (A): The centre of mass of a uniform body lies at its geometric centre.

Reason (R): For a body with uniform density the mass is symmetrically distributed about the geometric centre.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.

Q18 1 Mark

Assertion (A): Torque is the rotational analogue of force.

Reason (R): Just as force changes linear momentum torque changes angular momentum: τ = dL/dt.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.

Q19 1 Mark

Assertion (A): A solid sphere has lower moment of inertia than a hollow sphere of the same mass and radius about the same axis.

Reason (R): The mass of a solid sphere is distributed closer to the centre than for a hollow sphere — so the average r² is smaller.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.

Q20 1 Mark

Assertion (A): A figure skater spins faster when she pulls her arms in.

Reason (R): Conservation of angular momentum: as I decreases ω must increase to keep L = Iω constant.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.

Q21 1 Mark

Assertion (A): A solid cylinder rolls down a slope faster than a hollow cylinder.

Reason (R): The solid cylinder has lower I/MR² so a smaller fraction of total KE goes into rotation — more KE remains for translation.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.

Statement-Based Questions 5 questions

Q22 1 Mark

Statement 1: The COM of a system moves as if all external forces acted at it.

Statement 2: The COM of an isolated system moves with constant velocity.

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Correct answer: Option 1 — Both statements are true.

Q23 1 Mark

Statement 1: Torque is the rotational analogue of force.

Statement 2: τ = r × F where r is the position vector and F the applied force.

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Correct answer: Option 1 — Both statements are true.

Q24 1 Mark

Statement 1: Moment of inertia depends on the mass distribution and the axis.

Statement 2: A larger moment of inertia gives larger angular momentum at the same angular velocity.

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Correct answer: Option 1 — Both statements are true.

Q25 1 Mark

Statement 1: Angular momentum is the rotational analogue of linear momentum.

Statement 2: In the absence of external torque angular momentum is conserved.

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Correct answer: Option 1 — Both statements are true.

Q26 1 Mark

Statement 1: For pure rolling v = Rω.

Statement 2: Pure rolling occurs when the contact point is momentarily at rest with respect to the surface.

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Correct answer: Option 1 — Both statements are true.

Case Study / Passage Questions 3 questions

Q27 3 Marks

Two particles of masses 2 kg and 3 kg are placed at positions (1 m 0) and (4 m 0) respectively. The student wants to find the position of the centre of mass and compute the total momentum if the 2 kg particle moves at 5 m/s in +x direction and 3 kg at 2 m/s in −x direction.

The x-coordinate of the centre of mass equals:

A2 m

B2.5 m

C2.8 m

D3 m
The total momentum of the system (in +x direction) equals:

A4 kg·m/s

B8 kg·m/s

C12 kg·m/s

D16 kg·m/s
Compute the velocity of the centre of mass.

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1. Option 3 — 2.8 m

2. Option 1 — 4 kg·m/s

3. x_COM = (Σm_ix_i)/(Σm_i) = (2×1 + 3×4)/(2 + 3) = (2 + 12)/5 = 14/5 = 2.8 m. Total momentum p_total = 2(5) + 3(−2) = 10 − 6 = 4 kg·m/s in +x direction. Velocity of COM v_COM = p_total/M_total = 4/5 = 0.8 m/s in +x. Useful: in absence of external force the COM moves with constant velocity.

Q28 3 Marks

A figure skater with arms outstretched has moment of inertia 6 kg·m² and rotates at 1 rad/s. She pulls in her arms reducing her moment of inertia to 2 kg·m². The coach wants to find her new angular velocity the change in her rotational kinetic energy and explain where the extra energy comes from.

Her new angular velocity equals:

A1 rad/s

B2 rad/s

C3 rad/s

D6 rad/s
Her rotational kinetic energy:

ATripled

BSame

CHalved

DDoubled
Why does her kinetic energy increase when angular momentum is conserved?

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1. Option 3 — 3 rad/s

2. Option 1 — Tripled

3. By conservation of angular momentum (no external torque): I₁ω₁ = I₂ω₂ ⇒ 6(1) = 2(ω₂) ⇒ ω₂ = 3 rad/s. KE₁ = (1/2)(6)(1²) = 3 J. KE₂ = (1/2)(2)(3²) = 9 J. KE tripled. The extra energy (6 J) comes from the work the skater's muscles do pulling her arms in against the centrifugal pseudo-force experienced in her rotating frame.

Q29 3 Marks

A solid sphere a hollow sphere and a solid cylinder all of the same mass and radius are released from rest at the top of an incline of height h. They roll down without slipping. The student wants to determine the order of arrival at the bottom and the velocity of the solid sphere.

The velocity of the solid sphere at the bottom is:

A√(gh)

B√(2gh)

C√(10gh/7)

D√(6gh/5)
Order of arrival at the bottom (fastest first):

ASolid sphere first

BHollow sphere first

CSolid cylinder first

DAll same
Why does the solid sphere reach the bottom first?

