Chapter 6: Systems of Particles and Rotational Motion Long Questions | physics Class 11 CBSE

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Chapter 6: Systems of Particles and Rotational Motion Long Questions | physics Class 11 CBSE | ByteNotes.in

Long Answer

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Hard

Question 1

Long Form

Derive the expression for the position of the centre of mass of a system of n particles distributed in space and extend it to a rigid body treated as a continuous distribution of mass.

For n particles with masses m1, m2, …, mn at positions r1, r2, …, rn, the CM position vector is R = (Σmiri)/M, where M = Σmi is the total mass.
For the x-coordinate: X = Σmixi/M, and similarly Y = Σmiyi/M and Z = Σmizi/M, obtained by considering components separately.
For a rigid body treated as a continuous distribution, the discrete sum is replaced by integrals: X = (1/M)∫x dm, Y = (1/M)∫y dm, Z = (1/M)∫z dm.
In vector form R = (1/M)∫r dm; if the CM is the origin, then ∫r dm = 0.
For a homogeneous thin rod, taking the geometric centre as origin, reflection symmetry ensures that for every dm at +x there is an equal dm at −x, so ∫x dm = 0, confirming the CM is at the geometric centre.
This symmetry argument extends to all regular homogeneous shapes (rings, discs, spheres), so their CM always coincides with the geometric centre.

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Long Answer

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Question 2

Long Form

Derive the equation MA = Fext for a system of particles and explain how it applies to both rigid bodies and non-rigid systems.

Starting from R = (Σmiri)/M, differentiate with respect to time: M dR/dt = Σmi dri/dt, giving MV = Σmivi = P.
Differentiating again: M dV/dt = Σmi dvi/dt, giving MA = Σmi ai = ΣFi, where Fi is the total force on particle i.
The force on each particle includes external forces Fiext and internal forces Fiint; by Newton's third law, internal forces occur in equal and opposite pairs summing to zero.
Therefore Σ Fi = Σ Fiext = Fext, and MA = Fext — the CM accelerates as if all mass were concentrated there with all external forces acting at that point.
This applies to any system regardless of internal motion — rigid bodies (rotating or translating), deformable bodies, or systems like exploding projectiles and decaying nuclei.
For a rigid body in pure translation, this reduces to F = ma; for rotation, it gives only the translational component of motion — the equation governing the trajectory of the CM.

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Medium

Question 3

Long Form

Describe the properties of the vector product of two vectors. Derive the component form of a × b using unit vector cross products.

The vector product c = a × b has magnitude ab sinθ (θ is the angle between a and b), direction perpendicular to the plane of a and b by the right-hand screw rule, and is anti-commutative: a × b = −b × a.
Elementary results: i × i = j × j = k × k = 0 (since sin 0° = 0), and i × j = k, j × k = i, k × i = j (cyclic order gives positive results).
With a = axi + ayj + azk and b = bxi + byj + bzk, expand a × b using the distributive property over the nine pairwise cross products.
Collecting: a × b = (aybz − azby)i + (azbx − axbz)j + (axby − aybx)k.
This can be written as the 3×3 determinant with first row (i, j, k), second row (ax, ay, az), third row (bx, by, bz).
The vector product does not change sign under reflection (a × b → (−a)×(−b) = a × b), unlike individual vectors which change sign; this makes it a pseudo-vector.

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Hard

Question 4

Long Form

Derive the relation between angular momentum and torque for a system of particles, and establish the law of conservation of angular momentum.

Total angular momentum of a system of n particles is L = Σli = Σ(ri × pi), where pi = mivi is the linear momentum of the ith particle.
Differentiating: dL/dt = Σ dli/dt = Σ τi, where τi = ri × Fi is the torque on the ith particle due to the total force Fi.
The total force Fi includes external forces Fiext and internal forces Fiint; separating: dL/dt = τext + τint.
Assuming forces between any two particles are equal, opposite, and act along the line joining them (Newton's third law extended), each internal action-reaction torque pair sums to zero, so τint = 0.
Therefore dL/dt = τext — the time rate of change of total angular momentum equals the sum of external torques.
When τext = 0, dL/dt = 0, so L = constant — the total angular momentum of the system is conserved. This is the law of conservation of angular momentum.

