StudyOSStudyOS CBSE Study Material, Class 5–12

System of Particles and Rotational Motion

NCERT Class 11 · Physics Based on NCERT Class 11 Physics textbook · Free CBSE study kit

Everything you need to ace this chapter, in one place.
Notes · Mind Map · Quiz · Assertion-Reason · Case Studies · Question Bank · Past Papers

Get Full Study Kit →
Studying with friends? Send them this chapter.

Chapter Notes

Overview
This chapter extends the study of motion from a single particle to extended bodies and systems of particles. It introduces the concept of the centre of mass, which simplifies the analysis of translational motion of a system, and then delves into rotational motion about a fixed axis, covering angular velocity, torque, angular momentum, moment of inertia, and the laws governing rotation. The one big idea is that the motion of any rigid body can be separated into the translational motion of its centre of mass (governed by external forces) and rotational motion about the centre of mass (governed by torques).
Key Concepts (12)
Centre of mass
The centre of mass of a system of particles is that point which moves as if all the mass of the system were concentrated at that point and all external forces were applied at that point.
e.g. For a uniform L-shaped lamina made of three 1 kg squares, the centre of mass is found by treating each square's mass concentrated at its geometric centre and then calculating the weighted average.
Rigid body
A rigid body is a body with a perfectly definite and unchanging shape; the distances between all pairs of particles of such a body do not change.
e.g. A steel beam, a wheel, or a top can be treated as rigid bodies in many situations because their deformations are negligible.
Pure translational motion
In pure translational motion, at any instant of time, all particles of the body have the same velocity.
e.g. A rectangular block sliding down an inclined plane without any sidewise movement is in pure translational motion.
Rotation about a fixed axis
Rotation about a fixed axis is the motion of a rigid body in which every particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has its centre on the axis.
e.g. A ceiling fan rotating about its fixed central rod is an example of rotation about a fixed axis.
Angular velocity
Angular velocity is a vector quantity whose magnitude is the rate of change of angular displacement (dθ/dt) and whose direction is along the axis of rotation, given by the right-handed screw rule.
e.g. For a rotating ceiling fan, the angular velocity vector points along the axis of the fan, in the direction a right-handed screw would advance if turned with the fan.
Torque
Torque (or moment of a force) is the vector product of the position vector of the point of application of the force and the force vector.
e.g. When you open a door by applying a force perpendicular to the door at its edge, you are applying a torque about the hinges.
Angular momentum
Angular momentum of a particle is defined as the vector product of its position vector and its linear momentum.
e.g. A spinning top has angular momentum about its point of contact with the ground.
Moment of inertia
Moment of inertia is the rotational analogue of mass; it is the sum of the products of the mass of each particle and the square of its perpendicular distance from the axis of rotation.
e.g. A flywheel has a large moment of inertia, making it resist changes in its rotational motion.
Vector product (cross product)
The vector product of two vectors a and b is a vector c such that its magnitude is ab sinθ, it is perpendicular to the plane containing a and b, and its direction is given by the right-handed screw rule.
e.g. The torque τ = r × F is a vector product; if r and F are in the plane of the paper, τ is perpendicular to the paper.
Equilibrium of a rigid body
A rigid body is in equilibrium if both the net external force and the net external torque acting on it are zero.
e.g. A book lying on a table is in equilibrium because the weight is balanced by the normal force (net force zero) and there is no net torque.
Law of conservation of linear momentum
When the total external force acting on a system of particles is zero, the total linear momentum of the system is constant.
e.g. In the radioactive decay of a radium nucleus at rest, the radon nucleus and alpha particle fly off in opposite directions with equal and opposite momenta, keeping total momentum zero.
Precession
Precession is the movement of the axis of a spinning top around the vertical through its point of contact, sweeping out a cone.
e.g. A spinning top's axis slowly rotates around the vertical line through its tip.
Formulas & Equations (18)
X = (m1x1 + m2x2) / (m1 + m2)
Reads as: X equals m1 x1 plus m2 x2 divided by m1 plus m2
Variables: X = position of centre of mass (m), m1, m2 = masses of particles (kg), x1, x2 = positions of particles (m)
Use when: Finding the centre of mass of a two-particle system along a straight line.
⚠ Forgetting that X is measured from the same origin as x1 and x2, and not using the same units.
X = (Σ mi xi) / M