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1. Option 3 — √(10gh/7)

2. Option 1 — Solid sphere first

3. For rolling: mgh = (1/2)mv² + (1/2)Iω²; with v = Rω and I = β·MR²: mgh = (1/2)mv²(1 + β). v = √(2gh/(1 + β)). Higher β → slower rolling. Solid sphere (β = 2/5): v = √(10gh/7). Solid cylinder (β = 1/2): v = √(4gh/3). Hollow sphere (β = 2/3): v = √(6gh/5). Solid sphere has lowest β → fastest. Order: solid sphere > solid cylinder > hollow sphere.

Table-Based Questions 4 questions

Q30 3 Marks

Study moments of inertia of common shapes about their COM:

Body	Axis	I (about COM)
Thin ring	through centre perp	MR²
Solid disc	through centre perp	(1/2)MR²
Solid sphere	through centre	(2/5)MR²
Hollow sphere (thin shell)	through centre	(2/3)MR²
Thin rod (length L)	through centre perp	(1/12)ML²

The moment of inertia of a solid sphere about its diameter is:

AMR²

B(1/2)MR²

C(2/5)MR²

D(2/3)MR²
The moment of inertia of a uniform thin rod about its centre perpendicular to its length is:

A(1/3)ML²

B(1/12)ML²

C(1/4)ML²

DML²
Compare a solid disc and a hollow ring of equal mass and radius — which would roll faster down an incline?

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1. Option 3 — (2/5)MR²

2. Option 2 — (1/12)ML²

3. Moment of inertia depends on the mass distribution and the chosen axis. Solid bodies have lower I than hollow bodies of same mass and outer radius (mass concentrated near centre). The numeric coefficient (β = I/MR²) determines rotational dynamics: lower β = easier to spin up. Useful for predicting rolling speed flywheel performance and gyroscopic stability.

Q31 3 Marks

Study the analogy between linear and rotational quantities:

Linear	Rotational
Mass m	Moment of inertia I
Velocity v	Angular velocity ω
Force F	Torque τ
Linear momentum p = mv	Angular momentum L = Iω
F = ma	τ = Iα
KE = (1/2)mv²	KE = (1/2)Iω²

The rotational analogue of mass is:

AMass

BMoment of inertia

CForce

DTorque
The rotational analogue of F = ma is τ =:

AIα

Bmω

CIv

Dm·α
Write the rotational analogue of Newton's second law: F = dp/dt.

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1. Option 2 — Moment of inertia

2. Option 1 — Iα

3. Linear and rotational mechanics have direct analogies. Replacing m with I and v with ω (along with the substitution of force-quantities like F → τ p → L) lets us transcribe linear equations to rotational form. This is why the motion of a rolling sphere requires both translational and rotational equations to describe fully — the analogy is structural not just computational.

Q32 6 Marks

Compute the moment of inertia about the indicated axis for each shape (mass M, radius R or length L as relevant).

Body	Axis
Thin ring	Through centre perpendicular
Solid disc	Through centre perpendicular
Solid sphere	Diameter
Hollow sphere	Diameter
Thin rod (length L)	Through centre perpendicular
Thin rod (length L)	Through one end

Q33 6 Marks

A solid sphere a hollow sphere and a solid disc each of mass M and radius R roll without slipping down an incline of height h. Compute the velocity at the bottom of the incline for each.

Body	I about axis	β = I/(MR²)
Solid sphere	(2/5)MR²	2/5
Hollow sphere	(2/3)MR²	2/3
Solid disc	(1/2)MR²	1/2

Picture-Based Questions 1 question

Q34 3 Marks

Study the rolling sphere on the inclined plane and answer:

System of Particles and Rotational Motion figure

For pure rolling (without slipping), the velocity of the centre of mass and angular velocity are related by:

Av_cm = R ω

Bv_cm = 2R ω

Cv_cm = R ω²

Dv_cm = ω/R
The contact point of the rolling sphere with the incline is:

AMaximum velocity

BMaximum acceleration

CMomentarily at rest

DEqual to v_cm
Explain why the contact point is at rest during rolling and derive v_cm = Rω.

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1. Option 1 — v_cm = R ω

2. Option 3 — Momentarily at rest

3. In pure rolling: the contact point is momentarily at rest with respect to the surface. This is because the velocity of any point on the sphere = velocity of centre + velocity due to rotation. At the contact point: v_contact = v_cm + (ω × r) = v_cm + (Rω in the opposite direction) = 0. Hence v_cm = Rω. Total kinetic energy = translational KE + rotational KE = (1/2)Mv² + (1/2)Iω². For a solid sphere I = (2/5)MR², so KE_total = (1/2)Mv²(1 + 2/5) = (7/10)Mv².

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