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Question 5

Long Form

Distinguish between translational and rotational equilibrium. Explain partial equilibrium with suitable examples using the concept of a couple.

Translational equilibrium requires ΣFi = 0 — the vector sum of all external forces is zero, so linear momentum remains constant.
Rotational equilibrium requires Στi = 0 — the vector sum of all external torques is zero, so angular momentum remains constant.
Mechanical equilibrium requires both conditions simultaneously; a body may satisfy one without the other — this is called partial equilibrium.
Example of rotational equilibrium only: two equal parallel forces acting in the same direction at both ends of a rod produce zero net torque (they are opposite in moment direction) but nonzero net force, so the body translates but the rod has no angular acceleration initially.
A couple (two equal and opposite forces with different lines of action) gives zero net force (translational equilibrium) but nonzero torque — the body is in translational equilibrium but rotates, as when opening a bottle lid.
Key insight: the moment of a couple is independent of the origin chosen, since the moment = AB × F = (r2 − r1) × F depends only on the separation vector, not on the reference point.

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Question 6

Long Form

Define moment of inertia and explain how it depends on the distribution of mass. Calculate the moment of inertia of a thin circular ring and a solid cylinder about their respective axes.

Moment of inertia I = Σmiri² is a body's resistance to change in rotational motion; it depends on total mass, shape, and crucially on how mass is distributed relative to the axis.
Unlike mass, I is not a fixed property — changing the axis of rotation or redistributing mass changes I even if total mass stays the same.
For a thin ring of radius R and mass M rotating about its central axis: every particle of the ring is at the same distance R from the axis, so I = ΣmiR² = MR².
For a solid cylinder of mass M and radius R about its symmetry axis: I = MR²/2 (standard result from Table 6.1, derivation beyond scope).
The radius of gyration k is defined by I = Mk² — for the ring, k = R; for the cylinder, k = R/√2 — giving a geometric measure of effective mass distribution.
A flywheel exploits large I to resist sudden speed changes in engines, absorbing energy during high-torque strokes and releasing it during low-torque strokes for smooth operation.

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Medium

Question 7

Long Form

Develop the kinematic equations of rotational motion about a fixed axis for uniform angular acceleration and establish their analogy with equations of linear motion.

Since angular acceleration α = dω/dt is constant, integrating: ω = ω0 + αt, which is the first kinematic equation (analogous to v = v0 + at).
Since ω = dθ/dt, integrating the first equation: θ = θ0 + ω0t + (1/2)αt², analogous to x = x0 + v0t + (1/2)at².
Eliminating t between the first two equations gives ω² = ω0² + 2α(θ − θ0), analogous to v² = v0² + 2a(x − x0).
The three equations describe rotational motion completely when α is constant; they involve only one degree of freedom (angular displacement θ) since the axis is fixed.
Full analogy: x↔θ, v↔ω, a↔α, m↔I, F↔τ, p↔L; every translational formula has a rotational counterpart with these substitutions.
Example application: a motor wheel accelerated from 1200 rpm to 3120 rpm in 16 s has α = (104π − 40π)/16 = 4π rad s−2, and makes 576 revolutions in this time.

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Long Answer

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Hard

Question 8

Long Form

Derive the expression for work done by a torque on a rotating body about a fixed axis and hence establish τ = Iα.

Consider a particle at P at radius r1 from the axis; a force F1 in the perpendicular plane acts on it. In angular displacement dθ, the particle moves ds1 = r1 dθ tangentially.
Work done by F1: dW1 = F1 ds1 sinα1 = F1r1 sinα1 dθ = τ1 dθ, where τ1 = r1 F1 sinα1 is the torque due to F1 about the axis.
Summing over all particles: dW = (τ1 + τ2 + …)dθ = τdθ, since dθ is the same for all particles.
Power: P = dW/dt = τ dθ/dt = τω, analogous to P = Fv in linear motion.
The rate of increase of rotational kinetic energy is d(Iω²/2)/dt = Iω(dω/dt) = Iωα. Setting this equal to power: τω = Iωα.
Dividing by ω: τ = Iα, which is Newton's second law for rotational motion, showing torque produces angular acceleration inversely proportional to moment of inertia.