Reads as: X equals sum over i of m_i x_i divided by total mass M
Variables: X = x-coordinate of centre of mass (m), mi = mass of ith particle (kg), xi = x-coordinate of ith particle (m), M = total mass (kg)
Use when: Finding the centre of mass of a system of particles along a straight line.
⚠ Confusing the summation index or forgetting to divide by total mass M.
R = (1/M) Σ mi ri
Reads as: R equals one over M times the sum of m_i r_i
Variables: R = position vector of centre of mass (m), mi = mass of ith particle (kg), ri = position vector of ith particle (m), M = total mass (kg)
Use when: Finding the centre of mass of a system of particles in vector form.
⚠ Treating the vector sum as a scalar sum; forgetting that ri are vectors.
X = (1/M) ∫ x dm, Y = (1/M) ∫ y dm, Z = (1/M) ∫ z dm
Reads as: X equals one over M times the integral of x dm, Y equals one over M times the integral of y dm, Z equals one over M times the integral of z dm
Variables: X, Y, Z = coordinates of centre of mass (m), M = total mass (kg), dm = infinitesimal mass element (kg), x, y, z = coordinates of dm (m)
Use when: Finding the centre of mass of a continuous mass distribution.
⚠ Forgetting to use the correct limits of integration or not expressing dm in terms of the coordinate variable.
M V = Σ mi vi
Reads as: M V equals the sum over i of m_i v_i
Variables: M = total mass (kg), V = velocity of centre of mass (m/s), mi = mass of ith particle (kg), vi = velocity of ith particle (m/s)
Use when: Relating the total linear momentum of a system to the velocity of its centre of mass.
⚠ Treating V as the average of velocities without mass weighting.
M A = Σ mi ai = F_ext
Reads as: M A equals the sum over i of m_i a_i equals the net external force
Variables: M = total mass (kg), A = acceleration of centre of mass (m/s²), mi = mass of ith particle (kg), ai = acceleration of ith particle (m/s²), F_ext = net external force (N)
Use when: Applying Newton's second law to the centre of mass of a system.
⚠ Including internal forces in the sum; they cancel out.
P = M V
Reads as: P equals M V
Variables: P = total linear momentum of system (kg m/s), M = total mass (kg), V = velocity of centre of mass (m/s)
Use when: Calculating the total linear momentum of a system of particles.
⚠ Forgetting that P is a vector sum, not a scalar sum.
dP/dt = F_ext
Reads as: dP by dt equals F_ext
Variables: P = total linear momentum (kg m/s), t = time (s), F_ext = net external force (N)
Use when: Stating Newton's second law for a system of particles.
⚠ Writing dP/dt = 0 when external forces are present.
c = a × b, |c| = ab sinθ
Reads as: c equals a cross b, magnitude of c equals a b sine theta
Variables: c = vector product, a, b = magnitudes of vectors a and b, θ = angle between a and b
Use when: Defining the magnitude of the vector product.
⚠ Forgetting that the direction of c is perpendicular to both a and b, given by the right-hand rule.
a × b = (ay bz - az by) i + (az bx - ax bz) j + (ax by - ay bx) k
Reads as: a cross b equals (a_y b_z minus a_z b_y) i plus (a_z b_x minus a_x b_z) j plus (a_x b_y minus a_y b_x) k
Variables: ax, ay, az = components of a; bx, by, bz = components of b; i, j, k = unit vectors along x, y, z axes
Use when: Computing the cross product in component form.
⚠ Mixing up the order of subscripts; the cross product is anti-commutative.
v = ω r
Reads as: v equals omega r
Variables: v = linear velocity (m/s), ω = angular velocity (rad/s), r = perpendicular distance from axis (m)
Use when: Relating linear velocity of a particle in a rigid body to its angular velocity about a fixed axis.
⚠ Using the distance from the origin instead of the perpendicular distance from the axis.
v = ω × r
Reads as: v equals omega cross r
Variables: v = linear velocity vector (m/s), ω = angular velocity vector (rad/s), r = position vector from origin on axis (m)
Use when: Expressing the vector relation between linear and angular velocity for a rotating rigid body.
⚠ Forgetting that the origin must be on the axis of rotation for this relation to hold.
τ = r × F
Reads as: tau equals r cross F
Variables: τ = torque vector (N m), r = position vector of point of application of force (m), F = force vector (N)
Use when: Calculating the torque produced by a force about a point.
⚠ Using the magnitude of r instead of the perpendicular distance from the axis.
l = r × p
Reads as: l equals r cross p
Variables: l = angular momentum of a particle (kg m²/s), r = position vector of particle (m), p = linear momentum of particle (kg m/s)
Use when: Defining angular momentum of a particle about a point.
⚠ Confusing angular momentum with linear momentum.
L = I ω
Reads as: L equals I omega
Variables: L = angular momentum about fixed axis (kg m²/s), I = moment of inertia about that axis (kg m²), ω = angular velocity (rad/s)
Use when: Calculating angular momentum of a rigid body rotating about a fixed axis.
⚠ Forgetting that I depends on the axis of rotation.
τ = I α
Reads as: tau equals I alpha
Variables: τ = net torque about fixed axis (N m), I = moment of inertia about that axis (kg m²), α = angular acceleration (rad/s²)
Use when: Applying Newton's second law for rotational motion about a fixed axis.
⚠ Using the wrong moment of inertia (not about the axis of rotation).
I = Σ mi ri²
Reads as: I equals sum over i of m_i r_i squared
Variables: I = moment of inertia (kg m²), mi = mass of ith particle (kg), ri = perpendicular distance from axis (m)
Use when: Calculating moment of inertia for a system of discrete particles.
⚠ Using the distance from the origin instead of the perpendicular distance from the axis.
I = ∫ r² dm
Reads as: I equals the integral of r squared dm
Variables: I = moment of inertia (kg m²), r = perpendicular distance from axis (m), dm = infinitesimal mass element (kg)
Use when: Calculating moment of inertia for continuous mass distributions.
⚠ Forgetting to express dm in terms of the coordinate variable and using correct limits.
How It Works
Finding centre of mass of a system of particles
  1. Step 1: Choose a coordinate system and origin.
  2. Step 2: For each particle, note its mass and coordinates (xi, yi, zi).
  3. Step 3: Compute the weighted sum of coordinates: Σ mi xi, Σ mi yi, Σ mi zi.
  4. Step 4: Divide each sum by the total mass M = Σ mi to get (X, Y, Z).
💡 The centre of mass is a mass-weighted average; for equal masses, it is the arithmetic mean of positions.
Finding centre of mass of a continuous body
  1. Step 1: Subdivide the body into infinitesimal mass elements dm.
  2. Step 2: Express dm in terms of a coordinate variable (e.g., dm = λ dx for a rod).
  3. Step 3: Set up integrals: X = (1/M) ∫ x dm, Y = (1/M) ∫ y dm, Z = (1/M) ∫ z dm.
  4. Step 4: Evaluate the integrals with appropriate limits.
💡 For symmetric homogeneous bodies (ring, disc, sphere, rod), the centre of mass coincides with the geometric centre due to reflection symmetry.
Motion of centre of mass
  1. Step 1: The centre of mass moves as if all mass is concentrated there and all external forces act there.
  2. Step 2: The acceleration of the centre of mass is given by A = F_ext / M.
  3. Step 3: Internal forces cancel out and do not affect the motion of the centre of mass.
💡 Even if a system explodes or has internal motions, the centre of mass continues along its original trajectory (e.g., projectile exploding in mid-air).
Conservation of linear momentum for a system
  1. Step 1: If net external force F_ext = 0, then dP/dt = 0, so total linear momentum P is constant.
  2. Step 2: This implies the velocity of the centre of mass remains constant.
  3. Step 3: In radioactive decay or collisions, the total momentum before and after is the same if no external forces act.
💡 Internal forces do not change total momentum; they only redistribute it among particles.
Vector product (cross product) calculation
  1. Step 1: Write vectors a and b in component form: a = ax i + ay j + az k, b = bx i + by j + bz k.
  2. Step 2: Use the determinant: a × b = |i j k; ax ay az; bx by bz|.
  3. Step 3: Expand: (ay bz - az by)i + (az bx - ax bz)j + (ax by - ay bx)k.
  4. Step 4: The result is a vector perpendicular to both a and b, direction given by right-hand rule.
💡 Cross product is anti-commutative: a × b = - (b × a).
Relating linear and angular velocity in rotation
  1. Step 1: For a rigid body rotating about a fixed axis, every particle moves in a circle.
  2. Step 2: The linear velocity v of a particle is tangential to its circular path.
  3. Step 3: The magnitude is v = ω r, where r is the perpendicular distance from the axis.
  4. Step 4: In vector form: v = ω × r, with origin on the axis.
💡 All particles have the same angular velocity ω, but different linear velocities depending on their distance from the axis.
Torque and angular momentum
  1. Step 1: Torque τ = r × F; it is the rotational analogue of force.
  2. Step 2: Angular momentum l = r × p for a particle; for a rigid body about a fixed axis, L = I ω.
  3. Step 3: Newton's second law for rotation: τ_net = dL/dt = I α (for fixed axis with constant I).
💡 Torque depends on the point about which it is calculated; for a rigid body in equilibrium, both net force and net torque must be zero.
Equilibrium of a rigid body
  1. Step 1: For translational equilibrium, the vector sum of all external forces must be zero: Σ F = 0.
  2. Step 2: For rotational equilibrium, the vector sum of all external torques about any point must be zero: Σ τ = 0.
  3. Step 3: Both conditions must be satisfied simultaneously for the body to be in complete equilibrium.
💡 Choosing the point about which torques are calculated wisely can simplify calculations (e.g., choose a point where unknown forces act).
Exam Traps — Don't Make These Mistakes
⚠ Centre of mass is always inside the body
✗ The centre of mass of a body always lies within the material of the body.
→ The centre of mass can lie outside the body, e.g., for a ring or a hollow sphere, the centre of mass is at the geometric centre which is not within the material.
✓ The centre of mass may lie inside or outside the body; for a uniform ring, it is at the centre, which is outside the material.
⚠ Angular velocity is a scalar