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Long Answer

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Hard

Question 9

Long Form

Analyse the angular momentum of a rigid body rotating about a fixed axis. Under what conditions does L = Iω hold, and when does it not?

For a typical particle at P, l = r × p = (OC + CP) × mv; expanding gives l = (OC × mv) + (CP × mv).
CP × mv is parallel to the fixed axis (z-axis) with magnitude mr⊥²ω; OC × mv is perpendicular to the fixed axis.
Summing over all particles: Lz = Σ(mr⊥²ω) = Iω k̂, while L⊥ = Σ(OC × mivi) may not be zero.
For bodies symmetric about the axis of rotation, for each velocity vi there exists an equal mass particle with velocity −vi diametrically opposite; these pairs contribute zero to L⊥, so L = Lz = Iω directed along the axis.
For asymmetric bodies, L⊥ ≠ 0 and hence L ≠ Lz; the total angular momentum L is not along the axis and L ≠ Iω in general.
Unlike translational motion where p and v are always parallel, for rotational motion l and ω are only parallel for symmetric bodies or when the axis is a symmetry axis — this is a fundamental difference.

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Long Answer

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Medium

Question 10

Long Form

Explain the conservation of angular momentum for a rigid body rotating about a fixed axis. Discuss three practical applications where this principle is evident.

For rotation about a fixed axis, if external torque τext = 0, then d(Iω)/dt = 0, so Iω = constant — the angular momentum is conserved.
If the body is rigid (I constant), constant Iω means ω is constant too. If I changes (e.g., mass redistribution), ω must change inversely: I1ω1 = I2ω2.
Application 1 — Swivel chair experiment: A person sitting on a frictionless swivel chair with arms outstretched and set rotating then pulls arms in; this decreases I, so ω increases, and the person spins faster. Stretching arms again increases I and slows the rotation.
Application 2 — Ballet dancer / acrobat: A diver or acrobat tucks into a compact position to reduce I, greatly increasing ω to perform multiple somersaults, then extends the body to slow rotation before entering water.
Application 3 — Circus acrobat / figure skater: Skaters spinning on ice extend or contract limbs to control spin speed; classical Indian and western dancers performing pirouettes exploit the same principle to spin with precision.
The underlying physics is identical in all cases: in the absence of external torque, the product Iω is conserved, so a decrease in the moment of inertia is compensated by a proportional increase in angular velocity.

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Chapter 6: Systems of Particles and Rotational Motion | Long Questions

Derive the expression for the position of the centre of mass of a system of n particles distributed in space and extend it to a rigid body treated as a continuous distribution of mass.

Derive the equation MA = Fext for a system of particles and explain how it applies to both rigid bodies and non-rigid systems.

Describe the properties of the vector product of two vectors. Derive the component form of a × b using unit vector cross products.

Derive the relation between angular momentum and torque for a system of particles, and establish the law of conservation of angular momentum.

Distinguish between translational and rotational equilibrium. Explain partial equilibrium with suitable examples using the concept of a couple.

Define moment of inertia and explain how it depends on the distribution of mass. Calculate the moment of inertia of a thin circular ring and a solid cylinder about their respective axes.

Develop the kinematic equations of rotational motion about a fixed axis for uniform angular acceleration and establish their analogy with equations of linear motion.

Derive the expression for work done by a torque on a rotating body about a fixed axis and hence establish τ = Iα.

Analyse the angular momentum of a rigid body rotating about a fixed axis. Under what conditions does L = Iω hold, and when does it not?

Explain the conservation of angular momentum for a rigid body rotating about a fixed axis. Discuss three practical applications where this principle is evident.

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