✗ Angular velocity is a scalar quantity because it has magnitude and direction is not needed.
→ Angular velocity is a vector; its direction is along the axis of rotation as given by the right-hand screw rule, and it obeys vector addition.
✓ Angular velocity is a vector quantity; its direction is along the axis of rotation, determined by the right-hand rule.
⚠ Torque and force are the same thing
✗ Torque is just another name for force in rotational motion.
→ Torque is the rotational analogue of force, but they are different physical quantities. Torque depends on both force and the lever arm (distance from axis).
✓ Torque is the rotational effect of a force; it is given by τ = r × F, and its SI unit is N m, not N.
⚠ Moment of inertia is a constant for a body
✗ The moment of inertia of a rigid body is a fixed property of the body.
→ Moment of inertia depends on the axis of rotation; the same body can have different moments of inertia about different axes.
✓ Moment of inertia depends on the axis of rotation; it is not a fixed property of the body alone.
⚠ Conservation of angular momentum requires no external force
✗ Angular momentum is conserved when no external force acts on the system.
→ Angular momentum is conserved when the net external torque acting on the system is zero, not necessarily when net external force is zero.
✓ Angular momentum is conserved when the net external torque on the system is zero.
⚠ Linear velocity of all particles in a rotating body is the same
✗ In pure rotation, all particles of a rigid body have the same linear velocity.
→ In pure rotation, all particles have the same angular velocity, but linear velocity v = ω r depends on the distance from the axis.
✓ In pure rotation about a fixed axis, all particles have the same angular velocity, but linear velocities vary with distance from the axis.
Compare & Contrast
Nature of motion
Translational motion
Rotational motion
In pure translation, all particles have the same linear velocity at any instant; in pure rotation about a fixed axis, all particles have the same angular velocity but different linear velocities.
Definition and conservation
Linear momentum
Angular momentum
Linear momentum p = mv is conserved when net external force is zero; angular momentum L = Iω is conserved when net external torque is zero.
Role in motion
Mass
Moment of inertia
Mass is a measure of inertia in translational motion (resistance to change in linear velocity); moment of inertia is the rotational analogue, measuring resistance to change in angular velocity, and depends on the distribution of mass about the axis.
Effect on motion
Force
Torque
Force causes linear acceleration (F = ma); torque causes angular acceleration (τ = Iα). Torque is the vector product of position vector and force.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 11 marks
Most asked: Centre of mass of discrete and continuous systems · Moment of inertia for standard shapes (rod, ring, disc, sphere) · Torque and angular momentum · Conservation of angular momentum · Equilibrium of rigid bodies · Relation between linear and angular velocity
Don't skip: Numerical problems on centre of mass (especially for two or three particles); Derivation of moment of inertia for a thin rod or ring; Application of conservation of angular momentum (e.g., a dancer spinning, or a rotating platform); Problems combining torque and equilibrium (e.g., ladder problems)
One-Line Revision (read aloud before the exam)
  1. Centre of mass is the mass-weighted average position of all particles in a system.
  2. For a two-particle system, the centre of mass lies closer to the heavier particle.
  3. The centre of mass of a symmetric homogeneous body is at its geometric centre.
  4. The centre of mass of a system moves as if all mass is concentrated there and all external forces act there.
  5. Internal forces do not affect the motion of the centre of mass.
  6. Total linear momentum of a system is conserved if net external force is zero.
  7. Angular velocity is a vector along the axis of rotation, direction given by right-hand rule.
  8. Linear velocity v = ω × r for a particle in a rotating rigid body.
  9. Torque τ = r × F is the rotational analogue of force.
  10. Angular momentum L = I ω for a rigid body rotating about a fixed axis.
  11. Newton's second law for rotation: τ = I α.
  12. Moment of inertia depends on the axis of rotation and the distribution of mass.
  13. A rigid body is in equilibrium if net external force and net external torque are both zero.
  14. The vector product of two vectors is anti-commutative: a × b = - b × a.
  15. For a rigid body rotating about a fixed axis, all particles have the same angular velocity.
  16. The moment of inertia of a thin rod of length L about its centre is ML²/12.
  17. The moment of inertia of a ring about its centre is MR².
  18. The moment of inertia of a disc about its centre is MR²/2.
  19. The moment of inertia of a solid sphere about its diameter is 2MR²/5.
  20. The moment of inertia of a hollow sphere about its diameter is 2MR²/3.

MCQs — 10 Questions with Answers

Q1. Which of the following statements correctly defines a rigid body?

  • A. A body in which particles can move freely relative to each other
  • B. A body with perfectly definite shape where distances between all particles remain constant ✓
  • C. Any body that does not deform under applied forces
  • D. A body whose all particles must move with same velocity

Answer: B — A rigid body is defined by constant inter-particle distances; option B matches this definition exactly.

Q2. A block slides down an inclined plane without rotating. Which type of motion does it have?

  • A. Pure rotational motion about the axis perpendicular to plane
  • B. Pure translational motion since all particles have same velocity ✓
  • C. Combined translational and rotational motion
  • D. No motion—it remains stationary

Answer: B — In pure translation, every particle moves with identical velocity; the sliding block satisfies this condition.

Q3. In rotational motion about a fixed axis, consider particles P1 and P2 at distances r₁ and r₂ from the axis respectively (r₁ < r₂). Which statement is true?

  • A. Both particles have the same linear velocity
  • B. Particle P2 has greater linear velocity than P1 ✓
  • C. Particle P1 has greater linear velocity than P2
  • D. Both particles have zero velocity

Answer: B — Linear velocity in rotation is v = ωr; since r₂ > r₁ and ω is same for all particles, particle P2 has greater velocity.

Q4. A cylinder rolls down an inclined plane without slipping. What can be said about the velocity of the point in contact with the plane at any instant?

  • A. It equals the velocity of the centre of the cylinder
  • B. It is zero because it is instantaneously at rest ✓
  • C. It is maximum among all points of the cylinder
  • D. It equals the velocity of the top point of the cylinder

Answer: B — For rolling without slipping, the contact point is instantaneously fixed (v = 0) at that moment, which is the no-slip condition.

Q5. A spinning top maintains a fixed point of contact with the ground, but its axis of rotation moves around the vertical. This motion is called—

  • A. Pure rotation
  • B. Pure translation
  • C. Precession ✓
  • D. Rolling motion

Answer: C — Precession is the motion where one point is fixed but the axis itself sweeps out a cone, exactly as a spinning top does.

Q6. In rotational motion about a fixed axis, a particle located on the axis itself (at r = 0)—

  • A. Moves with maximum linear velocity
  • B. Remains stationary throughout the rotation ✓
  • C. Moves in a straight line parallel to the axis
  • D. Oscillates perpendicular to the axis

Answer: B — A particle on the axis has r = 0; since it must be on the fixed axis, it cannot move and remains stationary.

Q7. Which is NOT a correct statement about the difference between pure translation and pure rotation?

  • A. In translation, all particles have same velocity; in rotation, they do not
  • B. In translation, body moves from one place to another; in rotation, one line is fixed
  • C. In translation, particles move in straight lines; in rotation, they move in circles
  • D. In translation, the velocity is the same for all particles; in rotation all particles have the same angular displacement ✓

Answer: D — Option D is incorrect because in rotation, all particles have the same angular velocity (ω) but same angular displacement is not necessarily true for different time intervals when ω changes.

Q8. A ceiling fan rotates about a vertical fixed axis. Consider two points: one at the tip of a blade at distance R from axis, and another at distance R/2 from axis. If both points are rotating with same angular velocity ω, the ratio of their linear velocities is—

  • A. 1:1
  • B. 1:2
  • C. 2:1 ✓
  • D. 4:1

Answer: C — v = ωr; for point 1: v₁ = ωR; for point 2: v₂ = ω(R/2); ratio v₁:v₂ = ωR:ω(R/2) = 2:1.

Q9. An oscillating table fan has rotating blades about a vertical axis, but this vertical axis itself oscillates sidewise. At any instant, which point of the fan remains fixed?

  • A. The centre of a blade
  • B. The pivot point where the axis is attached to the stand ✓
  • C. Any point on the rotating blade
  • D. The entire axis of rotation

Answer: B — The pivot point (point O) is the fixed point about which the axis oscillates; this is the constraint keeping one point stationary.

Q10. Which situation demonstrates combined translational and rotational motion? (HOTS)

  • A. A stone thrown horizontally rotating as it falls
  • B. A cylinder rolling down an inclined plane ✓
  • C. A ceiling fan spinning about its axis
  • D. A block sliding down a frictionless incline

Answer: B — Rolling combines translation (cylinder's centre moves down slope) and rotation (cylinder spins); stone in option A primarily translates with possible rotation but does not maintain contact; ceiling fan and sliding block show only rotation and translation respectively.

Flashcards

What is a rigid body?

A body with perfectly definite and unchanging shape, where distances between all pairs of particles never change.

Define pure translational motion.

Motion in which all particles of a body have the same velocity at any instant of time.

In rotational motion about a fixed axis, what path does each particle follow?

Each particle moves in a circle lying in a plane perpendicular to the axis, with the circle's centre on the axis.

What is the axis of rotation?

The fixed straight line about which a rigid body rotates, through which all circular paths of particles have their centres.

Give one difference between rolling and pure translation.

In rolling, different particles have different velocities at the same instant, but in pure translation all particles have the same velocity.

What does 'precession' mean for a spinning top?

The movement of the top's axis of rotation around the vertical line passing through its fixed point of contact with the ground.

Why is a real block sliding down an incline considered in pure translation?

Because all particles of the block move with the same velocity at any instant and there is no rotation about any axis.

In rotational motion, what is true for a particle on the axis itself?

A particle on the axis has r = 0 (zero distance from axis) and remains stationary while the body rotates.

How is rolling motion different from pure rotation?

Rolling is a combination of translation (axis moving forward) and rotation (body spinning), whereas pure rotation keeps the axis fixed.

Name one real example where rotation occurs about a non-fixed axis.

A spinning top or oscillating pedestal fan, where one point is fixed but the axis itself moves and sweeps a cone or oscillates.

Important Board Questions

Define a rigid body and state one key difference between pure translational motion and pure rotational motion about a fixed axis. [2 marks]

Define rigid body using inter-particle distances. For difference: compare velocity conditions—in translation all particles have same v; in rotation each particle moves in circle with different v = ωr.

A rolling cylinder moves down an inclined plane without slipping. Explain why this motion is neither pure translation nor pure rotation. Use a velocity diagram to show the velocity of the centre of mass and the velocity of the contact point. [5 marks]

Show that different particles have different velocities at same instant (not pure translation) and axis moves (not pure rotation). Contact point velocity = 0 at instant of contact due to no-slip condition; use v = v_cm + v_rot at different points to prove combined motion.

A rigid body rotates about a fixed axis. Derive the relationship between the linear velocity v and angular velocity ω for a particle at perpendicular distance r from the axis. Then explain using this relationship why particles at different distances from the rotation axis have different linear velocities, even though they have the same angular velocity. Provide one real-life example where this relationship is applied. [6 marks]

Derive v = ωr from arc length s = rθ and angular velocity ω = dθ/dt; show that for fixed ω, v increases with r. Apply to ceiling fan: blade tip (large r) moves faster than inner point (small r). Explain centripetal acceleration also increases with r.

Assertion-Reason Questions for System of Particles and Rotational Motion

Q1. Assertion (A):

A rigid body undergoing pure translational motion has all its particles moving with the same acceleration at any instant.

Reason (R):

In pure translation, the distances between all pairs of particles remain constant, and the body moves as a single unit without rotation.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. If all particles have the same velocity in pure translation, then taking the time derivative, all particles must have the same acceleration. The reason correctly explains why this is true—constant particle distances prevent relative motion, so the body moves uniformly.

Q2. Assertion (A):

A point on the axis of rotation of a rigid body remains stationary during the rotation.

Reason (R):

The perpendicular distance of any point on the axis from the axis itself is zero.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason is the correct explanation. Since r = 0 for points on the axis, circular motion is impossible (v = ωr = 0), so such points remain stationary. This is the defining characteristic of a rotation axis.

Q3. Assertion (A):

In the rolling motion of a cylinder down an incline, the velocity of the contact point is always zero.

Reason (R):

Rolling without slipping is defined by the condition v_center = ω × R, where the contact point acts as the instantaneous center of rotation.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true—in rolling without slipping, the contact point is instantaneously at rest. However, the reason is not the correct explanation. The reason states a mathematical relationship but doesn't explain why the contact point is stationary; it's stationary because of the no-slip condition that relates center velocity to angular velocity geometrically.

Unlock more Assertion-Reason questions in Premium — unlock →

Case-Based Questions (CBQ) for System of Particles and Rotational Motion

Case Study:

A pottery wheel used by artisans in Indian villages is a classic example of rotational motion. The wheel is made of clay and has a radius of 30 cm. When a potter spins the wheel by hand, it rotates about a fixed vertical axis. An artisan marks two points on the wheel: Point A at the center (on the axis) and Point B at the rim, 30 cm from the axis. The wheel completes one full rotation in 2 seconds. The artisan observes the motion and makes measurements while shaping clay.

Q1. What is the angular velocity of the pottery wheel?

  • A. π rad/s ✓
  • B. 2π rad/s
  • C. 3π rad/s
  • D. 4π rad/s

Answer: A

Q2. What is the linear velocity of a point on the rim (Point B)?

  • A. 0.15π m/s
  • B. 0.30π m/s ✓
  • C. 0.60π m/s
  • D. 1.20π m/s

Answer: B

Q3. What is the linear velocity of the point at the center (Point A)?

  • A. π m/s
  • B. 0.5π m/s
  • C. 0 m/s ✓
  • D. 0.15π m/s

Answer: C

Q4. Which type of motion does the pottery wheel undergo?

  • A. Pure translation
  • B. Pure rotation about a fixed axis ✓
  • C. Combined rotation and translation
  • D. Oscillatory motion

Answer: B

Unlock more case studies in Premium — unlock →

Class 11 Physics Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

View the formula sheet →

Next chapterGravitation →
Found this useful? Share it with your class group.

You've seen the free preview. Open the full interactive study kit — mind map, instant-graded quiz, progress tracking — free, no sign-in required.

Open Full Study Kit — Free →

Premium unlocks the full Question Bank, Daily Practice Papers, Chapter Quiz and Full Mock Papers for every chapter.